Chapter 3 Linear Maps

Section 3A Vector Space of Linear Maps

Section 3B Null Spaces and Ranges

Section 3C Matrices

3.31 definition: matrix of a linear map, $ℳ(𝑇)$

Suppose $𝑇 ∈ ℒ(𝑉, 𝑊)$ and $v_1, \cdots, v_n$ is a basis of $𝑉$ and $w_1, \cdots, w_m$ is a basis of $𝑊$.

The matrix of $𝑇$ with respect to these bases is the $𝑚$-by-$𝑛$ matrix $ℳ(𝑇)$ whose entries $A_{i, j}$ are defined by $$𝑇𝑣_𝑘 = A_{1, k} w_1 + \cdots + A_{m, k} w_m \\ $$

3.35 matrix of the sum of linear maps

Suppose $S, T \in \mathcal{L}(V, W).$ Then $\mathcal{M}(S+T) = \mathcal{M}(S) + \mathcal{M}(T)$.

Proof:

Assume $\mathcal{M}(S) = [A_{i, j}], \mathcal{M}(T) = [B_{i, j}]$.

$$Tv_k = A_{1, k} w_1 + \cdots + A_{m, k} w_m \\ Sv_k = B_{1, k} w_1 + \cdots + B_{m, k} w_m \\ (T+S)v_k = Tv_k + Sv_k \\ =(A_{1, k} w_1 + \cdots + A_{m, k} w_m) + \\ (B_{1, k} w_1 + \cdots + B_{m, k} w_m) \\ = (A_{1, k}+B_{1, k}) w_1 + \cdots + (A_{m, k}+B_{m, k}) w_m$$

So $\mathcal{M}(S+T) = [A_{i, j} + B_{i, j}] = [A_{i, j}] + [B_{i, j}] = \mathcal{M}(S) + \mathcal{M}(T)$.

$\square$

3.38 the matrix of a scalar times a linear map

Suppose $𝜆 ∈ 𝐅$ and $𝑇 ∈ ℒ(𝑉, 𝑊)$. Then $ℳ(𝜆𝑇) = 𝜆ℳ(𝑇)$.

Proof:

Assume $\mathcal{M}(T) = [A_{i, j}]$, then

$$(𝜆T)(v_k) = 𝜆(Tv_k) \\ = 𝜆 (A_{1, k} w_1 + \cdots + A_{m, k} w_m) \\ = (𝜆A_{1, k}) w_1 + (𝜆A_{m, k}) w_m \\ $$

So

$$\mathcal{M}(𝜆T) = [𝜆 A_{i, j}] = 𝜆 [A_{i, j}] = 𝜆 \mathcal{M}(T)$$

$\square$

3.44 notation: $𝐴_{𝑗,⋅}$ , $𝐴_{⋅,𝑘}$

Suppose $𝐴$ is an $𝑚$-by-$𝑛$ matrix.

• If $1 ≤ 𝑗 ≤ 𝑚$, then $A_{j, .}$ denotes the $1$-by-$𝑛$ matrix consisting of row $𝑗$ of $𝐴$.

• If $1 ≤ 𝑘 ≤ 𝑛$, then $A_{., k}$ denotes the $𝑚$-by-$1$ matrix consisting of column $𝑘$ of $𝐴$.

3.50 linear combination of columns

Suppose $𝐴$ is an $𝑚$-by-$𝑛$ matrix and $b = \begin{pmatrix} b_1 \\ \vdots \\ b_n \\ \end{pmatrix}$ is an 𝑛-by-1 matrix. Then

$$Ab = b_1 A_{., 1} + \cdots + b_n A_{., n}$$

In other words, 𝐴𝑏 is a linear combination of the columns of 𝐴, with the scalars that multiply the columns coming from 𝑏.

Proof:

The $k$th row on the left is

$$Ab_{k, 1} = A_{1, k} b_1 + \cdots + A_{n, k} b_n$$

Which is the same as the $k$th row on the right.

