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Let $\text{tr}A$ be the trace of the matrix $A \in M_n(\mathbb{R})$.

  • I realize that $\text{tr}A: M_n(\mathbb{R}) \to \mathbb{R}$ is obviously linear (but how can I write down a formal proof?). However, I am confused about how I should calculate $\text{dim}(\text{Im(tr)})$ and $\text{dim}(\text{Ker(tr)})$ and a basis for each of these subspace according to the value of $n$.
  • Also, I don’t know how to prove that $\text{tr}(AB)= \text{tr}(BA)$, and I was wondering if it is true that $\text{tr}(AB)= \text{tr}(A)\text{tr}(B)$.
  • Finally, I wish to prove that $g(A,B)=\text{tr}(AB)$ is a positive defined scalar product if $A,B$ are symmetric; and also $g(A,B)=-\text{tr}(AB)$ is a scalar product if $A,B$ are antisymmetric. Can you show me how one can proceed to do this?

I would really appreciate some guidance and help in clarifying the doubts and questions above. Thank you.

mathlearner
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2 Answers2

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Look at the trace of $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Do you see why $tr(AB) \neq tr(A)tr(B)$?

Proof of $tr(AB) = tr(BA)$ can be found here: How to prove tr AB = tr BA?.

To prove the trace is linear we have to check that $Tr(c(A + B)) = cTr(A) + cTr(B)$.

Community
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Nescrio
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The image is very easy. All you really need is the fact that there is some $A$ with $\text{tr}\; A \ne 0$. As for the kernel: note that given all but one entry of the $n \times n$ matrix $A$ (if that one is ...), you can choose a value for that one to make the trace $0$.

Robert Israel
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