This deck shows how an ordered basis turns abstract vectors into coordinate tuples, how a linear map becomes a matrix, why matrix multiplication is exactly composition, and how a change of basis - that is, similarity - re-coordinates one and the same map. It targets four real errors: confusing a vector with its coordinate tuple, multiplying matrices in the wrong order for a composition, mixing up P and P-inverse in the change-of-basis formula, and thinking that similar matrices are equal.
Subject: Foundations of Higher Mathematics · 114 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you will be able to:
1. Write the coordinate vector of any vector once an ordered basis is fixed.
2. Build the matrix of a linear map from its action on a basis.
3. Explain why matrix multiplication is exactly composition of maps.
4. Change basis for both vectors and maps, and recognize similar matrices as one map in disguise.
Warm-up
Discussion prompt
Before we open Matrices, Coordinates & Change of Basis: without looking back, what was the main idea of The Rank–Nullity Theorem, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck defines rank and nullity as the dimensions of the image and the kernel, then states and proves the Rank-Nullity Theorem by basis extension, computes both quantities by row reduction, and covers the finite-dimensional miracle that for an endomorphism injective and surjective mean the same thing. It targets the misconceptions of adding to the dimension of the codomain, confusing rank with the codomain, and expecting injective to force surjective in infinite dimensions.
Concept
Abstract vectors — polynomials, functions, arrows — are awkward to compute with directly. A basis fixes that.
Once you pick a basis, every vector becomes a unique list of scalars: how much of each basis vector you need.
\[ v = c_1 b_1 + c_2 b_2 + \cdots + c_n b_n \]
Counterexample
Discussion prompt
Abstract vectors — polynomials, functions, arrows — are awkward to compute with directly. A basis fixes that.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Once you pick a basis, every vector becomes a unique list of scalars: how much of each basis vector you need.
Intuition
Think of a city grid. A basis is the choice of streets and avenues; a vector's coordinates are its address on that grid.
The building does not move when you renumber the streets — but its address changes. Hold that thought: it is the entire deck.
Analogy
Discussion prompt
Explain A basis is an address system by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of a city grid. A basis is the choice of streets and avenues; a vector's coordinates are its address on that grid.
Concept
In a coordinate space itself, the standard basis makes coordinates invisible: a vector's coordinates are just its components.
\[ [(a,b,c)]_E = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \]
That is why we usually skip mentioning the basis. But it is still a choice, and other bases give other numbers.
Explain it
Discussion prompt
Explain The standard basis is the default to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
In a coordinate space itself, the standard basis makes coordinates invisible: a vector's coordinates are just its components.
Concept
Coordinates need the basis vectors in a fixed order — first, second, third.
ordered basis — A basis together with a chosen order of its vectors, written as a tuple B = (b1, b2, ..., bn). Reordering the basis permutes every coordinate vector the same way.
Swap two basis vectors and every coordinate tuple has those two entries swapped. Order is part of the data.
Ranking
Put in order
Put the moves of Coordinates of a polynomial into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. p is already written as a combination of 1, x and x-squared, so the coefficients are exactly the coordinates.
Worked example
Work in the space of polynomials of degree at most two, with its standard ordered basis.
\[ B = (1,\; x,\; x^2) \]
Find the coordinate vector of this polynomial:
\[ p(x) = 7 - 2x + 5x^2 \]
Read off the coefficient of each basis vector
Why: p is already written as a combination of 1, x and x-squared, so the coefficients are exactly the coordinates.
\[ p = 7\cdot 1 + (-2)\cdot x + 5\cdot x^2 \]
Stack the coefficients in basis order
Why: Coordinate entries follow the order of B: constant term first, then x, then x-squared.
\[ [p]_B = \begin{bmatrix} 7 \\ -2 \\ 5 \end{bmatrix} \]
Verify by rebuilding p from its coordinates
Why: Multiply each basis vector by its coordinate and add; we must recover p exactly.
\[ 7\cdot 1 + (-2)\,x + 5\,x^2 = 7 - 2x + 5x^2 = p(x) \]
Picture it
Animation
Shows: Each line of the worked example "Coordinates of a polynomial", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiply each basis vector by its coordinate and add; we must recover p exactly.
Step zero
Discussion prompt
Reordering the basis permutes the coordinates — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Original order
Answer:
Worked example
Coordinates depend on the order of the basis, not just the set of vectors. Reorder and watch.
\[ p(x) = 7 - 2x + 5x^2 \]
Original order
Why: The basis in the order constant, x, x-squared reads the coefficients in that order.
\[ B = (1,\, x,\, x^2): \quad [p]_B = \begin{bmatrix} 7 \\ -2 \\ 5 \end{bmatrix} \]
Reorder the basis
Why: Put the quadratic term first, then the constant, then x; the coordinates permute to match.
\[ B_2 = (x^2,\, 1,\, x): \quad [p]_{B_2} = \begin{bmatrix} 5 \\ 7 \\ -2 \end{bmatrix} \]
Verify both rebuild p
Why: Each column, combined with its own ordered basis, must return p.
\[ 5x^2 + 7\cdot 1 + (-2)x = 7 - 2x + 5x^2 = p \]
Picture it
Animation
Shows: Each line of the worked example "Reordering the basis permutes the coordinates", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each column, combined with its own ordered basis, must return p.