$\square$

3.56 column–row factorization

Suppose $𝐴$ is an $𝑚$-by-$𝑛$ matrix with entries in $𝐅$ and column rank $𝑐 ≥ 1$. Then there exist an $𝑚$-by-$𝑐$ matrix $𝐶$ and a $𝑐$-by-$𝑛$ matrix $𝑅$, both with entries in $𝐅$, such that $𝐴 = 𝐶𝑅$.

Proof:

The column $A_{., 1}, \cdots, A_{., n}$ can be reduced $c$ columns which is a basis of the span of the columns of $A$. Put these $c$ columns together to make $C$.

Each columns of $A$ is the linear combination of the columns of $C$. Then make the coefficients into column of $R$.

Then $A = CR$.

3.57 column rank equals row rank

Suppose $𝐴 ∈ 𝐅^{𝑚, 𝑛}$ . Then the column rank of 𝐴 equals the row rank of 𝐴.

Proof: Assume column rank is $c$, then $A = CR$. $C$ is $m \times c$ and $R$ is $c \times n$.

Each row of $A$ is the linear combination of the row of $R$, so row rank of $A \leq c$.

To prove the other direction, consider $A^T$,

$$\text{ column rank of } 𝐴 = \\ \text{ row rank of } 𝐴t ≤ \\ \text{ column rank of } 𝐴t = \\ \text{ row rank of } 𝐴.$$

So column rank and row rank are equal.

$\square$

Section 3F Duality

3.108 definition: linear functional

A linear functional on $𝑉$ is a linear map from $𝑉$ to $𝐅$. In other words, a linear functional is an element of $ℒ(𝑉, 𝐅)$.

3.109 example: linear functionals

• Define 𝜑 ∶ 𝒫(𝐑) → 𝐑 by

$$\varphi(𝑝) = 3𝑝″ (5) + 7𝑝(4).$$

Then 𝜑 is a linear functional on 𝒫(𝐑).

• Define 𝜑 ∶ 𝒫(𝐑) → 𝐑 by

$$\varphi(p) = \int_{0}^{1} p$$

for each 𝑝 ∈ 𝒫(𝐑). Then 𝜑 is a linear functional on 𝒫(𝐑).

3.110 definition: dual space, 𝑉′

The dual space of 𝑉, denoted by 𝑉′, is the vector space of all linear functionals on 𝑉. In other words, 𝑉′ = ℒ(𝑉, 𝐅).

3.111 dim 𝑉′ = dim 𝑉

Suppose $𝑉$ is finite-dimensional. Then $𝑉′$ is also finite-dimensional and

$$\dim V' = \dim V$$

Proof:

$$\dim V' = \dim V \times \dim F = \dim V$$

$\square$

3.127 condition for the annihilator to equal {0} or the whole space

Suppose $𝑉$ is finite-dimensional and $𝑈$ is a subspace of $𝑉$. Then

(a) $U^0 = \{0\} \Longleftrightarrow U = V$

Proof:

$$\begin{align*} U^0 = \{0\} &\Longleftrightarrow \dim U^0 = 0 \\ &\Longleftrightarrow \dim V - \dim U = 0 &\text{ (3.125 dimension of the annihilator) }\\ &\Longleftrightarrow V = U &\text{ (2.39) }\\ \end{align*}$$

$\square$

(b) $U^0 = V' \Longleftrightarrow U = 0$

Proof:

$$\begin{align*} U^0 = V' &\Longleftrightarrow \dim U^0 = \dim V' &\text{ (2.39) } \\ &\Longleftrightarrow \dim U^0 = \dim V &\text{ (3.111) }\\ &\Longleftrightarrow U = 0 &\text{ (3.125) }\\ \end{align*}$$

$\square$

3.128 the null space of 𝑇′

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊). Then

(a) $\text{null } T' = (\text{range } T)^0$

Proof:

If $\psi \in \text{null } T'$, then $T'(\psi ) = 0$. Also

$$(T'(\psi))(v) = (\psi \circ T)(v) = \psi (T(v)) = 0$$

So $\psi \in (\text{range } T)^0$.

If $\psi \in (\text{range } T)^0$, then $\psi (T(v)) = 0$, then $(T'(\psi))(v) = 0, \forall v.$ So $T'(\psi) = 0$.

