Lecture 3

Sep 20, 2026

Rank–Nullity and Spectral Theory

Reminder

Let

A = (\begin{matrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & - 1 \\ 1 & 1 & 2 & 1 \end{matrix}) : ℝ^{4} ⟶ ℝ^{3} .

Row reduction gives

A \sim (\begin{matrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & - 1 \\ 0 & 0 & 0 & 0 \end{matrix}) .

Solving $A x = 0$ , we get

x_{1} = - x_{3} - 2 x_{4}, x_{2} = - x_{3} + x_{4} .

With $x_{3} = s$ and $x_{4} = t$ ,

x = s (\begin{matrix} - 1 \\ - 1 \\ 1 \\ 0 \end{matrix}) + t (\begin{matrix} - 2 \\ 1 \\ 0 \\ 1 \end{matrix}) .

Therefore,

Ker A = span {(\begin{matrix} - 1 \\ - 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 2 \\ 1 \\ 0 \\ 1 \end{matrix})} .

These two vectors form a basis for $Ker A$ . Geometrically, $Ker A$ is a two-dimensional plane in $ℝ^{4}$ whose vectors are mapped to the origin in $ℝ^{3}$ .

The range is

\begin{matrix} Ran A & = {A x : x \in ℝ^{4}} \\ = span {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}), (\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix})} \\ = span {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix})} . \end{matrix}

The last two displayed vectors form a basis for $Ran A$ , which is the plane

Ran A = {(a, b, a + b) : a, b \in ℝ} .

We have

\dim ℝ^{4} = \dim Ran A + \dim Ker A = 2 + 2 = 4.

This is the rank–nullity theorem.

Reminder: Direct sum

Let

V_{1} = span {(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix})} .

Then

ℝ^{4} = Ker A \oplus︎ V_{1}, A |_{V_{1}} : V_{1} ⟶ Ran A is bijective .

The rank–nullity theorem

Theorem (Rank–nullity)

Let $T : V \to W$ be a linear map, where $V$ is finite-dimensional. Then

\dim V = \dim Ker T + \dim Ran T .

More concretely, let ${b_{1}, \dots, b_{r}}$ be a basis for $Ran T$ , where $r = rank T$ , and choose $a_{1}, \dots, a_{r} \in V$ such that

T a_{i} = b_{i}, i = 1, \dots, r .

Then $a_{1}, \dots, a_{r}$ are linearly independent and

V = Ker T \oplus︎ V_{1}, V_{1} = span {a_{1}, \dots, a_{r}} .

Credit: Sheldon Axler, Linear Algebra Done Right.

Proof (Sheldon Axler)

Suppose that $V$ is a finite-dimensional vector space and $T \in ℒ (V, W)$ . Let $(u_{1}, \dots, u_{m})$ be a basis of $null T$ ; thus $\dim null T = m$ . The linearly independent list $(u_{1}, \dots, u_{m})$ can be extended to a basis $(u_{1}, \dots, u_{m}, w_{1}, \dots, w_{n})$ of $V$ . Thus $\dim V = m + n$ , and to complete the proof, we need only show that $range T$ is finite dimensional and $\dim range T = n$ . We will do this by proving that $(T w_{1}, \dots, T w_{n})$ is a basis of $range T$ .

Let $v \in V$ . Because $(u_{1}, \dots, u_{m}, w_{1}, \dots, w_{n})$ spans $V$ , we can write

v = a_{1} u_{1} + \dots + a_{m} u_{m} + b_{1} w_{1} + \dots + b_{n} w_{n},

where the $a$ ’s and $b$ ’s are in $𝕂$ . Applying $T$ to both sides of this equation, we get

T v = b_{1} T w_{1} + \dots + b_{n} T w_{n},

where the terms of the form $T u_{j}$ disappeared because each $u_{j} \in null T$ . The last equation implies that $(T w_{1}, \dots, T w_{n})$ spans $range T$ . In particular, $range T$ is finite dimensional.

To show that $(T w_{1}, \dots, T w_{n})$ is linearly independent, suppose that $c_{1}, \dots, c_{n} \in 𝕂$ and

c_{1} T w_{1} + \dots + c_{n} T w_{n} = 0.

Then

T (c_{1} w_{1} + \dots + c_{n} w_{n}) = 0,

and hence

c_{1} w_{1} + \dots + c_{n} w_{n} \in null T .

Because $(u_{1}, \dots, u_{m})$ spans $null T$ , we can write

c_{1} w_{1} + \dots + c_{n} w_{n} = d_{1} u_{1} + \dots + d_{m} u_{m},

where the $d$ ’s are in $𝕂$ . This equation implies that all the $c$ ’s (and $d$ ’s) are $0$ (because $(u_{1}, \dots, u_{m}, w_{1}, \dots, w_{n})$ is linearly independent). Thus $(T w_{1}, \dots, T w_{n})$ is linearly independent and hence is a basis of $range T$ , as desired.