Concept
Fixing a basis B on an n-dimensional space V gives a function that sends each vector to its coordinate column.
\[ [\,\cdot\,]_B : V \to F^n, \qquad v \mapsto [v]_B \]
This turns abstract vectors into ordinary columns of numbers you can compute with.
Intuition
If a vector had two different coordinate tuples, subtracting them would give a nontrivial combination of basis vectors equal to zero.
But basis vectors are independent, so that cannot happen. Every vector has exactly one address.
Concept
The coordinate map is linear: the coordinates of a sum are the sum of the coordinates, and scaling a vector scales its coordinates.
\[ [u+v]_B = [u]_B + [v]_B, \qquad [c\,v]_B = c\,[v]_B \]
It is also a bijection, so every n-dimensional space over F is structurally identical to the column space — a genuine isomorphism.
\[ V \;\cong\; F^n \]
Estimation
Predict first
In the plane, coordinates are obvious only in the standard basis. Try a tilted one.
Commit before you compute: what does Coordinates in a tilted basis come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by recombining
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Plug the coordinates back into the basis and confirm we land on v.
Worked example
In the plane, coordinates are obvious only in the standard basis. Try a tilted one.
\[ B = \big((2,1),\,(1,1)\big), \qquad v = (5,3) \]
Set up the defining equation
Why: The coordinates are the scalars that combine the basis vectors into v.
\[ c_1 (2,1) + c_2 (1,1) = (5,3) \]
Turn it into a linear system
Why: Match the first and second components separately.
\[ \begin{cases} 2c_1 + c_2 = 5 \\ c_1 + c_2 = 3 \end{cases} \]
Subtract the equations
Why: Subtracting the second from the first eliminates c2 and isolates c1.
\[ (2c_1 + c_2) - (c_1 + c_2) = 5 - 3 \;\Rightarrow\; c_1 = 2 \]
Back-substitute
Why: Use c1 = 2 in the second equation to get c2.
\[ 2 + c_2 = 3 \;\Rightarrow\; c_2 = 1, \qquad [v]_B = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]
Verify by recombining
Why: Plug the coordinates back into the basis and confirm we land on v.
\[ 2(2,1) + 1(1,1) = (4,2)+(1,1) = (5,3) = v \]
Picture it
Animation
Shows: Each line of the worked example "Coordinates in a tilted basis", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plug the coordinates back into the basis and confirm we land on v.
Missing information
Discussion prompt
Keep the same vector, change only the basis, and watch its coordinates change.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
In the standard basis the coordinates are just the components of the vector.
Worked example
Keep the same vector, change only the basis, and watch its coordinates change.
Standard basis coordinates
Why: In the standard basis the coordinates are just the components of the vector.
\[ [v]_E = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \]
Tilted basis coordinates
Why: We solved this system in the previous example.
\[ [v]_B = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]
Two different columns, one and the same vector. The numbers are relative; the vector is not.
Verify both describe v
Why: Each coordinate column, recombined with its own basis, returns the vector (5,3).
\[ 5e_1 + 3e_2 = (5,3) = 2(2,1)+1(1,1) \]
Picture it
Animation
Shows: Each line of the worked example "Same vector, two bases", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each coordinate column, recombined with its own basis, returns the vector (5,3).
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting mistake: treat the coordinate column as if it were the vector, with no reference to the basis.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The column (2,1) are B-coordinates; the actual vector is (5,3).
Always carry the basis with the coordinates. The tuple only means something relative to its basis.
Why: The column (2,1) are B-coordinates; the actual vector is (5,3). Reading them as standard components is wrong.
Trap
Tempting mistake: treat the coordinate column as if it were the vector, with no reference to the basis.
\[ \text{claim: } [v]_B = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \text{ means } v = (2,1) \]
This confuses address with location
Why: The column (2,1) are B-coordinates; the actual vector is (5,3). Reading them as standard components is wrong.
\[ 2(2,1)+1(1,1) = (5,3) \neq (2,1) \]
Always carry the basis with the coordinates. The tuple only means something relative to its basis.
State coordinates together with their basis
Why: The vector is (5,3); its coordinate column depends on which basis you chose.
\[ [v]_E = \begin{bmatrix}5\\3\end{bmatrix}, \quad [v]_B = \begin{bmatrix}2\\1\end{bmatrix}, \quad v = (5,3) \]
Notation
Annotate
From Trap: a vector is not its coordinate tuple — read this one piece at a time. What is each part doing?
On: \( \text{claim: } [v]_B = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \text{ means } v = (2,1) \)
Concept
Because the coordinate map is linear, you can do all the vector algebra in coordinates and translate back only at the end.
Add vectors by adding their coordinate columns; scale a vector by scaling its column.
\[ [3u - 2v]_B = 3[u]_B - 2[v]_B \]
Step zero
Discussion prompt
Coordinates of a sum — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Coordinates of each polynomial
Answer:
Worked example
Confirm the coordinate map turns a polynomial sum into a column sum. Work in degree-at-most-two polynomials with basis constant, x, x-squared.