So $\psi \in \text{null } T'$.

$\square$

(b) $\dim \text{null } 𝑇' = \dim \text{null } 𝑇 + \dim 𝑊 − \dim 𝑉 .$

Proof:

$$\begin{align*} \dim \text{null } 𝑇' &= \dim (\text{range } T)^0 &\text{ (part (a)) }\\ &= \dim W - \dim \text{range } T &\text{ (3.125) }\\ &= \dim W - (\dim V - \text{dim null } T) &\text{ (3.21) }\\ &= \dim \text{null } 𝑇 + \dim 𝑊 − \dim 𝑉 \end{align*}$$

$\square$

3.129 𝑇 surjective is equivalent to 𝑇' injective

Suppose 𝑉 and 𝑊 are finite-dimensional and 𝑇 ∈ ℒ(𝑉, 𝑊). Then

$$𝑇 \text{ is surjective} ⟺ 𝑇' \text{ is injective}.$$

Proof:

$$\begin{align*} 𝑇 \text{ is surjective} &\Longleftrightarrow \text{range } T = W \\ &\Longleftrightarrow (\text{range } T)^0 = \{0\} &\text{ (3.127 (a))} \\ &\Longleftrightarrow \text{null } T' = \{0\} &\text{ (3.128 (a))} \\ &\Longleftrightarrow 𝑇' \text{ is injective} \end{align*}$$

$\square$

3.130 the range of $𝑇'$

Suppose $𝑉$ and $𝑊$ are finite-dimensional and $𝑇 ∈ ℒ(𝑉, 𝑊)$. Then

(a) $\text{dim range } T = \text{dim range } T'$;

Proof:

$$\text{dim range } T' = \dim W - \text{dim null } T' \\ = \dim W - (\text{dim null } T + \dim W − \dim V) \\ = \dim V - \text{dim null } T \\ = \text{dim range } T$$

Another way

$$\begin{align*} \text{dim range } T' &= \dim W' - \text{dim null } T' &\text{ (fundamental theorem of linear maps) } \\ &= \dim W' - \dim (\text{range } T)^0 &\text{ (3.128 (a)) }\\ &= \dim W - \dim (\text{range } T)^0 &\text{ (3.11) }\\ &= \dim W - (\dim W - \dim (\text{range } T) ) &\text{ (3.125 dimension of the annihilator) }\\ &= \dim (\text{range } T) \end{align*}$$

(b) $\text{range } T' = (\text{null } T)^0$

Proof:

We first prove $\text{range } T' \subseteq (\text{null } T)^0$.

Assume $\varphi \in \text{range } T'$, then we can find $\psi \in W'$, such that $T'(\psi) = \varphi$.

Then if $u \in \text{null } T$

$$\varphi (u) = (T'(\psi))(u) = (\psi \circ T)(u) = \psi (T(u)) = 0$$

So $\varphi(u) = 0, \varphi \in (\text{null } T)^0$.

Next, we compute this

$$\begin{align*} \text{dim range } T' &= \text{dim range } T & \text{(part a)} \\ &= \dim T - \text{dim null } T &\text{ (3.21 fundamental theorem of linear maps) } \\ &= \dim (\text{null } T)^0 &\text{(3.125 dimension of the annihilator)} \\ \end{align*}$$

3.131 $𝑇$ injective is equivalent to $𝑇'$ surjective

Suppose $𝑉$ and $𝑊$ are finite-dimensional and $𝑇 ∈ ℒ(𝑉, 𝑊)$. Then

$$T \text{ is injective } \Longleftrightarrow T' \text{ is surjective }$$

Proof:

$$\begin{align*} T \text{ is injective } &\Longleftrightarrow \text{null } T = \{0\}\\ &\Longleftrightarrow (\text{null } T)^0 = V' &\text{ (3.127 (b)) }\\ &\Longleftrightarrow \text{range } T' = V' &\text{ (3.130 (b)) }\\ &\Longleftrightarrow T' \text{ is surjective } \\ \end{align*}$$

$\square$