This 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.
Subject: Foundations of Higher Mathematics · 102 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. Define rank and nullity as the dimensions of the image and the kernel.
2. State and prove the Rank–Nullity Theorem by extending a basis of the kernel.
3. Compute rank and nullity of an explicit matrix by row reduction.
4. Use the theorem to decide when a linear map can or cannot be injective or surjective.
5. Read the solution set of a linear system as one particular solution plus the kernel.
Warm-up
Discussion prompt
Before we open The Rank–Nullity Theorem: without looking back, what was the main idea of Linear Maps, 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 presents linear maps as the structure-preserving morphisms between vector spaces. It gives the two axioms, explains why the origin is fixed, shows that a basis determines the whole map, and covers the kernel and image as subspaces and injectivity as the kernel being trivial. It targets the classic traps of calling affine or squaring maps linear, thinking that one vector pins a map down, locating the kernel in the wrong space, and assuming that injective forces surjective.
Concept
For a linear map, the image (or range) is the set of every output it can produce.
\[ \operatorname{im} T = \{\, T(v) : v \in V \,\} \subseteq W \]
It is a subspace of the codomain, not just a subset. It measures how much of the target the map actually reaches.
Counterexample
Discussion prompt
For a linear map, the image (or range) is the set of every output it can produce.
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.
Concept
The rank of a linear map is the dimension of its image.
rank — The dimension of the image (range) of a linear map. It counts how many independent output directions the map can hit.
\[ \operatorname{rank}(T) = \dim(\operatorname{im} T) \]
Analogy
Discussion prompt
Explain Rank is the dimension of the image 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:
The rank of a linear map is the dimension of its image.
Concept
The kernel (or null space) is the set of every input the map sends to the zero vector.
\[ \ker T = \{\, v \in V : T(v) = 0 \,\} \subseteq V \]
It too is a subspace, this time of the domain. It measures how much the map collapses.
Explain it
Discussion prompt
Explain Recall: the kernel of a linear 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:
The kernel (or null space) is the set of every input the map sends to the zero vector.
Concept
The nullity of a linear map is the dimension of its kernel.
nullity — The dimension of the kernel (null space) of a linear map. It counts how many independent input directions get crushed to zero.
\[ \operatorname{nullity}(T) = \dim(\ker T) \]
Definition probe
Sort into buckets
Every line below is part of the definition of rank or of nullity — one or the other, never both. Put each where it belongs.
Intuition
Picture the domain as a bundle of independent directions feeding into the map.
Some directions get crushed to zero: those live in the kernel, and there are exactly nullity of them.
The rest survive as independent output directions: there are exactly rank of them.
Every input direction does one or the other. That accounting is the whole theorem.
Ranking
Put in order
Put the moves of Worked example: a coordinate flattening map 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 kernel is every input the map sends to the zero vector.
Worked example
Take the map on the plane that keeps the first coordinate and zeroes the second.
\[ T(x,y) = (x,\,0) \]
Find the kernel
Why: The kernel is every input the map sends to the zero vector.
\[ T(x,y) = (0,0) \iff x = 0 \]
Describe the kernel
Why: Only the first coordinate must vanish; the second is unconstrained.
\[ \ker T = \{(0,y) : y \in \mathbb{R}\} \]
This is the y-axis, a one-dimensional subspace, so the nullity is 1.
Find the image
Why: The outputs are exactly the vectors whose second coordinate is zero.
\[ \operatorname{im} T = \{(x,0) : x \in \mathbb{R}\} \]
This is the x-axis, one-dimensional, so the rank is 1.
Read off rank and nullity
Why: Rank is the dimension of the image; nullity the dimension of the kernel.
\[ \operatorname{rank}(T) = 1, \qquad \operatorname{nullity}(T) = 1 \]
Verify the accounting
Why: Rank plus nullity should equal the dimension of the domain.
\[ 1 + 1 = 2 = \dim \mathbb{R}^2 \]
Picture it
Animation
Shows: Each line of the worked example "a coordinate flattening map", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Rank plus nullity should equal the dimension of the domain.
Concept
Rank and nullity are dimensions of subspaces, so they do not depend on any choice of basis or coordinates.
You can compute them with whatever matrix representation is convenient; the numbers you get are intrinsic to the map.
That is why the theorem we are about to state is a statement about the map itself, not about one matrix for it.
Concept
Here is the central result. Let the domain be finite-dimensional.
\[ T : V \to W, \quad \dim V < \infty \]
\[ \dim V = \operatorname{rank}(T) + \operatorname{nullity}(T) \]
In words: the dimension of the domain splits exactly into the dimension of the image plus the dimension of the kernel.