Coordinates of each polynomial
Why: Read the coefficients in basis order.
\[ u = 1 + x \Rightarrow [u]_B = \begin{bmatrix}1\\1\\0\end{bmatrix}, \quad v = x^2 - x \Rightarrow [v]_B = \begin{bmatrix}0\\-1\\1\end{bmatrix} \]
Add the coordinate columns
Why: Linearity predicts that the coordinates of u+v are the sum of the two columns.
\[ [u]_B + [v]_B = \begin{bmatrix}1\\0\\1\end{bmatrix} \]
Verify by adding the polynomials directly
Why: Compute u+v as polynomials and read off its coordinates; they must match the column sum.
\[ u+v = (1+x)+(x^2 - x) = 1 + x^2 \Rightarrow \begin{bmatrix}1\\0\\1\end{bmatrix} \]
Reverse engineer
Discussion prompt
Work backwards. The example finished here:
Verify by adding the polynomials directly
What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.
Hint: Every quantity in the result had to enter somewhere. Account for each one.
Answer:
Confirm the coordinate map turns a polynomial sum into a column sum. Work in degree-at-most-two polynomials with basis constant, x, x-squared.
Concept
A linear map is completely determined by what it does to a basis. Record those outputs as coordinate columns and you get its matrix.
matrix of a linear map — For T from V to W with bases B and C, the matrix has, as its j-th column, the C-coordinates of T applied to the j-th basis vector of B.
\[ [T]_{B,C} = \Big[\, [Tb_1]_C \;\big|\; [Tb_2]_C \;\big|\; \cdots \;\big|\; [Tb_n]_C \,\Big] \]
Intuition
You never need T on every vector — just on the basis. Linearity fills in all the rest.
Each column answers one question: where does this basis vector land, written in the target basis? Stack the answers side by side.
Concept
Write a vector in the basis, then apply T. Linearity spreads T across the combination.
\[ T(c_1 b_1 + \cdots + c_n b_n) = c_1 T(b_1) + \cdots + c_n T(b_n) \]
In coordinates that is exactly the matrix whose columns are the images of the basis vectors, times the column of scalars.
Worked example
Differentiation is linear. Build its matrix on the degree-at-most-two polynomials in the standard basis.
\[ D(p) = p'(x), \qquad B = (1,\, x,\, x^2) \]
Differentiate each basis vector
Why: The columns of the matrix are the coordinate images of the basis vectors under D.
\[ D(1) = 0, \quad D(x) = 1, \quad D(x^2) = 2x \]
Write each image in coordinates
Why: Express each derivative in the basis constant, x, x-squared.
\[ [0]_B=\begin{bmatrix}0\\0\\0\end{bmatrix},\; [1]_B=\begin{bmatrix}1\\0\\0\end{bmatrix},\; [2x]_B=\begin{bmatrix}0\\2\\0\end{bmatrix} \]
Assemble the columns
Why: Place the three coordinate images side by side in basis order.
\[ [D]_B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]
Verify on the polynomial 7 minus 2x plus 5 x-squared
Why: Multiply the matrix by that polynomial's coordinate column and check it gives the coordinates of the derivative.
\[ [D]_B \begin{bmatrix}7\\-2\\5\end{bmatrix} = \begin{bmatrix}-2\\10\\0\end{bmatrix}, \quad p'(x) = -2 + 10x \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Matrix of differentiation on the quadratics", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiply the matrix by that polynomial's coordinate column and check it gives the coordinates of the derivative.
Estimation
Predict first
Use the derivative matrix the way a computer would: encode, multiply, decode.
Commit before you compute: what does End to end with the derivative matrix come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against direct differentiation
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Differentiate q by hand and compare with the decoded result.
Worked example
Use the derivative matrix the way a computer would: encode, multiply, decode.
\[ q(x) = 2 - x + 4x^2, \quad [D]_B = \begin{bmatrix}0&1&0\\0&0&2\\0&0&0\end{bmatrix} \]
Encode q as a coordinate column
Why: Read its coefficients in basis order.
\[ [q]_B = \begin{bmatrix}2\\-1\\4\end{bmatrix} \]
Multiply by the matrix
Why: This computes the coordinates of the derivative with pure arithmetic, no symbolic calculus.
\[ [D]_B[q]_B = \begin{bmatrix}-1\\8\\0\end{bmatrix} \]
Decode back to a polynomial
Why: Turn the output column back into a polynomial in the basis.
\[ \begin{bmatrix}-1\\8\\0\end{bmatrix} \to -1 + 8x \]
Verify against direct differentiation
Why: Differentiate q by hand and compare with the decoded result.
\[ q'(x) = 8x - 1 = -1 + 8x \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "End to end with the derivative matrix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiate q by hand and compare with the decoded result.
Concept
The domain and codomain can carry different bases. The matrix depends on both choices.
\[ [Tv]_C = [T]_{B,C}\,[v]_B \]
Read the subscript as input basis B, output basis C. Change either basis and the matrix changes.
Concept
This single equation is the whole point of matrices: applying the map and then taking coordinates equals multiplying by the matrix.
\[ [\,Tv\,]_C = [T]_{B,C}\,[\,v\,]_B \]
Left side: act first, then coordinatize. Right side: coordinatize first, then multiply. They always agree.
Picture it
Figure (svg): A commuting square: V maps to W by T along the top, coordinate maps run down the two sides, and matrix multiplication runs along the bottom between the coordinate spaces.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture a square. The top edge is the abstract map from V to W. Each side is take-coordinates. The bottom edge is multiply-by-the-matrix.
Intuition
Picture a square. The top edge is the abstract map from V to W. Each side is take-coordinates. The bottom edge is multiply-by-the-matrix.
