I have in my course that in the $n\times n$ matrix space with real entries, we can provide a scalar product $\left<A,B\right>=Tr(A^TB)$ where $Tr$ is the trace function and $A^T$ the transpose of $T$. I can prove that but I don't understand why $\varphi (A,B)=Tr(AB)$ is not a scalar product. Also, is the scalar product $\left<A,B\right>=Tr(A^TB)$ the unique one on $\mathcal M_{n\times n}(\mathbb R)$ ?
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2Do you want $\langle A,A\rangle\ge0$? – Angina Seng Jul 08 '19 at 10:04
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$Tr(AB)$ is a bilinear form. An inner product is a bilinear form satisfying $\langle A,A\rangle\ge 0$ so $\sqrt{\langle A,A\rangle}$ is a (semi)norm. $\langle A,B\rangle=Tr(A^T B)$ is an inner product which induces a norm, thus all the other inner products are of the form $(A,B) = \langle f(A),f(B)\rangle$ for some linear map $f$. – reuns Jul 08 '19 at 10:29
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Take $n=2$ and $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then $$\varphi (A,A)=\text{Tr}(A^2)=0,$$ but $A\neq 0$, so $\varphi (A,A)=0$ doesn't implies $A=0$.
There are indeed many other scalar product. Any $n^2\times n^2$ positive definite matrix will provide a scalar product on $\mathbb {n^2}$ and thus, by identification you can easily construct a scalar product on $\mathcal M_{n\times n}(\mathbb R)$. At the end $\mathcal M_{n\times n}(\mathbb R)$ is nothing else than $\mathbb R^{n^2}$ (as vector space).
Surb
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2Your “alternative” scalar product is just $\operatorname{Tr}(A^TB)$ written in components. – celtschk Jul 08 '19 at 11:42
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1@celtschk: Indeed, you right. I didn't really expected that. Good to know :-) – Surb Jul 08 '19 at 13:40
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In $2 \times 2$ dimension $A^2=-I$ has infinitely many solutions (or for example you can take this solution $A^2=-4I$).
For all these matrices $A$ you would have $\left<A,A\right> < 0$.
Widawensen
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