From Serge Lang Linear Algebra:
Show that the association $A \rightarrow g_A$ is an isomorphism between the space of $m \times n$ matrices, and the space of bilinear maps of $\mathbb{K}^m \times \mathbb{K}^n$ into $\mathbb{K}$.
Note: In calculus, if $f$ is a function of $n$ variables, one associates with $f$ a matrix of second partial derivatives ($\frac{\partial^2 f}{\partial x_i \partial x_j}$), which is symmetric (in most of the "real-life" cases). This matrix represents the second derivative, which is a bilinear map.
$g_A$ simply implies that the bilinear map $g$ has a matrix $A$ associated with it, such that $g_A(X, Y)=X^TAY$.
This is an extension of my previous question about understanding the association above.
It is easily visible, that the association above can be represented by a linear map $L$:
$$L: \textrm{Mat}_{m \times n}(\mathbb{K}) \rightarrow \textrm{Bil}_{\mathbb{K}}({\mathbb{K^m} \times \mathbb{K^n}})$$
where $\textrm{Mat}_{m \times n}(\mathbb{K})$ is a matrix space of all $m \times n$ matrices over a field $\mathbb{K}$.
Hence:
$$L(A)(X, Y)=g_{A}(X, Y)=X^TAY$$
Problems with showing an isomorphism:
(1) First method to show that the transformation $L$ is an isomorphism, is by showing that matrix associated with it ($A$) is nonsingular (i.e such that there exists some matrix $B$ where $AB=I$, if $m = n$). But I can not ensure that matrix $A$ is necessarily squared, therefore I can not ensure that it is invertible.
(2) Another simple method would be to show that the transformation is injectve, which by rank-nullity should also guarantee surjectivity. This can be done by showing that the kernel is trivial, i.e:
$$\textrm{Ker}(g_{A}) = \{X = O \in \mathbb{K}^m \mid g_{A}(X, Y)=0 \, \, \forall \, \, Y \in \mathbb{K}^n \}$$
But in this case, $X$ does not need to be a zero vector, because if $Y=O$, then:
$$X^TAY=X^TA(O)=0$$
(3) From the examples above, I think I'm trying to show that the bilinear maps in the image under $g_{A}$ are isomorphic. If so, maybe I need to show that:
$$\textrm{Ker}(L)=\{\ A=O \in \textrm{Mat}_{m \times n}(\mathbb{K}) \mid L(A)=O \}$$
But how can I show this? I assume there is some invertible matrix $B$ associated with $L$ such that $BA=L(A)$, but what is this matrix $B$?
Question:
How can I show that the association $A \rightarrow g_A$ is an isomorphism? Is the problem (3) above on the correct path (if so how can it be extended)? Or should I follow the approach that I've utilized in problems (2), (1)?
Thank you!
I think this exercise may well be easier than it seems to .
First, both $\;V:=M_{m\times n}(\Bbb K)\;,\;\;W:=Bil_{\Bbb K}(\Bbb K^m\times\Bbb K^n)\;$ are vector spaces over the same field $\;\Bbb K\;$ and of the same dimension: $\;\dim_{\Bbb K}V=\dim_{\Bbb K}W=nm\;$ , and since this dimension is finite, the map $\;L\;$ is an isomorphism iff it is injective iff $\;\ker L=\{0\}\;$ , but
$$A\in\ker L\implies LA=g_A\equiv 0\iff gA(X,Y)=X^tAY=0\,,\,\,\forall\,X\in\Bbb K^m\,,\,Y\in\Bbb K^n$$
Let us write $\;A=(a_{ij})\,,\,\,a_{ij}\in\Bbb K\;,\;\;1\le i\le m\,,\,\,1\le j\le n\;$ , and suppose
$$\;A\neq 0\implies \exists\,1\le i\le m\,,\,\,1\le j\le n\;\;s.t.\;\;a\:=a_{ij}\neq0$$
Let us choose now
$$X=(0,0...,0,\overbrace{1}^i,0,...0)^t\in\Bbb K^m\;,\;\;Y=(0,...,0,\overbrace{i}^j,0,...0)^t\in\Bbb K^n$$
then $\;AY\;$ is an $\;m\times 1\;$ (column) vector with $\;a\cdot 1=a\;$ in the $\;i\,-$ th entry , and thus $\;X^tAY\;$ is the number $\;a\neq0\;$ ...! This contradicts the fact that $\;g_A\equiv0\;$ . Thus, $\;A=0\implies \ker L=\{0\}\;$ and we're done.
Now, the "note" on the Hessian matrix is hard for me to understand in this context....