Figure (svg): A commuting square: V maps to W by T along the top, coordinate maps run down the two sides, and matrix multiplication runs along the bottom between the coordinate spaces.
Both routes from top-left to bottom-right give the same column. The matrix is just T seen through coordinates.
Ranking
Put in order
Put the moves of Matrix of a map on the plane, then apply it into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The columns are the images of the two standard basis vectors.
Worked example
Build the standard matrix of a concrete map on the plane, then use it.
\[ T(x,y) = (x+2y,\; 3x) \]
Apply T to each standard basis vector
Why: The columns are the images of the two standard basis vectors.
\[ T(1,0) = (1,3), \qquad T(0,1) = (2,0) \]
Assemble the matrix
Why: First column is the image of the first basis vector, second column the image of the second.
\[ [T]_E = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix} \]
Apply the matrix to a test vector
Why: Multiply by the coordinates of the vector (4,1).
\[ \begin{bmatrix}1&2\\3&0\end{bmatrix}\begin{bmatrix}4\\1\end{bmatrix} = \begin{bmatrix}6\\12\end{bmatrix} \]
Verify against the formula
Why: Compute the value of T at (4,1) directly and compare with the matrix output.
\[ T(4,1) = (4+2,\; 12) = (6,12) \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Matrix of a map on the plane, then apply it", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Compute the value of T at (4,1) directly and compare with the matrix output.
Hypothesis
Predict first
A non-square matrix: three-space to the plane is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Image of each standard basis vector
Why: Three input basis vectors give three columns; each image lives in the plane.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Domain and codomain can even have different dimensions. Here T maps three-space to the plane.
\[ T(x,y,z) = (x - y,\; 2y + z) \]
Image of each standard basis vector
Why: Three input basis vectors give three columns; each image lives in the plane.
\[ T(1,0,0)=(1,0),\; T(0,1,0)=(-1,2),\; T(0,0,1)=(0,1) \]
Assemble a two-by-three matrix
Why: Two output coordinates make two rows; three basis vectors make three columns.
\[ [T] = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \]
Verify on the vector (2,1,3)
Why: Multiply the matrix by that column and compare with T computed directly.
\[ \begin{bmatrix}1&-1&0\\0&2&1\end{bmatrix}\begin{bmatrix}2\\1\\3\end{bmatrix} = \begin{bmatrix}1\\5\end{bmatrix}, \quad T(2,1,3)=(1,5)\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A non-square matrix: three-space to the plane", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiply the matrix by that column and compare with T computed directly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting mistake: lay the images of the basis vectors down as the rows of the matrix.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Multiply by the first basis vector: you get (1,2), but the true image is (1,3).
Images of the basis vectors are the columns of the matrix.
Why: Multiply by the first basis vector: you get (1,2), but the true image is (1,3). Rows give the wrong map.
Trap
Tempting mistake: lay the images of the basis vectors down as the rows of the matrix.
\[ \text{wrong: } \begin{bmatrix} 1 & 3 \\ 2 & 0 \end{bmatrix} \]
Test it and it fails
Why: Multiply by the first basis vector: you get (1,2), but the true image is (1,3). Rows give the wrong map.
\[ \begin{bmatrix}1&3\\2&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}1\\2\end{bmatrix} \neq \begin{bmatrix}1\\3\end{bmatrix} \]
Images of the basis vectors are the columns of the matrix.
\[ [T]_E = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix} \]
Now the test passes
Why: Multiplying by the first basis vector selects the first column, which is exactly its image.
\[ \begin{bmatrix}1&2\\3&0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}1\\3\end{bmatrix} = T(e_1)\;\checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Multiplying by the first basis vector selects the first column, which is exactly its image.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Multiply by the first basis vector: you get (1,2), but the true image is (1,3). Rows give the wrong map.
Concept
If T maps U to V and S maps V to W, then doing T first and S second is a single map from U to W.
\[ S \circ T : U \to W, \qquad (S\circ T)(u) = S(T(u)) \]
The composite is again linear. The question is: what is its matrix?
Concept
The matrix of a composite is the product of the matrices, in the same right-to-left order as the maps.
\[ [S \circ T] = [S]\,[T] \]
This is not a definition pulled from nowhere. It is forced by the fundamental identity.
Intuition
Apply the fundamental identity twice: push a coordinate column through the matrix of T, then through the matrix of S.
\[ [S]\big([T][v]\big) = [S][T]\,[v] = [S\circ T]\,[v] \]
Row-times-column multiplication is exactly the bookkeeping that composes the two maps. That is where the rule comes from.
