# 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 $\left({e_1, \ldots, e_n}\right)$ and $\left({f_1, \ldots, f_m}\right)$ 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:

$\displaystyle \sum_{i \mathop = 1}^n a_i e_i \mapsto \left({a_1, \ldots, a_n}\right)$

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

$\displaystyle \sum_{i \mathop = 1}^m b_i f_i \mapsto \left({b_1, \ldots, b_m}\right)$

Let $N \left({A}\right) = \left\{ {x \in \R^n: A x = 0}\right\}$ be the null space of $A$.

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

Then:

$f \left[{\ker \phi}\right] = N \left({A}\right)$

and

$f^{-1} \left[{N \left({A}\right)}\right] = \ker \phi$

where $f \left[{X}\right]$ denotes the image set of a subset $X$ of the domain of $f$.

## Proof

By the definition of the matrix $A$:

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

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

$A f \left({x}\right) = g \left({\phi \left({x}\right)}\right) = g \left({0}\right) = 0$

This shows that:

$f \left[{\ker \phi}\right] \subseteq N \left({A}\right)$

Now let $x = \left({x_1, \ldots, x_n}\right) \in N \left({A}\right)$.

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

We have:

$g \circ \phi \left({y}\right) = A \circ f \left({y}\right) = A \left({x_1, \ldots, x_n}\right)^T = 0$

so $\phi \left({y}\right) = 0 f_1 + \cdots + 0 f_m = 0$.

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

Since $x = f \left({y}\right)$, we have shown that $N \left({A}\right) \subseteq f \left({\ker \phi}\right)$.

Therefore $f \left({\ker \phi}\right) = N \left({A}\right)$.

We deduce immediately from the definitions that $\ker \phi \subseteq f^{-1} \left[{N \left({A}\right)}\right]$.

Now suppose that $x \in f^{-1} \left[{N \left({A}\right)}\right]$.

Then:

$f \left({x}\right) \in N \left({A}\right)$

Therefore:

$g \circ \phi \left({x}\right) = A \circ f \left({x}\right) = 0$

so $\phi \left({x}\right) = 0 f_1 + \cdots + 0 f_m = 0$.

This shows that:

$f^{-1} \left[{N \left({A}\right)}\right] \subseteq \ker \phi$

as required.

$\blacksquare$