[Linear Algebra] Linear Transformation

Linear Transformation

$\mathbf{Definition.}$ Linear Transformation
Let $V$ and $W$ be vector spaces over a field $F$.
1. A map $T: V \to W$ is called a linear transformation, or $\mathbf{F}$-linear map if

\[\begin{align*} T(kx + y) = k T(x) + T(y) \; \forall x, y \in V, \forall k \in F \end{align*}\]

2. The set $\text{ker}(T) \; (\text{null}(T))$ is called the kernel, or null space of $T$ if

\[\begin{align*} \text{ker}(T) \; (\text{null}(T)) = \{ x \in V | T(x) = 0 \} \end{align*}\]

3. The set $\text{Im}(T) \; (\text{rank}(T))$ is called the image, or rank of $T$ if

\[\begin{align*} \text{Im}(T) \; (\text{rank}(T)) = \{ x \in V | T(x) \} \end{align*}\]

$\mathbf{Remark.}$
(1) $T(0) = 0$

(2) $T(-x) = - T(x)$

(3) $\text{ker}(T)$ is a subspace of $V$:

\[\begin{align*} x, y \in \text{ker}(T) \; \Rightarrow \; T(x) = T(y) = 0 \\ &\Rightarrow T(kx + y) = 0 \\ &\Rightarrow kx + y \in \text{ker}(T) \forall k \in F \end{align*}\]

Similarly, $\text{Im}(T)$ is also a subspace of $V$.

$\mathbf{Examples.}$

1. $T: V \to W$ such that $T(x) = 0 \forall x \in V$ is a linear transformation, and

\[\begin{align*} &\text{ker}(T) = V, \\ &\text{Im}(T) = \{ 0 \} \end{align*}\]

2. For a field $0 \neq k \in F$, $T: V \to V$ such that $T(x) = kx$ is a linear transformation, and

\[\begin{align*} &\text{ker}(T) = 0, \\ &\text{Im}(T) = V \end{align*}\]

3. $T: M_{m \times n} (F) \to M_{n \times m} (F)$ by $T(A) = A^\top$ is a linear transformation, and

\[\begin{align*} &\text{ker}(T) = \mathbb{0}, \\ &\text{Im}(T) = M_{n \times m} (F) \end{align*}\]

4. A formal derivative $T: F[x] \to F[x]$ by $T(a_n x^n + \cdots + a_0) = n a_n x^{n-1} + \cdots + a_1$ is a linear transformation, and

\[\begin{align*} &\text{ker}(T) = F, \\ &\text{Im}(T) = F[x] \end{align*}\]

5. $T: F^n \to F^m$ by $T(x) = Ax$ is a linear transformation, and

\[\begin{align*} &\text{ker}(T) = \text{null}(A), \\ &\text{Im}(T) = \text{col}(A) \end{align*}\]

Rank–nullity theorem

$\mathbf{Thm\ 1.}$ Let $T: V \to W$ be a linear transformation with $\text{dim } V < \infty$. Then

\[\begin{align*} \text{dim } V = \text{dim}(\text{ker}(T)) + \text{dim}(\text{Im}(T)) \end{align*}\]

$\mathbf{Proof.}$

Let { $\mathbf{v}_1, \cdots, \mathbf{v}_m$ } be a basis of $\text{ker}(T)$, and note that $\text{ker}(T)$ is a subspace of $V$. And, let’s extend this to a basis { $\mathbf{v}_1, \cdots, \mathbf{v}_m, \cdots, \mathbf{v}_n$ } of $V$.

We claim that $S =$ { $T(\mathbf{v}_{m+1}, \cdots, T(\mathbf{v}_n)$ } is a basis of $\text{Im}(T)$.