Step zero
Discussion prompt
Composite equals product — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write down each matrix
Answer:
Worked example
Two maps on the plane. Compose them two ways and confirm they agree.
\[ T(x,y) = (x+y,\, y), \quad S(x,y) = (2x,\, x+y) \]
Write down each matrix
Why: Columns are the images of the standard basis vectors under each map.
\[ [T] = \begin{bmatrix}1&1\\0&1\end{bmatrix}, \quad [S] = \begin{bmatrix}2&0\\1&1\end{bmatrix} \]
Multiply the matrices, S on the left
Why: T acts first, so its matrix sits on the right of the product.
\[ [S][T] = \begin{bmatrix}2&0\\1&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix} = \begin{bmatrix}2&2\\1&2\end{bmatrix} \]
Compose the maps directly
Why: Substitute T into S and simplify to get the composite's formula.
\[ S(T(x,y)) = S(x+y,\, y) = (2(x+y),\; (x+y)+y) = (2x+2y,\; x+2y) \]
Verify the formula matches the product
Why: Read the matrix of the composite off its formula and compare with the matrix product.
\[ (2x+2y,\; x+2y) \;\leftrightarrow\; \begin{bmatrix}2&2\\1&2\end{bmatrix} = [S][T]\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Composite equals product", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Read the matrix of the composite off its formula and compare with the matrix product.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting mistake: multiply in reading order, T then S, so the matrix of T on the left.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It equals S-first-then-T, not T-first-then-S.
The map applied first goes on the right. T first means the matrix of S times the matrix of T.
Why: It equals S-first-then-T, not T-first-then-S. Test on (1,0): the true composite gives (2,1), but this gives (3,1).
Trap
Tempting mistake: multiply in reading order, T then S, so the matrix of T on the left.
\[ \text{wrong: } [T][S] = \begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix} = \begin{bmatrix}3&1\\1&1\end{bmatrix} \]
This is a different map
Why: It equals S-first-then-T, not T-first-then-S. Test on (1,0): the true composite gives (2,1), but this gives (3,1).
\[ \begin{bmatrix}3&1\\1&1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}3\\1\end{bmatrix} \neq \begin{bmatrix}2\\1\end{bmatrix} \]
The map applied first goes on the right. T first means the matrix of S times the matrix of T.
\[ [S\circ T] = [S][T] = \begin{bmatrix}2&2\\1&2\end{bmatrix} \]
This one passes the test
Why: Applying T then S to (1,0) gives (2,1), matching the product with S on the left.
\[ \begin{bmatrix}2&2\\1&2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}2\\1\end{bmatrix}\;\checkmark \]
Translation
\( \text{wrong: } [T][S] = \begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix} = \begin{bmatrix}3&1\\1&1\end{bmatrix} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Now the central move: keep the vector fixed, switch the basis, and track how the coordinate column transforms.
We need a machine that converts a vector's coordinates in one basis into its coordinates in another.
Concept
Enlarge the basis vectors and the coordinates of a fixed vector shrink. The coordinate column moves opposite to the basis.
This is why converting a vector's coordinates uses the inverse of the matrix that changes the basis. Keep this asymmetry in mind — it returns in the change-of-basis formula for maps.
Concept
The change-of-basis matrix has, as its columns, the old basis vectors written in the new basis.
\[ P = \Big[\, [b_1]_{B'} \;\big|\; \cdots \;\big|\; [b_n]_{B'} \,\Big], \qquad [v]_{B'} = P\,[v]_B \]
Multiplying by P takes B-coordinates to B-prime-coordinates.
Intuition
The change-of-basis matrix is nothing but the matrix of the identity map, read with input basis B and output basis B-prime.
\[ P = [\,\mathrm{id}\,]_{B,\,B'} \]
The identity moves no vector; it only relabels its address. That is exactly a change of coordinates.
Missing information
Discussion prompt
Convert coordinates between the standard basis and a tilted basis, in both directions.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Written in standard coordinates, these columns convert B-coordinates into standard coordinates.
Worked example
Convert coordinates between the standard basis and a tilted basis, in both directions.
\[ B = \big((1,1),\,(1,-1)\big), \quad E = \text{standard} \]
Build P with the B-vectors as columns
Why: Written in standard coordinates, these columns convert B-coordinates into standard coordinates.
\[ P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \]
Convert a B-coordinate column to standard
Why: Multiply the B-coordinates (3,1) by P.
\[ [v]_E = P\begin{bmatrix}3\\1\end{bmatrix} = \begin{bmatrix}4\\2\end{bmatrix} \]
Invert P for the reverse direction
Why: The inverse converts standard coordinates back into B-coordinates; the determinant here is negative two.
\[ P^{-1} = \begin{bmatrix} \tfrac12 & \tfrac12 \\ \tfrac12 & -\tfrac12 \end{bmatrix} \]
Convert back
Why: Multiply the standard column (4,2) by the inverse; we should recover (3,1).
\[ P^{-1}\begin{bmatrix}4\\2\end{bmatrix} = \begin{bmatrix}3\\1\end{bmatrix} \]
Verify against the actual vector
Why: The B-coordinates (3,1) should rebuild the standard vector (4,2).
\[ 3(1,1) + 1(1,-1) = (4,2) = v \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Building P and converting both ways", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The B-coordinates (3,1) should rebuild the standard vector (4,2).
Concept
A change of basis can always be undone. The columns of P are a basis, hence independent, so P is invertible.
\[ P^{-1} = [\,\mathrm{id}\,]_{B',\,B} \]
The inverse is just the change-of-basis matrix going the other way.
Anomaly
Predict first
A student writes this, and it looks reasonable:
You have a vector's standard coordinates (4,2) and want its B-coordinates. Tempting: multiply by P.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The true B-coordinates are (3,1); the output (6,2) does not rebuild (4,2), since 6(1,1)+2(1,-1) = (8,4).
P sends B-coordinates to standard, so going the other way needs the inverse.