Intuition
Think of it as a conservation law for degrees of freedom.
You feed in a fixed number of independent directions, namely the dimension of the domain.
The map can either lose a direction (send it into the kernel) or transmit it (into the image). Nothing else can happen, and nothing is double-counted.
So the input budget is partitioned, never created or destroyed: lost plus transmitted equals total.
Picture it
Figure (svg): A horizontal bar of total length labelled dim V, split into a left segment labelled rank (image) and a right segment labelled nullity (kernel).
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:
The domain's dimension is a fixed-length bar. The map splits it into two pieces.
Concept
The domain's dimension is a fixed-length bar. The map splits it into two pieces.
Figure (svg): A horizontal bar of total length labelled dim V, split into a left segment labelled rank (image) and a right segment labelled nullity (kernel).
Grow the kernel and the image must shrink by exactly as much. The total length never changes.
Step zero
Discussion prompt
Worked example: flattening three-space — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the kernel
Answer:
Worked example
Now project three-space onto its horizontal plane by zeroing the third coordinate.
\[ T(x,y,z) = (x,\,y,\,0) \]
Find the kernel
Why: Both of the first two outputs must be zero; the third input is already discarded.
\[ \ker T = \{(0,0,z) : z \in \mathbb{R}\} \]
The z-axis: nullity is 1.
Find the image
Why: Outputs are every vector with third coordinate zero.
\[ \operatorname{im} T = \{(x,y,0)\} \]
The horizontal plane: rank is 2.
Verify the theorem
Why: Rank plus nullity must recover the dimension of the domain.
\[ 2 + 1 = 3 = \dim \mathbb{R}^3 \]
Picture it
Animation
Shows: Each line of the worked example "flattening three-space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Rank plus nullity must recover the dimension of the domain.
Concept
When the map is a matrix, row reduction hands you both numbers at once.
The rank equals the number of pivot columns. The nullity equals the number of free columns (columns without a pivot).
Since every column is either a pivot column or a free column, the two counts add up to the total number of columns, which is the dimension of the domain.
\[ (\#\text{pivots}) + (\#\text{free}) = (\#\text{columns}) = \dim V \]
Estimation
Predict first
Compute rank and nullity of this matrix acting on four-dimensional space.
Commit before you compute: what does Worked example: rank and nullity of a 3-by-4 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 the theorem
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. The domain is four-dimensional, so the two counts must sum to 4.
Worked example
Compute rank and nullity of this matrix acting on four-dimensional space.
\[ A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 1 & 2 & 1 & 4 \end{pmatrix} \]
Eliminate below the first pivot
Why: Subtract row 1 from row 3 to clear the leading 1 in the third row.
\[ R_3 \to R_3 - R_1 : \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 1 & 3 \end{pmatrix} \]
Eliminate the duplicate row
Why: Row 3 now equals row 2, so subtracting kills it.
\[ R_3 \to R_3 - R_2 : \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
Count pivots
Why: Pivots sit in columns 1 and 3; columns 2 and 4 are free.
\[ \operatorname{rank}(A) = 2, \qquad \operatorname{nullity}(A) = 2 \]
Verify the theorem
Why: The domain is four-dimensional, so the two counts must sum to 4.
\[ 2 + 2 = 4 = \dim \mathbb{R}^4 \]
Picture it
Animation
Shows: Each line of the worked example "rank and nullity of a 3-by-4 matrix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The domain is four-dimensional, so the two counts must sum to 4.
Concept
The free columns tell you not just the nullity but an actual basis of the kernel.
Set one free variable to 1 and the others to 0, solve for the pivot variables, and record the resulting vector. One such vector per free column.
Those vectors are automatically independent (each has a lone 1 in its own free slot), and they span the kernel. So there are exactly nullity of them.
Trap
A tempting misread of the theorem: add rank and nullity to get the dimension of the codomain.
\[ \text{claim: } \operatorname{rank}(T) + \operatorname{nullity}(T) = \dim W \;(?) \]
Test it on the flattening map from three-space to three-space with rank 2 and nullity 1.
\[ 2 + 1 = 3 = \dim W \;\text{— looks fine, but only because } \dim V = \dim W \]
Now break the coincidence with a map into a bigger codomain.
\[ T:\mathbb{R}^2 \to \mathbb{R}^5,\ T(x,y)=(x,y,0,0,0) \]
\[ \operatorname{rank}=2,\ \operatorname{nullity}=0,\ 2+0 = 2 \neq 5 = \dim W \]
The sum always recovers the dimension of the domain, never the codomain.
\[ \operatorname{rank}(T) + \operatorname{nullity}(T) = \dim V \]
For the same map into five-space:
\[ 2 + 0 = 2 = \dim \mathbb{R}^2 \;\checkmark \]
The codomain only sets an upper bound on the rank; it plays no part in the sum.
