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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!

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    $\begingroup$ Do not use single letters to write the name of a book, and even less to denote the writer of the book! Is "S.L. Linear Algebra" the book in linear algebra by Serge Lang? $\endgroup$ – DonAntonio May 12 at 16:34
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    $\begingroup$ The matrix in your "Note" is symmetric if the second order mixed partial derivatives are equal...otherwise it is not symmetric. $\endgroup$ – DonAntonio May 12 at 16:35
  • $\begingroup$ @DonAntonio I heavily apologize for that, I've seen Serge Lang has been often referred to as S.L and unfortunately took that habit. But I've fixed it now, thank you! As for note, does this mean it should be more specific? $\endgroup$ – ShellRox May 12 at 16:40
  • $\begingroup$ I think it must. Clairaut's Theorem states that if the mixed second order partial derivatives exists **and are continuous"" in an open set in $\;\Bbb R^n\;$ , then the are equal (over the same variables, of course). I think the condition can be weakened to "all but one" mixed partial derivatives are continuous"...Anyway. something more must be said, imo. $\endgroup$ – DonAntonio May 12 at 16:52
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    $\begingroup$ @DonAntonio Thank you! Updated the note with brief hyperlink to a Wikipedia page which contains a theorem on equality of mixed partials. $\endgroup$ – ShellRox May 12 at 17:45
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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....

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  • $\begingroup$ Thank you for the great answer! Perhaps author of the book has made a mistake when assuming the general symmetry of hessian matrix, does it make sense if the author simply assumes that ($\frac{\partial^2 f}{\partial x_i \partial x_j}$) is a $1 \times 1$ matrix (belonging to the image of some bilinear map)? $\endgroup$ – ShellRox May 12 at 18:59
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    $\begingroup$ I think Lang uses the Hessian just as an example in exercise 4 there. I suppose he's assuming the functions has mixed partial derivatives that are continuous and then...etc. Not a big deal, imo. $\endgroup$ – DonAntonio May 12 at 19:32

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