# Kernel of Linear Transformation is Null Space of Matrix Representation

## Theorem

Let $V$ and $W$ be finite dimensional vector spaces.

Let $\phi: V \to W$ be a linear transformation from $V$ to $W$.

Let $\tuple {e_1, \ldots, e_n}$ and $\tuple {f_1, \ldots, f_m}$ be ordered bases of $V$ and $W$ respectively.

Let $A$ be the matrix of $\phi$ in these bases.

Define $f: V \to \R^n$ by:

- $\ds \sum_{i \mathop = 1}^n a_i e_i \mapsto \tuple {a_1, \ldots, a_n}$

and $g : W \to \R^m$ by:

- $\ds \sum_{i \mathop = 1}^m b_i f_i \mapsto \tuple {b_1, \ldots, b_m}$

Let $\map N A = \set {x \in \R^n: A x = 0}$ be the null space of $A$.

Let $\map \ker \phi = \set {x \in V: \phi x = 0}$ be the kernel of $\phi$.

Then:

- $f \sqbrk {\map \ker \phi} = \map N A$

and

- $f^{-1} \sqbrk {\map N A} = \map \ker \phi$

where $f \sqbrk X$ denotes the image set of a subset $X$ of the domain of $f$.

## Proof

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By the definition of the matrix $A$:

- $A \circ f = g \circ \phi$

Therefore if $x \in \map \ker \phi$ we have:

- $A \map f x = \map g {\map \phi x} = \map g 0 = 0$

This shows that:

- $f \sqbrk {\map \ker \phi} \subseteq \map N A$

Now let $x = \tuple {x_1, \ldots, x_n} \in \map N A$.

Let $y = x_1 e_1 + \cdots + x_n e_n \in V$.

We have:

- $\map {g \circ \phi} y = \map {A \circ f} y = \map A {x_1, \ldots, x_n}^T = 0$

so:

- $\map \phi y = 0 f_1 + \cdots + 0 f_m = 0$

This shows that $y \in \map \ker \phi$.

Since $x = \map f y$, we have shown that:

- $\map N A \subseteq \map f {\map \ker \phi}$

Therefore:

- $\map f {\map \ker \phi} = \map N A$

We deduce immediately from the definitions that:

- $\map \ker \phi \subseteq f^{-1} \sqbrk {\map N A}$

Now suppose that $x \in f^{-1} \sqbrk {\map N A}$.

Then:

- $\map f x \in \map N A$

Therefore:

- $\map {g \circ \phi} x = \map {A \circ f} x = 0$

so:

- $\map \phi x = 0 f_1 + \cdots + 0 f_m = 0$

This shows that:

- $f^{-1} \sqbrk {\map N A} \subseteq \map \ker \phi$

as required.

$\blacksquare$