MAT240H1/Unit 05
Unit 05 · Algebra I

Rank–Nullity & Matrices

The accounting identity of linear algebra: dimension in equals dimension collapsed plus dimension out. And once bases are fixed, every linear map becomes a matrix — with composition of maps becoming multiplication of matrices.

By the end you can
Why it matters & where it comes from
Why it exists

The Rank–Nullity theorem ties together how much a map crushes to zero and how much it produces. It is the bookkeeping that explains when systems have solutions and how many.

In the real world

Rank measures the true information content of a data matrix; low-rank structure is the basis of compression, recommender systems, and PCA. Nullity is the space of "free" solution directions.

Story & open questions

Representing a linear map by a matrix in coordinates — and recognizing rank as basis-independent — unified the previously separate arts of solving equations and studying transformations.

The Rank–Nullity theorem

Definition

$\operatorname{rank}(T)=\dim R(T)$ and $\operatorname{nullity}(T)=\dim N(T)$.

Theorem Rank–Nullity (dimension theorem)

For linear $T:V\to W$ with $\dim V=n$ finite, $$\operatorname{rank}(T)+\operatorname{nullity}(T)=n.$$

Proof
Pick a basis $u_1,\dots,u_k$ of $N(T)$ and extend it (Unit 3) to a basis $u_1,\dots,u_k,v_1,\dots,v_{n-k}$ of $V$.
Claim: $T(v_1),\dots,T(v_{n-k})$ is a basis of $R(T)$. Spanning: $T$ kills every $u_i$, so the image of any vector is a combination of the $T(v_j)$ alone.
Independence: if $\sum c_jT(v_j)=0$, then $T\big(\sum c_jv_j\big)=0$, so $\sum c_jv_j\in N(T)$ is a combination of the $u_i$ — but the full list is a basis of $V$, so all coefficients vanish, giving every $c_j=0$.
Hence $\operatorname{rank}=n-k$ and $\operatorname{nullity}=k$. ∎
Corollary Operators: one-sided is two-sided

If $\dim V$ is finite and $T:V\to V$ is linear, then $T$ injective $\Leftrightarrow$ $T$ surjective $\Leftrightarrow$ $T$ bijective. (Injective means nullity $0$, so rank $=\dim V$, so $R(T)=V$ — and back.) False in infinite dimensions: the shift on sequences is injective but not onto.

The matrix of a linear map

Definition

Fix ordered bases $\beta=\{v_1,\dots,v_n\}$ of $V$ and $\gamma$ of $W$. The matrix $[T]_\beta^\gamma$ has as its $j$th column the $\gamma$-coordinates of $T(v_j)$. Then applying the map is matrix multiplication on coordinates: $[T(v)]_\gamma=[T]_\beta^\gamma\,[v]_\beta$, and composition of maps is the product of their matrices.

Example Differentiation on $P_2$

In the basis $1,x,x^2$: $D(1)=0$, $D(x)=1$, $D(x^2)=2x$, so the columns are $(0,0,0)$, $(1,0,0)$, $(0,2,0)$ and $[D]=\left(\begin{smallmatrix}0&1&0\\0&0&2\\0&0&0\end{smallmatrix}\right)$. Rank $2$ (range $=P_1$), nullity $1$ (the constants) — and $2+1=3=\dim P_2$. ✓

Columns are images The whole dictionary in one line: column $j$ of the matrix is where basis vector $j$ goes. Every matrix identity in this course decodes back to a statement about images of basis vectors.

What rank settles instantly

Facts

$\operatorname{rank}(T)\le\min(\dim V,\dim W)$. So a map $\mathbb{R}^2\to\mathbb{R}^5$ is never onto (rank $\le2$), and a map $\mathbb{R}^5\to\mathbb{R}^3$ is never one-to-one (nullity $\ge5-3=2$). For systems: $Ax=b$ is solvable iff $b$ lies in the column space of $A$, and then the solution set is a translate of the null space.