Missing information
Discussion prompt
Return to the reduced 3-by-4 matrix. The free variables were the second and fourth.
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:
Rows 1 and 2 give the pivot variables in terms of the free ones.
Worked example
Return to the reduced 3-by-4 matrix. The free variables were the second and fourth.
\[ \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]
Solve for the pivot variables
Why: Rows 1 and 2 give the pivot variables in terms of the free ones.
\[ x_1 = -2x_2 - x_4, \qquad x_3 = -3x_4 \]
Toggle the first free variable on
Why: Set the second variable to 1 and the fourth to 0 to isolate one kernel direction.
\[ k_1 = (-2,\,1,\,0,\,0) \]
Toggle the second free variable on
Why: Set the fourth variable to 1 and the second to 0 to isolate the other.
\[ k_2 = (-1,\,0,\,-3,\,1) \]
Verify both lie in the kernel
Why: Each should be sent to the zero vector, confirming nullity is 2.
\[ A k_1 = 0, \quad A k_2 = 0, \quad \dim(\ker A) = 2 \]
Picture it
Animation
Shows: Each line of the worked example "a basis for that kernel", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each should be sent to the zero vector, confirming nullity is 2.
Concept
For a matrix map, the image has a concrete description: it is the span of the columns.
\[ \operatorname{im} A = \operatorname{span}\{\, A e_1, \dots, A e_n \,\} = \operatorname{span}(\text{columns of } A) \]
So the rank is the number of independent columns, which is exactly the number of pivot columns you found by reduction.
Intuition
Read rank as bandwidth: how many independent signals make it through the map.
A high nullity map is lossy, folding many inputs onto the same output. A zero-nullity map loses nothing and is injective.
The theorem says loss plus throughput is a fixed constant, set by the size of the input.
Concept
The proof is a single clean idea executed in three moves.
Move 1. Take a basis of the kernel.
Move 2. Extend it to a basis of the whole domain by adding some new vectors.
Move 3. Show the images of exactly those new vectors form a basis of the image. Counting then gives the theorem.
Concept
Fix the dimensions and bases before proving anything.
\[ \dim V = n, \quad \ker T \text{ has basis } (u_1,\dots,u_m), \quad m = \operatorname{nullity}(T) \]
Extend that kernel basis to a basis of the full domain by adjoining new vectors.
\[ (u_1,\dots,u_m,\; w_1,\dots,w_p) \text{ is a basis of } V, \quad n = m + p \]
The goal reduces to one claim: the images of the adjoined vectors form a basis of the image, so the rank equals the count of new vectors.
\[ \text{claim: } (T w_1,\dots,T w_p) \text{ is a basis of } \operatorname{im} T \]
Step zero
Discussion prompt
Worked example: the proof, part 1 (spanning) — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take a generic image element
Answer:
Worked example
Show the images of the adjoined vectors span the whole image.
Take a generic image element
Why: Every element of the image is the map applied to some domain vector.
\[ y = T(v), \quad v \in V \]
Expand v in the basis
Why: The combined list is a basis of the domain, so v is a unique combination of it.
\[ v = \textstyle\sum_i a_i u_i + \sum_j b_j w_j \]
Apply T and kill the kernel terms
Why: Each u_i lies in the kernel, so its image is zero; only the w-terms survive.
\[ T(v) = \sum_i a_i \underbrace{T(u_i)}_{=\,0} + \sum_j b_j T(w_j) = \sum_j b_j T(w_j) \]
Conclude spanning
Why: An arbitrary image element is a combination of the adjoined images.
\[ \operatorname{im} T = \operatorname{span}(T w_1,\dots,T w_p) \]
Verify the kernel step
Why: Confirm the vanishing that made the argument work: each kernel-basis vector really maps to zero.
\[ u_i \in \ker T \implies T(u_i) = 0 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the proof, part 1 (spanning)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Confirm the vanishing that made the argument work: each kernel-basis vector really maps to zero.
Hypothesis
Predict first
Worked example: the proof, part 2 (independence) 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: Assume a vanishing combination
Why: Independence means the only way to get zero is with all-zero coefficients.
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
Now show those same images are linearly independent.