Why: The true B-coordinates are (3,1); the output (6,2) does not rebuild (4,2), since 6(1,1)+2(1,-1) = (8,4).
Trap
You have a vector's standard coordinates (4,2) and want its B-coordinates. Tempting: multiply by P.
\[ \text{wrong: } P\begin{bmatrix}4\\2\end{bmatrix} = \begin{bmatrix}6\\2\end{bmatrix} \]
Check it and it is wrong
Why: The true B-coordinates are (3,1); the output (6,2) does not rebuild (4,2), since 6(1,1)+2(1,-1) = (8,4).
\[ 6(1,1)+2(1,-1) = (8,4) \neq (4,2) \]
P sends B-coordinates to standard, so going the other way needs the inverse.
\[ [v]_B = P^{-1}\begin{bmatrix}4\\2\end{bmatrix} = \begin{bmatrix}3\\1\end{bmatrix} \]
Confirm the direction
Why: Now the B-coordinates rebuild the vector, since 3(1,1)+1(1,-1) = (4,2).
\[ 3(1,1)+1(1,-1) = (4,2)\;\checkmark \]
Notation
Annotate
From Trap: P or P-inverse? — read this one piece at a time. What is each part doing?
On: \( 3(1,1)+1(1,-1) = (4,2)\;\checkmark \)
Concept
Changing basis also changes the matrix of a linear map. The new matrix is a sandwich of the old one between the inverse of P and P.
\[ [T]_{B'} = P^{-1}\,[T]_B\,P \]
Here P is the change-of-basis matrix that converts B-prime-coordinates into B-coordinates.
Intuition
Read the sandwich right to left: P converts a new-basis column to the old basis, the middle matrix applies the map, and the inverse of P converts the result back to the new basis.
It is the same map — but every vector is spoken to and answered in the new coordinate language.
Estimation
Predict first
Take a symmetric map and re-express it in a cleverly chosen basis.
Commit before you compute: what does Diagonalizing by change of basis come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the new basis vectors are eigenvectors
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Apply the original matrix to each new basis vector and read off the eigenvalues that appear on the diagonal.
Worked example
Take a symmetric map and re-express it in a cleverly chosen basis.
\[ [T]_E = \begin{bmatrix}3&1\\1&3\end{bmatrix}, \quad B' = \big((1,1),(1,-1)\big) \]
P has the new basis vectors as columns
Why: P converts B-prime-coordinates to standard coordinates; its inverse we computed earlier.
\[ P = \begin{bmatrix}1&1\\1&-1\end{bmatrix}, \quad P^{-1} = \begin{bmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{bmatrix} \]
Multiply the matrix by P
Why: Apply P on the right first.
\[ [T]_E\,P = \begin{bmatrix}3&1\\1&3\end{bmatrix}\begin{bmatrix}1&1\\1&-1\end{bmatrix} = \begin{bmatrix}4&2\\4&-2\end{bmatrix} \]
Multiply on the left by the inverse of P
Why: Complete the sandwich to get the matrix in the new basis.
\[ P^{-1}\begin{bmatrix}4&2\\4&-2\end{bmatrix} = \begin{bmatrix}4&0\\0&2\end{bmatrix} \]
In the new basis the matrix is diagonal — the basis vectors are eigenvectors.
Verify the new basis vectors are eigenvectors
Why: Apply the original matrix to each new basis vector and read off the eigenvalues that appear on the diagonal.
\[ A(1,1) = (4,4) = 4(1,1), \quad A(1,-1) = (2,-2) = 2(1,-1)\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Diagonalizing by change of basis", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Apply the original matrix to each new basis vector and read off the eigenvalues that appear on the diagonal.
Intuition
The diagonal result was not luck. When the basis vectors are eigenvectors, the map simply scales each one, so the matrix is diagonal.
Choosing the right basis is how you make a matrix as simple as possible. That search is the theme of the next unit.
Step zero
Discussion prompt
A rotation in a new basis — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Multiply the rotation matrix by P
Answer:
Worked example
Change basis for a quarter-turn rotation. The matrix will look different, but the map is unchanged.
\[ [R]_E = \begin{bmatrix}0&-1\\1&0\end{bmatrix}, \quad P = \begin{bmatrix}1&1\\1&-1\end{bmatrix} \]
Multiply the rotation matrix by P
Why: Apply P on the right to start the similarity sandwich.
\[ [R]_E\,P = \begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}1&1\\1&-1\end{bmatrix} = \begin{bmatrix}-1&1\\1&1\end{bmatrix} \]
Multiply on the left by the inverse of P
Why: Finish the sandwich to read the rotation in the new basis.
\[ [R]_{B'} = P^{-1}\begin{bmatrix}-1&1\\1&1\end{bmatrix} = \begin{bmatrix}0&1\\-1&0\end{bmatrix} \]
Different entries, but this is the same rotation — only the coordinate labels changed.
Verify the similarity invariants match
Why: Both matrices have the same trace and determinant, as similar matrices must.
\[ \operatorname{tr} = 0 = 0, \quad \det = 1 = 1\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A rotation in a new basis", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both matrices have the same trace and determinant, as similar matrices must.
Concept
similar matrices — Two square matrices A and A-prime are similar if A-prime equals the inverse of P times A times P, for some invertible P. They represent the same linear map in different bases.
\[ A' = P^{-1} A P \]
Definition probe
Sort into buckets
Every line below is part of the definition of ordered basis or of similar matrices — one or the other, never both. Put each where it belongs.