$□$

Example (Differentiation on polynomials)

Let $𝒫 = ℝ [x]$ be the vector space of all real polynomials, and let

D : 𝒫 ⟶ 𝒫, D p = p^{'} .

Find

Ker D, rank D, Ker D^{2}, rank D^{2},

and determine

\dim Ker (D (D + 2 I)) and \dim Ker (D + 2 I) .

Here $I$ is the identity map on $𝒫$ , $D^{2} = D \circ D$ , and $D (D + 2 I) = D \circ (D + 2 I)$ .

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1. Kernel and rank of $D$ . A polynomial satisfies $p^{'} = 0$ exactly when it is constant. Therefore,

Ker D = span {1}, \dim Ker D = 1.

Every polynomial has a polynomial antiderivative:

D (\sum_{k = 0}^{n} \frac{a_{k}}{k + 1} x^{k + 1}) = \sum_{k = 0}^{n} a_{k} x^{k} .

Hence

Ran D = 𝒫, rank D = \infty .

2. Kernel and rank of $D^{2}$ . We have $D^{2} p = p^{''} = 0$ exactly when $p (x) = a + b x$ . Therefore,

Ker D^{2} = span {1, x}, \dim Ker D^{2} = 2.

Every polynomial also has a polynomial second antiderivative:

D^{2} (\sum_{k = 0}^{n} \frac{a_{k}}{(k + 1) (k + 2)} x^{k + 2}) = \sum_{k = 0}^{n} a_{k} x^{k} .

Hence

Ran D^{2} = 𝒫, rank D^{2} = \infty .

3. Kernel of $D + 2 I$ . The equation $(D + 2 I) p = 0$ means

p^{'} + 2 p = 0.

If $p$ is a nonzero polynomial of degree $n$ , then $2 p$ has a nonzero $x^{n}$ term, while $p^{'}$ has no $x^{n}$ term. They cannot cancel. Therefore,

Ker (D + 2 I) = {0}, \dim Ker (D + 2 I) = 0.

4. Kernel of $D (D + 2 I)$ . We have

D (D + 2 I) p = p^{''} + 2 p^{'} = (D + 2 I) (p^{'}) .

Since $Ker (D + 2 I) = {0}$ , this vanishes exactly when $p^{'} = 0$ . Therefore,

Ker (D (D + 2 I)) = span {1}, \dim Ker (D (D + 2 I)) = 1.

Spectral theory

Credit: Sheldon Axler, Linear Algebra Done Right.

Let $V$ be a finite-dimensional vector space over $𝔽$ and let $T : V \to V$ be linear. For $u \neq 0$ , consider

U = span {u} = {a u : a \in 𝔽} .

If $U$ is invariant under $T$ , then $T u \in U$ , so

T u = λ u

for some $λ \in 𝔽$ . Conversely, if $T u = λ u$ , then

T (a u) = a T u = a λ u \in U,

so $U$ is invariant under $T$ .

Definition (Eigenvalue and eigenvector)

A scalar $λ \in 𝔽$ is an eigenvalue of $T$ if there is a nonzero vector $u \in V$ such that

T u = λ u .

The vector $u$ is an eigenvector corresponding to $λ$ .

The eigenvalue problem is to find these scalars $λ$ and nonzero vectors $u$ . We have

T u = λ u ⟺ (T - λ I) u = 0.

In finite dimensions,

\begin{matrix} λ is an eigenvalue of T \\ ⟺ Ker (T - λ I) \neq {0} \\ ⟺ T - λ I is not injective \\ ⟺ T - λ I is not invertible \\ ⟺ T - λ I is not surjective . \end{matrix}

The corresponding eigenspace is

E_{λ} = Ker (T - λ I) .

Its nonzero vectors are the eigenvectors corresponding to $λ$ .

Example

Let

B = (\begin{matrix} 2 & 1 \\ 0 & 3 \end{matrix}) .

The equation $B u = λ u$ , with $u = (x, y) \neq (0,0)$ , gives

{\begin{matrix} (2 - λ) x + y = 0, \\ (3 - λ) y = 0. \end{matrix}

If $y = 0$ , then $x \neq 0$ and $λ = 2$ . If $y \neq 0$ , then $λ = 3$ and $x = y$ . Therefore the eigenvalues are $2$ and $3$ , with eigenspaces

E_{2} = span {(\begin{matrix} 1 \\ 0 \end{matrix})}, E_{3} = span {(\begin{matrix} 1 \\ 1 \end{matrix})} .

Indeed,

B (\begin{matrix} 1 \\ 0 \end{matrix}) = 2 (\begin{matrix} 1 \\ 0 \end{matrix}), B (\begin{matrix} 1 \\ 1 \end{matrix}) = 3 (\begin{matrix} 1 \\ 1 \end{matrix}) .