Let $T(v) = w \in \text{Im}(T)$. Write $v = a_1 v_1 + \cdots + a_n v_n$. Then,

\[\begin{align*} w = T(v) = a_{m + 1} T(\mathbf{v}_{m+1} + \cdots + a_n T(\mathbf{v}_n) \end{align*}\]

So, $S$ spans $\text{Im}(T)$. And, set

\[\begin{align*} b_{m + 1} T(\mathbf{v}_{m+1} + \cdots + b_n T(\mathbf{v}_n) = T (b_{m + 1} \mathbf{v}_{m+1} + \cdots + b_n \mathbf{v}_n) = 0 \end{align*}\]

Then,

\[\begin{align*} b_{m + 1} \mathbf{v}_{m+1} + \cdots + b_n \mathbf{v}_n \in \text{ker}(T) \end{align*}\]

Hence,

\[\begin{align*} b_{m + 1} \mathbf{v}_{m+1} + \cdots + b_n \mathbf{v}_n = c_1 \mathbf{v}_1 + \cdots + c_m \mathbf{v}_m \end{align*}\]

for some $c_1, \cdots, c_m \in F$.

Since $\mathbf{v_1}, \cdots, \mathbf{v_n}$ are linearly independent, $b_{m + 1} = \cdots = b_n = 0._\blacksquare$

For intuition, we can think the theorem as a conservation of projection by linear map. There are two possible ways for dimension: it is compressed to 0 if the element is in kernel space, or it is preserved. Thus, the dimension of domain, $\text{dim } V$ will be divided into the dimension of kernel space and image of linear map.

Rank-nullity theorem provides some useful fact:

$\mathbf{Remark.}$ If $V = F^n, W = F^m$, and $T(x) = Ax$ for some $m \times n$ matrix $A$, then $n = \text{dim}(\text{null}(A)) + \text{rank}(A)$.

$\mathbf{Remark.}$ If an $m \times n$ matrix $A$ has more columns than rows, i.e., $m < n$, $A \mathbf{x} = \mathbf{0}$ has a non-zero solution as $\text{rank}(A) \leq m < n$.

Uniqueness of Linear Transformation

$\mathbf{Thm\ 2.}$ Let { $\mathbf{v_1}, \cdots, \mathbf{v_n}$ } be a basis of a finite dimensional vector space $V$, and let $\mathbf{w_1}, \cdots, \mathbf{w_n}$ be vectors in a vector space $W$. Then, there exists an unique linear map $T: V \to W$ such that

\[\begin{align*} T(\mathbf{v}_i) = \mathbf{w}_i \; \forall 1 \leq i \leq n. \end{align*}\]

$\mathbf{Proof.}$

Define $T: V \to W$ by

\[\begin{align*} x = \sum_{i=1}^n a_i \mathbf{v}_i \mapsto \sum_{i=1}^n a_i \mathbf{w}_i \; (a_i \in F) \end{align*}\]

Then, $T(kx + y) = k T(x) + T(y)$. Assume that there exists another linear map $S: V \to W$ with $S(\mathbf{v_i}) = \mathbf{w_i} \; \forall i$. Then,

\[\begin{align*} \forall x = \sum_{i=1}^n a_i \mathbf{v}_i \Rightarrow S(x) = T(x) \end{align*}\]

Thus $S = T._\blacksquare$

So, the linear map is completely determined by its values of a basis of domain.

$\mathbf{Remark.}$ If $S, T: V \to W$ linear maps such that $T(\mathbf{v_i}) = S (\mathbf{v_i}) \; \forall 1 \leq i \leq n$ where the set of all $\mathbf{v_i}$s is a basis of $V$, then $S = T$.

Reference

[1] K. Hoffman, and R. Kunze. Linear Algebra PHI Learning, Second edition, 2004
[2] Mathematics Stack Exchange, Intuitive explanation of why $\text{dim}(\text{Im}(T)) + \text{dim}(\text{Ker}(T)) = \text{dim } V$
[3] Mathematics Stack Exchange, A theorem concerning unique linear mapping between vector spaces: What does it say?