Intuition
Being similar is reflexive, symmetric, and transitive, so it partitions all square matrices into classes.
Each class is one abstract map, and the matrices in it are its portraits in different bases — a direct echo of the equivalence relations from earlier in the course.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Tempting mistake: since they describe the same map, the two matrices must be equal.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Same map, but the entries depend on the basis; the top-left entries alone are three versus four.
Similar means related by a change of basis, not identical. The map is the same; the coordinates are not.
Why: Same map, but the entries depend on the basis; the top-left entries alone are three versus four.
Trap
Tempting mistake: since they describe the same map, the two matrices must be equal.
\[ \begin{bmatrix}3&1\\1&3\end{bmatrix} \stackrel{?}{=} \begin{bmatrix}4&0\\0&2\end{bmatrix} \]
Entry by entry they differ
Why: Same map, but the entries depend on the basis; the top-left entries alone are three versus four.
\[ 3 \neq 4 \]
Similar means related by a change of basis, not identical. The map is the same; the coordinates are not.
What actually matches are the invariants
Why: Trace, determinant, rank and eigenvalues agree even though the individual entries do not.
\[ \operatorname{tr} = 6, \quad \det = 8 \text{ for both} \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
Some numbers computed from a matrix do not change under a change of basis. They are properties of the map itself.
Rank is one: it is the dimension of the image, which does not care how you coordinatize.
\[ \operatorname{rank}(P^{-1}AP) = \operatorname{rank}(A) \]
Concept
The trace — the sum of the diagonal entries — is unchanged by similarity, thanks to the cyclic property of the trace.
\[ \operatorname{tr}(P^{-1}AP) = \operatorname{tr}(APP^{-1}) = \operatorname{tr}(A) \]
Ranking
Put in order
Put the moves of Checking trace and determinant invariance into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Sum the diagonal entries of each matrix.
Worked example
Verify numerically that trace and determinant survive the change of basis from the diagonalizing example.
\[ A = \begin{bmatrix}3&1\\1&3\end{bmatrix}, \quad A' = \begin{bmatrix}4&0\\0&2\end{bmatrix} \]
Compare the traces
Why: Sum the diagonal entries of each matrix.
\[ \operatorname{tr}(A) = 3+3 = 6, \quad \operatorname{tr}(A') = 4+2 = 6 \]
Compare the determinants
Why: Compute both two-by-two determinants.
\[ \det(A) = 9 - 1 = 8, \quad \det(A') = 8 - 0 = 8 \]
Verify the eigenvalues agree
Why: The diagonal of A-prime shows the eigenvalues; their sum is the trace and their product is the determinant.
\[ 4+2 = 6 = \operatorname{tr}, \quad 4\cdot 2 = 8 = \det\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Checking trace and determinant invariance", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The diagonal of A-prime shows the eigenvalues; their sum is the trace and their product is the determinant.
Concept
The determinant multiplies through the sandwich and cancels; the eigenvalues are roots of the characteristic polynomial, which is itself invariant.
\[ \det(P^{-1}AP) = \det(P)^{-1}\det(A)\det(P) = \det(A) \]
So trace, determinant, rank and the whole spectrum are features of the map, not of any one matrix that represents it.
Intuition
A matrix is a photograph of a map taken from the angle of a chosen basis. Change the angle and the pixels change.
The invariants are what every photograph shares — the true shape of the object standing behind them.
Explain it
Discussion prompt
Explain Invariants belong to the map to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A matrix is a photograph of a map taken from the angle of a chosen basis. Change the angle and the pixels change.
Intuition
Every matrix in a similarity class is a different description of one map. Some descriptions are messy; a few are beautifully simple.
The art of linear algebra is picking the basis in which your map is easiest to read: diagonal, triangular, or block form.
Analogy
Discussion prompt
Explain A good basis simplifies the matrix by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every matrix in a similarity class is a different description of one map. Some descriptions are messy; a few are beautifully simple.
Concept
In computing you constantly separate a value from its representation: an integer versus its bytes, a set versus the hash table that stores it.
Linear algebra is the same discipline. The linear map is the object; the matrix is one encoding of it in a chosen basis. Keep your eye on the object.
Counterexample
Discussion prompt
In computing you constantly separate a value from its representation: an integer versus its bytes, a set versus the hash table that stores it.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Linear algebra is the same discipline. The linear map is the object; the matrix is one encoding of it in a chosen basis. Keep your eye on the object.
Pattern
1. Fix ordered bases
Why: Choose a basis B for the domain and C for the codomain; every coordinate and every matrix entry depends on this choice.
2. Coordinatize vectors
Why: Solve for the scalars that build v from the basis; that column of scalars is the coordinate vector.
3. Build the matrix by columns
Why: The j-th column is the C-coordinates of T applied to the j-th basis vector of B.
4. Compute by multiplying
Why: Apply a map in coordinates via the fundamental identity; compose maps by multiplying matrices in right-to-left order.
5. Change basis with a sandwich
Why: For vectors, multiply by P or its inverse; for maps, use the inverse of P times the matrix times P. Check invariants like trace, determinant and rank to catch errors.
