# Characterizing linear maps with matrices

Bosch Linear Algebra page 92

We want to prove that the map

$$\psi:\operatorname{Hom}_K(V,W)\rightarrow K^{m\times n}$$ with $$f\mapsto > A_{f,X,Y}$$ is an isomorphism.

I have not understood why we can conclude from a Prior Theorem that this map is injective and surjective.

This is the Theorem from which we conclude the injectivity and the surjectivity:

If $$V$$ is a $$F$$-vector space with generating system $$a_1,....,a_n$$ and $$a'_1,...,a'_n$$ are vectors from a different $$F$$-vector space then:

$$(i)$$ There exists at most one linear map $$f:V\rightarrow V'$$ such that $$f(a_i)=a'_i$$

$$(ii)$$ If $$a_1,..,a_n$$ is a basis then there exists exactly one linear map with $$f(a_i)=a'_i$$

This is the definition of $$A_{f,X,Y}$$

Let $$x_1,...,x_n=X$$ be a basis of $$V$$ and $$y_1,...,y_m=Y$$ a basis of $$W$$. And $$f:V\rightarrow W$$ is a linear map then $$A_{f,X,Y}=\big{(}(k_Y\circ f)(x_1),...,(k_Y\circ f)(x_n)\big{)}\in K^{m\times n}$$ (The filevectornotation).

Where $$k_Y$$ was defined as an isomorphism $$W\rightarrow K^m$$, where each Vector of $$W$$ is described in its coordinates over the basis $$Y$$.

The linearity of $$\psi$$ is understood. The surjectivity and injectivity not yet, how exactly is the Prior Theorem applied here?

• Item $i)$ is injectivity (any linear map corresponds to AT MOST one matrix in a given basis). Item $ii)$ is surjectivity (any matrix corresponds to the action of a linear map on your given basis). Commented Aug 24, 2019 at 16:59
• How do you derive from $i)$ that any linear map corresponds to at most one matrix? Commented Aug 24, 2019 at 17:25

Here is how you show that $$\psi$$ is injective: First we can check that $$\psi$$ is $$K$$-linear, so it is enough to show that $$\ker(\psi)=\{0\}$$.$$\\$$
Let $$f=0$$. Then $$(k_Y \circ f)(x_i) = k_y(0(x_i)) = k_y(0) = 0$$, since $$k_y$$ is a $$K$$-linear map. So we have that $$\{0\} \subseteq \ker(\psi)$$. Now let $$g \in \ker(\psi)$$, we have $$\psi (g) = A_{g,X,Y} = 0$$, $$A_{g,X,Y}=\big{(}(k_Y\circ f)(x_1),...,(k_Y\circ f)(x_n)\big{)} = \big(0, \ldots, 0\big)$$, so for all $$i \in \{1,\ldots,n\}$$ $$(k_y \circ g)(x_i) =0.$$ But since $$k_y$$ is an isomoprhism, in particular injective, we have that $$k_y(g(x_i))=0$$ iff $$g(x_i)=0$$ for all $$i$$. Now since $$(x_i)_i$$ is a basis, $$g$$ must be the zero map. I'll leave the surejectivity to you.
• in the last paragraph you mean $k_Y$ and not $\psi$ right ? Commented Aug 24, 2019 at 17:35