Common pitfalls

• Rank–nullity counts the dimension of the domain, not the codomain.
• The matrix changes with the bases; rank and nullity do not.
• "Injective iff surjective" needs the same finite-dimensional space on both sides — not two different spaces, not infinite dimensions.
• Columns of $[T]$ are coordinates in the codomain basis — forgetting that is the classic wrong-matrix error.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Counting

A linear map $\mathbb{R}^4\to\mathbb{R}^3$ has rank $3$. Find its nullity.

Answer
$4-3=1$.
Practice 2 · Counting

If a $3\times3$ matrix has nullity $0$, what is its rank, and what does that say about the matrix?

Answer
Rank $3$ — full rank; the map is bijective, so the matrix is invertible.
Practice 3 · Impossible maps

Can a linear map $\mathbb{R}^2\to\mathbb{R}^5$ be onto? Can a linear map $\mathbb{R}^5\to\mathbb{R}^3$ be one-to-one?

Answer
No and no: rank $\le2<5$ for the first; nullity $\ge5-3=2>0$ for the second.
Practice 4 · The corollary

Prove: if $\dim V$ is finite and $T:V\to V$ is injective, then $T$ is surjective.

Solution
Injective means $\operatorname{nullity}=0$, so $\operatorname{rank}=\dim V$ by Rank–Nullity. A subspace $R(T)\subseteq V$ of full dimension equals $V$ (Unit 3, Practice 8).
Practice 5 · Build the matrix

Write the matrix of $T(x,y)=(x+y,\ x-y,\ 2y)$ in the standard bases, and find its rank and nullity.

Solution
Columns $T(1,0)=(1,1,0)$ and $T(0,1)=(1,-1,2)$: $\left(\begin{smallmatrix}1&1\\1&-1\\0&2\end{smallmatrix}\right)$. The columns are independent, so rank $2$, nullity $0$ — $T$ is injective but not onto $\mathbb{R}^3$.
Practice 6 · Second derivative

Let $D^2:P_3\to P_3$ send $p\mapsto p''$. Find rank and nullity and check the theorem.

Solution
$D^2$ kills $1$ and $x$, and sends $x^2\mapsto2$, $x^3\mapsto6x$: range $=\operatorname{span}\{1,x\}$. Rank $2$, nullity $2$, and $2+2=4=\dim P_3$. ✓
Practice 7 · Independence survives injections

Prove: if $T$ is linear and injective and $v_1,\dots,v_k$ are independent, then $T(v_1),\dots,T(v_k)$ are independent.

Solution
If $\sum c_iT(v_i)=0$ then $T(\sum c_iv_i)=0$; injectivity gives $\sum c_iv_i=0$, and independence of the $v_i$ forces every $c_i=0$.
Practice 8 · Composition

Suppose $T:\mathbb{R}^3\to\mathbb{R}^3$ has rank $2$ and $S:\mathbb{R}^3\to\mathbb{R}^3$ has rank $1$. What are the possible values of $\operatorname{rank}(S\circ T)$?

Answer
$0$ or $1$: the rank of a composition is at most each factor's rank ($R(S\circ T)\subseteq R(S)$), and it can be $0$ if $R(T)\subseteq N(S)$ — both happen for suitable $S,T$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT240H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT240H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The matrix spine is: choose bases, write columns from images, row-reduce to expose pivots, then read rank and nullity as dimensions.

Study routine. Keep the abstract map and its coordinate matrix separate; a matrix represents a map only after bases are fixed.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Rank-nullity

If $T:F^5 o F^3$ has nullity $2$, what is its rank?

Guide
Rank-nullity gives rank $=5-2=3$.
Extension 2 · Column meaning

What is the first column of a matrix representation of $T$?

Guide
It is the coordinate vector of $T$ applied to the first domain basis vector, expressed in the codomain basis.
Extension 3 · Injective or surjective

Can a linear map $F^5 o F^3$ be injective?

Guide
No. Nullity is at least $2$ by rank-nullity because rank is at most $3$.
Extension 4 · Kernel system

How do you find the kernel of a matrix $A$?

Guide
Solve $Ax=0$ by row reduction; the free variables parametrize a basis for the null space.
← Unit 04Linear transformations Unit 06 →Change of basis & similarity