Assume a vanishing combination
Why: Independence means the only way to get zero is with all-zero coefficients.
\[ \sum_j c_j\, T(w_j) = 0 \]
Pull the sum inside T
Why: Linearity lets you move the combination through the map.
\[ T\!\left(\sum_j c_j w_j\right) = 0 \implies \sum_j c_j w_j \in \ker T \]
Rewrite via the kernel basis
Why: Anything in the kernel is a combination of the kernel basis vectors.
\[ \sum_j c_j w_j = \sum_i d_i u_i \]
Use independence of the full basis
Why: Move everything to one side; the combined list is a basis, hence independent, so every coefficient is zero.
\[ \sum_j c_j w_j - \sum_i d_i u_i = 0 \implies c_j = 0,\ d_i = 0 \ \forall i,j \]
Verify the count closes the theorem
Why: The adjoined images are a basis of the image, so the rank is p; add the nullity m and recover n.
\[ \operatorname{rank}(T) = p, \quad p + m = n = \dim V \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the proof, part 2 (independence)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The adjoined images are a basis of the image, so the rank is p; add the nullity m and recover n.
Intuition
The kernel basis names the directions that die. The vectors you add to complete the basis are precisely the directions that live.
The map carries those survivors faithfully to an independent set downstream, and that downstream set is a basis of the image.
So the proof is not a trick; it is the bookkeeping of the crushed-versus-surviving picture made exact.
Concept
The image sits inside the codomain and is built from the domain, so the rank is squeezed from both sides.
\[ \operatorname{rank}(T) \le \dim V \quad\text{and}\quad \operatorname{rank}(T) \le \dim W \]
\[ \operatorname{rank}(T) \le \min(\dim V,\ \dim W) \]
These tiny inequalities, combined with the theorem, decide most injectivity and surjectivity questions instantly.
Fill the middle
Fill in the blanks
From Worked example: differentiation on cubics — finish the line. Write what belongs on the right of the equals sign before you look.
D : P_3 \to P_3, \quad D(a + bx + cx^2 + dx^3) = b + 2cx + 3dx^2
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. A polynomial has zero derivative exactly when it is constant.
Worked example
Let the map be differentiation on polynomials of degree at most 3, a four-dimensional space.
\[ D : P_3 \to P_3, \quad D(a + bx + cx^2 + dx^3) = b + 2cx + 3dx^2 \]
Find the kernel
Why: A polynomial has zero derivative exactly when it is constant.
\[ \ker D = \{\text{constants}\} = \operatorname{span}(1), \quad \operatorname{nullity}(D) = 1 \]
Find the image
Why: Derivatives of cubics are exactly the polynomials of degree at most 2.
\[ \operatorname{im} D = P_2 = \operatorname{span}(1,x,x^2), \quad \operatorname{rank}(D) = 3 \]
Verify the theorem
Why: Rank plus nullity must equal the dimension of the domain.
\[ 3 + 1 = 4 = \dim P_3 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "differentiation on cubics", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Rank plus nullity must equal the dimension of the domain.
Ranking
Put in order
Put the moves of Worked example: rotation loses nothing 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. A rotation moves every nonzero vector to another nonzero vector of the same length.
Worked example
Rotate the plane by a fixed angle about the origin.
\[ R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \]
Find the kernel
Why: A rotation moves every nonzero vector to another nonzero vector of the same length.
\[ R_\theta v = 0 \iff v = 0, \quad \ker R_\theta = \{0\} \]
Read off nullity and rank
Why: The trivial kernel gives nullity zero; the theorem then forces full rank.
\[ \operatorname{nullity} = 0 \implies \operatorname{rank} = 2 - 0 = 2 \]
Verify against the codomain
Why: Full rank equal to the codomain dimension means the map is onto the plane as well.
\[ \operatorname{rank} = 2 = \dim \mathbb{R}^2 \implies \text{surjective and injective} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "rotation loses nothing", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Full rank equal to the codomain dimension means the map is onto the plane as well.
Trap
A common slip: since the map lands in the codomain, treat the rank as the dimension of the codomain.
\[ T:\mathbb{R}^4 \to \mathbb{R}^3, \quad \text{``}\operatorname{rank}(T) = 3\text{ because the target is } \mathbb{R}^3\text{''} \]
But a map can fail to fill its codomain. Consider one whose image is a single line.
\[ T(x,y,z,w) = (x+y+z+w,\ 0,\ 0) \]
\[ \text{actual } \operatorname{rank}(T) = 1, \ \text{not } 3 \]
Rank is the dimension of the image, which can be far smaller than the codomain.
\[ \operatorname{rank}(T) = \dim(\operatorname{im} T) \le \dim W \]
The codomain is only a container. To find the rank you must measure how much of it the map actually reaches.
\[ \operatorname{im} T = \operatorname{span}\{(1,0,0)\} \implies \operatorname{rank} = 1 \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Rank is the dimension of the image, which can be far smaller than the codomain.
