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?

  • 1
    $\begingroup$ 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). $\endgroup$ Commented Aug 24, 2019 at 16:59
  • $\begingroup$ How do you derive from $i)$ that any linear map corresponds to at most one matrix? $\endgroup$
    – New2Math
    Commented Aug 24, 2019 at 17:25

1 Answer 1


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.

  • $\begingroup$ in the last paragraph you mean $k_Y$ and not $\psi$ right ? $\endgroup$
    – New2Math
    Commented Aug 24, 2019 at 17:35
  • $\begingroup$ yes, well spotted. $\endgroup$
    – Riquelme
    Commented Aug 24, 2019 at 17:38

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