Real world
Discussion prompt
Outside this lesson: where does Matrices, Coordinates & Change of Basis actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of The master recipe is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck shows how an ordered basis turns abstract vectors into coordinate tuples, how a linear map becomes a matrix, why matrix multiplication is exactly composition, and how a change of basis - that is, similarity - re-coordinates one and the same map. It targets four real errors: confusing a vector with its coordinate tuple, multiplying matrices in the wrong order for a composition, mixing up P and P-inverse in the change-of-basis formula, and thinking that similar matrices are equal.
Elimination
Eliminate the wrong options
Let B = ((1,0),(1,1)) be an ordered basis of the plane and v = (3,5). What is the coordinate column of v in the basis B?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Solve c1(1,0)+c2(1,1)=(3,5). The second component gives c2=5, then c1+c2=3 forces c1=-2, so the column is (-2,5). Rebuilding: -2(1,0)+5(1,1)=(3,5).
Check
Solve the system, then choose.
Check your understanding
Let B = ((1,0),(1,1)) be an ordered basis of the plane and v = (3,5). What is the coordinate column of v in the basis B?
Answer: A
Why: Solve c1(1,0)+c2(1,1)=(3,5). The second component gives c2=5, then c1+c2=3 forces c1=-2, so the column is (-2,5). Rebuilding: -2(1,0)+5(1,1)=(3,5).
Prediction
Predict first
T has matrix A = [[1,1],[0,1]] and S has matrix B = [[2,0],[1,1]]. Apply T first, then S. Which matrix represents the composite S after T?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: [[2, 2], [1, 2]]
Why: T acts first, so its matrix goes on the right: the composite is B times A. Multiplying B by A gives rows [2,2] and [1,2], the matrix [[2,2],[1,2]].
Check
Mind the order before you multiply.
Check your understanding
T has matrix A = [[1,1],[0,1]] and S has matrix B = [[2,0],[1,1]]. Apply T first, then S. Which matrix represents the composite S after T?
Answer: A
Why: T acts first, so its matrix goes on the right: the composite is B times A. Multiplying B by A gives rows [2,2] and [1,2], the matrix [[2,2],[1,2]].
Check
Watch the direction of the conversion.
Check your understanding
The matrix P = [[2,1],[1,1]] has the basis-B vectors as its columns, so it converts B-coordinates to standard coordinates. A vector has B-coordinates (3,2). What are its standard coordinates?
Answer: A
Why: Since P converts B-coordinates to standard coordinates, multiply: P times (3,2) equals (23+12, 13+12) = (8,5). Rebuilding, 3(2,1)+2(1,1)=(8,5).
Commit first
Predict first
Matrices A and A-prime are similar, meaning A-prime equals the inverse of P times A times P for some invertible P. Which statement is FALSE?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: A and A-prime are equal entry by entry.
Why: Similar matrices share invariants such as trace, determinant, and rank, but their individual entries depend on the basis and generally differ, so the claim that they are equal entry by entry is the false one.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
One of these statements is false.
Check your understanding
Matrices A and A-prime are similar, meaning A-prime equals the inverse of P times A times P for some invertible P. Which statement is FALSE?
Answer: C
Why: Similar matrices share invariants such as trace, determinant, and rank, but their individual entries depend on the basis and generally differ, so the claim that they are equal entry by entry is the false one.
Prediction
Predict first
In the plane, the basis B = (b1, b2) has b1 = (1,2) and b2 = (0,3). A vector w has coordinate column (2, -1) in this basis. What is w in standard coordinates?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: (2, 1)
Why: Rebuild w by combining the basis with its coordinates: 2(1,2) + (-1)(0,3) = (2, 4 minus 3) = (2,1). The coordinate column is not the vector itself.
Check
Recombine the basis with the coordinates.
Check your understanding
In the plane, the basis B = (b1, b2) has b1 = (1,2) and b2 = (0,3). A vector w has coordinate column (2, -1) in this basis. What is w in standard coordinates?
Answer: A
Why: Rebuild w by combining the basis with its coordinates: 2(1,2) + (-1)(0,3) = (2, 4 minus 3) = (2,1). The coordinate column is not the vector itself.
Elimination
Eliminate the wrong options
A linear map T on the plane satisfies T(1,0) = (2,5) and T(0,1) = (3,1). What is the standard matrix of T?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The columns of the standard matrix are the images of the basis vectors: first column T(1,0) = (2,5), second column T(0,1) = (3,1), giving the matrix [[2,3],[5,1]].
Check
Remember where the images go.
Check your understanding
A linear map T on the plane satisfies T(1,0) = (2,5) and T(0,1) = (3,1). What is the standard matrix of T?
Answer: A
Why: The columns of the standard matrix are the images of the basis vectors: first column T(1,0) = (2,5), second column T(0,1) = (3,1), giving the matrix [[2,3],[5,1]].
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The master recipe · Coordinates: naming vectors with numbers · A basis is an address system · The standard basis is the default · An ordered basis. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now move fluently between abstract vectors and their coordinate columns, and between linear maps and their matrices.
| Object | Its coordinate form |
|---|---|
| a vector v | the column of coordinates in a basis |
| a linear map T | the matrix whose columns are basis images |
| composition, T then S | the product with S on the left |
| a change of basis | the similarity sandwich with P and its inverse |
The through-line: a matrix is a representation in a chosen basis; the linear map is the real object. Invariants like trace, determinant and rank belong to the map, not to any single matrix for it.
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