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.
Concept
Suppose the domain has bigger dimension than the codomain. Then the rank is capped by the smaller codomain, and the theorem forces a positive nullity.
\[ \dim V > \dim W \implies \operatorname{nullity}(T) = \dim V - \operatorname{rank}(T) \ge \dim V - \dim W > 0 \]
A positive nullity means a nonzero kernel, so the map cannot be injective. Too many inputs, too little room.
Estimation
Predict first
Show that no linear map from three-space to the plane can be injective.
Commit before you compute: what does Worked example: three-space into the plane 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 injectivity conclusion
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. A nontrivial kernel is exactly the obstruction to injectivity.
Worked example
Show that no linear map from three-space to the plane can be injective.
\[ T : \mathbb{R}^3 \to \mathbb{R}^2 \]
Bound the rank
Why: The image lives in the plane, so the rank is at most 2.
\[ \operatorname{rank}(T) \le \dim \mathbb{R}^2 = 2 \]
Apply the theorem
Why: Nullity is the domain dimension minus the rank.
\[ \operatorname{nullity}(T) = 3 - \operatorname{rank}(T) \ge 3 - 2 = 1 \]
Interpret the nullity
Why: A kernel of dimension at least 1 contains nonzero vectors sent to zero.
\[ \dim(\ker T) \ge 1 \implies \ker T \neq \{0\} \]
Verify the injectivity conclusion
Why: A nontrivial kernel is exactly the obstruction to injectivity.
\[ \ker T \neq \{0\} \implies T \text{ is not injective} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "three-space into the plane", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A nontrivial kernel is exactly the obstruction to injectivity.
Intuition
This is the linear-algebra echo of the pigeonhole principle from the functions unit.
If you try to pack a big space into a small one linearly, some independent directions have nowhere to go, so they must be crushed together.
Where pigeonhole counts elements, rank-nullity counts dimensions. Same spirit, sharper tool.
Concept
When the domain and codomain have the same finite dimension, three properties collapse into one.
\[ T : V \to V, \quad \dim V < \infty \]
\[ T \text{ injective} \iff T \text{ surjective} \iff T \text{ bijective} \]
Checking any one of the three settles all three. This is the workhorse behind so much of linear algebra.
Concept
A linear map is injective exactly when its kernel is trivial.
\[ T \text{ injective} \iff \ker T = \{0\} \iff \operatorname{nullity}(T) = 0 \]
Because the map is linear, collapsing anything to zero is the only way two inputs can share an output. No kernel, no collision.
Concept
A linear map is surjective exactly when its image fills the codomain, that is, when the rank equals the codomain dimension.
\[ T \text{ surjective} \iff \operatorname{im} T = W \iff \operatorname{rank}(T) = \dim W \]
For an endomorphism the codomain equals the domain, so this reads: surjective exactly when the rank equals the dimension of the space.
Step zero
Discussion prompt
Worked example: injective forces surjective — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Translate injectivity into nullity
Answer:
Worked example
Prove that an injective linear map from a finite-dimensional space to itself is automatically surjective.
\[ T : V \to V, \quad \dim V = n, \quad T \text{ injective} \]
Translate injectivity into nullity
Why: Injective means the kernel is trivial, which is nullity zero.
\[ T \text{ injective} \implies \operatorname{nullity}(T) = 0 \]
Solve for the rank
Why: Rank-nullity turns the known nullity into the rank.
\[ \operatorname{rank}(T) = n - 0 = n \]
Compare image to the whole space
Why: A subspace whose dimension equals the ambient dimension is the whole space.
\[ \operatorname{im} T \subseteq V, \ \dim(\operatorname{im} T) = n = \dim V \implies \operatorname{im} T = V \]
Verify surjectivity
Why: Image equal to the codomain is exactly the definition of surjective.
\[ \operatorname{im} T = V \implies T \text{ surjective} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "injective forces surjective", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Image equal to the codomain is exactly the definition of surjective.
Missing information
Discussion prompt
Now run the same machine backwards on a surjective endomorphism.
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:
Surjective means the image is all of V, so the rank is the full dimension.
