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

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

Then we have:
 * $f\left(\ker\phi\right) = N\left(A\right)$

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

Proof
By the definition of the matrix $A$, we have $A \circ f = g \circ \phi$.

Therefore if $x \in \ker\phi$ we have:
 * $A f(x) = g(\phi(x)) = g(0) = 0$

This shows that $f\left(\ker\phi\right) \subseteq N(A)$.

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

Let $y = x_1e_1 + \cdots + x_ne_n \in V$.

We have:
 * $g\circ \phi(y) = A\circ f(y) = A(x_1,\ldots,x_n)^T = 0$

so $\phi(y) = 0 f_1 + \cdots + 0 f_m = 0$.

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

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

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

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

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

Then $f(x) \in N(A)$. Therefore,
 * $g \circ \phi(x) = A \circ f(x) = 0$

so $\phi(x) = 0f_1 + \cdots + 0 f_m = 0$.

This shows that $f^{-1}N\left(A\right) \subseteq \ker\phi$.