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.
$\blacksquare$