Worked example
Now run the same machine backwards on a surjective endomorphism.
\[ T : V \to V, \quad \dim V = n, \quad T \text{ surjective} \]
Translate surjectivity into rank
Why: Surjective means the image is all of V, so the rank is the full dimension.
\[ T \text{ surjective} \implies \operatorname{rank}(T) = n \]
Solve for the nullity
Why: Rank-nullity turns the known rank into the nullity.
\[ \operatorname{nullity}(T) = n - n = 0 \]
Verify injectivity
Why: A trivial kernel is exactly injectivity.
\[ \ker T = \{0\} \implies T \text{ injective} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "surjective forces injective", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A trivial kernel is exactly injectivity.
Trap
Believing that injective always implies surjective, even in infinite-dimensional spaces.
Consider the right-shift on the space of real sequences, which is infinite-dimensional.
\[ S(a_1,a_2,a_3,\dots) = (0,a_1,a_2,\dots) \]
It is injective: distinct sequences shift to distinct sequences, so the kernel is trivial. Yet nothing maps to a sequence starting with 1.
\[ (1,0,0,\dots) \notin \operatorname{im} S \implies S \text{ injective but not surjective} \]
The equivalence of injective and surjective is a finite-dimensional phenomenon. It relies on rank-nullity, which needs a finite domain dimension to even make sense.
\[ \dim V = \operatorname{rank}(T) + \operatorname{nullity}(T) \quad\text{requires } \dim V < \infty \]
When the dimension is infinite, both sides are infinite and the bookkeeping that trades rank against nullity breaks down. The shift is the standard counterexample.
Concept
A system of linear equations is just one linear map applied to an unknown vector.
\[ A x = b, \quad A : \mathbb{R}^n \to \mathbb{R}^m \]
Solving the system means finding every input the map sends to the specific output on the right. Rank-nullity governs the shape of that solution set.
Concept
The homogeneous system, with zero on the right, asks exactly for the kernel.
\[ A x = 0 \iff x \in \ker A \]
So the homogeneous solution set is a subspace, and its dimension is the nullity: the number of free variables.
Concept
If the system is consistent, fix any one solution. Every other solution differs from it by something in the kernel.
\[ A x = b,\ A p = b \implies A(x - p) = 0 \implies x - p \in \ker A \]
\[ \{\text{solutions}\} = p + \ker A \]
The solution set is a translate of the kernel: one particular solution shifted by every homogeneous solution. Same shape and dimension as the kernel, just moved off the origin.
Estimation
Predict first
Solve the system using the 3-by-4 matrix from before, with a right-hand side chosen to be consistent.
Commit before you compute: what does Worked example: the full solution set 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 particular solution
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 p back into the original system and check all three equations hold.
Worked example
Solve the system using the 3-by-4 matrix from before, with a right-hand side chosen to be consistent.
\[ A x = b, \quad A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 1 & 2 & 1 & 4 \end{pmatrix}, \quad b = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} \]
Find one particular solution
Why: Set both free variables to zero and solve the pivot equations.
\[ x_2 = x_4 = 0 \implies x_1 = 1,\ x_3 = 3, \quad p = (1,0,3,0) \]
Attach the kernel
Why: The homogeneous solutions are the kernel basis found earlier, of dimension equal to the nullity 2.
\[ \ker A = \operatorname{span}\{(-2,1,0,0),\ (-1,0,-3,1)\} \]
Write the general solution
Why: Every solution is the particular one plus an arbitrary kernel element.
\[ x = (1,0,3,0) + s(-2,1,0,0) + t(-1,0,-3,1) \]
Verify the particular solution
Why: Plug p back into the original system and check all three equations hold.
\[ A(1,0,3,0)^{\mathsf T} = (1,\,3,\,1+3) = (1,3,4) = b \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the full solution set", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plug p back into the original system and check all three equations hold.
Concept
The number of free variables in a system equals the nullity of its matrix.
\[ \#(\text{free variables}) = n - \operatorname{rank}(A) = \operatorname{nullity}(A) \]
Each free variable is one degree of freedom in the answer: a knob you can turn while staying a solution. Zero free variables means at most one solution.
Concept
More unknowns than equations is called underdetermined. There the rank cannot reach the number of unknowns, so the nullity is positive.
\[ n > m \implies \operatorname{rank}(A) \le m < n \implies \operatorname{nullity}(A) \ge n - m > 0 \]
So a consistent underdetermined system always has infinitely many solutions, never a unique one. Overdetermined systems, with more equations than unknowns, instead risk being inconsistent.
Trap
Hoping a wide matrix (more columns than rows) could have only the zero vector in its kernel, making its columns independent.
\[ A : \mathbb{R}^5 \to \mathbb{R}^3, \quad \text{``}\ker A = \{0\}\text{''} \;(?) \]
But the rank is capped by the number of rows, so the nullity cannot be zero.
\[ \operatorname{rank}(A) \le 3 \implies \operatorname{nullity}(A) = 5 - \operatorname{rank}(A) \ge 2 \]
A wide matrix always has a nontrivial kernel, so its columns are always dependent.
\[ n > m \implies \operatorname{nullity}(A) \ge n - m > 0 \implies \ker A \neq \{0\} \]
Five vectors in three-dimensional space cannot be independent. The kernel dimension of at least 2 records exactly how much dependence is forced.
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
For a square matrix, rank-nullity ties invertibility to a single number.
\[ A \ (n \times n): \quad \text{invertible} \iff \operatorname{nullity}(A) = 0 \iff \operatorname{rank}(A) = n \]
Full rank, trivial kernel, injective, surjective, invertible: for a square matrix these are all the same statement. This is the heart of the invertible-matrix theorem.
Step zero
Discussion prompt
Worked example: invertibility by rank — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Count pivots
Answer:
Worked example
Decide whether this square matrix is invertible using only rank and nullity.
\[ C = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix} \]
Count pivots
Why: The matrix is already upper triangular with nonzero diagonal, so each column has a pivot.
\[ \operatorname{rank}(C) = 3 \]
Get the nullity
Why: Rank-nullity with a three-dimensional domain.
\[ \operatorname{nullity}(C) = 3 - 3 = 0 \]
Verify invertibility
Why: Trivial kernel and full rank on a square matrix mean invertible.
\[ \operatorname{nullity}(C) = 0,\ \operatorname{rank}(C) = 3 \implies C \text{ invertible} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "invertibility by rank", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Trivial kernel and full rank on a square matrix mean invertible.
Intuition
Engineers meet rank-nullity as a degrees-of-freedom count.
You have as many input knobs as the domain dimension. The constraints imposed by a linear map use up rank of them, leaving nullity free to vary.
Solving a design problem often means arranging for a specific nullity: enough freedom to satisfy every requirement, but no wasted, unconstrained motion.
Ranking
Put in order
Put the moves of Worked example: a wide 2-by-4 matrix 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. Leading ones sit in columns 1 and 2; columns 3 and 4 are free.
Worked example
Read rank and nullity straight off an already-reduced wide matrix.
\[ B = \begin{pmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 3 \end{pmatrix} \]
Identify pivots
Why: Leading ones sit in columns 1 and 2; columns 3 and 4 are free.
\[ \operatorname{rank}(B) = 2, \qquad \operatorname{nullity}(B) = 2 \]
Build the kernel basis
Why: Solve the pivot variables in terms of the free ones, toggling each free variable in turn.
\[ (-2,\,1,\,1,\,0) \ \text{and} \ (1,\,-3,\,0,\,1) \]
Verify the theorem
Why: The two counts must add to the four columns of the domain.
\[ 2 + 2 = 4 = \dim \mathbb{R}^4 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a wide 2-by-4 matrix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two counts must add to the four columns of the domain.
Concept
A quietly remarkable fact: the number of independent rows of a matrix equals the number of independent columns.
\[ \operatorname{rank}(A) = \operatorname{rank}(A^{\mathsf T}) \]
That is why row reduction, which manipulates rows, correctly computes the rank we defined through the image, which lives in the columns. One number, two guises.
Concept
Chaining two linear maps cannot increase rank; a bottleneck anywhere limits the whole pipeline.
\[ \operatorname{rank}(S \circ T) \le \min\{\operatorname{rank}(S),\ \operatorname{rank}(T)\} \]
Once information is crushed into the kernel of one stage, no later stage can recover it. Rank is monotone down a composition.
Explain it
Discussion prompt
Explain Rank under composition 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:
Chaining two linear maps cannot increase rank; a bottleneck anywhere limits the whole pipeline.
Intuition
Almost every consequence in this deck is the same equation read differently.
Injectivity is the nullity being zero. Surjectivity is the rank being full. Solvability of a system is the rank meeting the right-hand side. Invertibility is both at once.
Learn to translate any question about a linear map into a statement about its rank or nullity, and the theorem answers it.
Analogy
Discussion prompt
Explain One theorem, many faces 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:
Almost every consequence in this deck is the same equation read differently.
Concept
Rank-nullity is the linear-algebra member of a family you have already met.
It plays the same structural role as the counting behind cosets and Lagrange, and it feeds directly into the classification of finite-dimensional spaces and the first isomorphism theorem: a map's domain is accounted for by what it kills plus what it preserves.
Kernel measures collapse; image measures survival; their dimensions add back to the source. That template recurs across the whole course.
Counterexample
Discussion prompt
Rank-nullity is the linear-algebra member of a family you have already met.
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:
Kernel measures collapse; image measures survival; their dimensions add back to the source. That template recurs across the whole course.
Pattern
1. Fix the domain dimension
Why: The theorem's total is the dimension of the domain, never the codomain. Pin it down first.
2. Get whichever of rank or nullity is easier
Why: Row reduce and count pivots for the rank, or describe the kernel directly for the nullity. You only need one.
3. Solve for the other
Why: Rank plus nullity equals the domain dimension, so the remaining number is forced.
4. Translate to the question asked
Why: Nullity zero means injective; rank equal to the codomain dimension means surjective; both on a square map means invertible.
5. Sanity-check the bounds
Why: Rank cannot exceed either dimension; a domain bigger than the codomain forces a positive nullity. If your numbers violate these, recheck.
Real world
Discussion prompt
Outside this lesson: where does The Rank–Nullity Theorem 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 Pattern: the rank-nullity toolkit 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 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.
Check
Use the theorem, being careful about which dimension is the total.
\[ T : \mathbb{R}^7 \to \mathbb{R}^4, \quad \operatorname{rank}(T) = 3 \]
Check your understanding
What is the nullity of T?
Answer: A
Why: Nullity equals the domain dimension minus the rank, and the domain is seven-dimensional, so nullity = 7 - 3 = 4. The codomain dimension 4 plays no role in the sum.
Check
Decide whether injectivity is even possible here.
\[ T : \mathbb{R}^5 \to \mathbb{R}^3 \]
Check your understanding
Can such a linear map be injective?
Answer: A
Why: The rank is capped by the codomain dimension 3, so the nullity is at least 5 minus 3, which is 2. A positive nullity means a nontrivial kernel, so the map cannot be injective.
Check
Apply the theorem to differentiation on quartic polynomials, a five-dimensional space.
\[ D : P_4 \to P_4 \]
Check your understanding
What is the nullity of D?
Answer: A
Why: The kernel of differentiation is the constant polynomials, a one-dimensional space, so the nullity is 1. The image is the degree-at-most-three polynomials, giving rank 4, and 4 plus 1 is 5.
Check
A system is consistent and its matrix has nullity 2.
\[ A x = b \ \text{consistent}, \quad \operatorname{nullity}(A) = 2 \]
Check your understanding
What does the solution set look like?
Answer: A
Why: The solution set is one particular solution plus the kernel, and the kernel is a 2-dimensional subspace, so the answer is a translate of a plane, described by two free parameters.
Check
A square matrix map is known to be injective.
\[ A : \mathbb{R}^n \to \mathbb{R}^n, \quad A \text{ injective} \]
Check your understanding
Which conclusion is forced?
Answer: A
Why: Injective means nullity zero, so the rank is n and the map is surjective as well by the finite-dimensional miracle. A square map that is both injective and surjective is invertible.
Elimination
Eliminate the wrong options
What is the smallest possible nullity of A?
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 rank is at most 3 because there are only 3 rows, so the nullity is 5 minus the rank, which is at least 5 minus 3, equal to 2. The minimum nullity is therefore 2.
Check
Consider a wide matrix with more columns than rows.
\[ A \ \text{is } 3 \times 5 \]
Check your understanding
What is the smallest possible nullity of A?
Answer: A
Why: The rank is at most 3 because there are only 3 rows, so the nullity is 5 minus the rank, which is at least 5 minus 3, equal to 2. The minimum nullity is therefore 2.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: the rank-nullity toolkit · Recall: the image of a linear map · Rank is the dimension of the image · Recall: the kernel of a linear map · Nullity is the dimension of the kernel. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Rank is the dimension of the image; nullity is the dimension of the kernel. The Rank–Nullity Theorem says the domain's dimension is their sum.
\[ \dim V = \operatorname{rank}(T) + \operatorname{nullity}(T) \]
The proof is a single idea: take a basis of the kernel, extend it to the domain, and the images of the extension vectors form a basis of the image.
Translate any question about a linear map into a statement about rank or nullity, and read off the answer:
| Question about T | Answered by |
|---|---|
| Is T injective? | nullity is 0 |
| Is T surjective? | rank equals dim of codomain |
| Is a square T invertible? | nullity is 0 (equivalently full rank) |
| How many free variables? | the nullity |
And the finite-dimensional miracle: for a map of a space to itself, injective, surjective, and bijective are one and the same.
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