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The original problem is this

If $g: M_2(\mathbb{R}) \to \mathbb{R}$ satisfies $g(AB) = g(BA)$, then there is a constant $k$ such that $g(A) = ktr(A)$ where $tr$ is the trace.

So after reading a bit, this identification is given when messing a bit with their kernels

Is there a more obvious solution? It wasn't clear to me to look at this general case.

Lemon
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  • I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$? – user9077 Dec 13 '18 at 00:14
  • You mean the trace? G doesn’t have an explicit map, so I can’t imagine how – Lemon Dec 13 '18 at 04:59
  • I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace. – user9077 Dec 13 '18 at 07:29

2 Answers2

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Let $E_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},E_2=\begin{pmatrix}0&1\\0&0\end{pmatrix},E_3=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ and $E_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

If we can show that $$g(E_1)=g(E_4)=k \text{ and } g(E_2)=g(E_3)=0,\quad (*)$$ then $$g\begin{pmatrix}a&b\\c&d\end{pmatrix}=ag(E_1)+bg(E_2)+cg(E_3)+dg(E_4)=k(a+d)=k\cdot\text{tr}(A)$$ as we want. Hence it is enough to show that $g$ satisfies $(*)$.

Note that by the condition $g(AB)=g(BA)$ we have $$g(E_2)=g(E_1E_2)=g(E_2E_1)=g(0)=0$$ and $$g(E_3)=g(E_4E_3)=g(E_3E_4)=g(0)=0 $$ Now $(E_1+E_3)(E_1+E_2)=E_1+E_2+E_3+E_4$ and $(E_1+E_2)(E_1+E_3)=2E_1$. Then $$g(2E_1)=g(E_1+E_2+E_3+E_4)=g(E_1)+g(E_4)$$ which implies that $$g(E_1)=g(E_4)=k \text{ for some } k\in \mathbb{R}.$$ Hence we are done.

user9077
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  • So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly. – Lemon Dec 12 '18 at 11:50
  • Oh no. Maybe we can use the same idea? – user9077 Dec 12 '18 at 12:04
  • Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable. – Lemon Dec 12 '18 at 12:25
  • Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$. – user9077 Dec 12 '18 at 12:53
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We can use the same idea if $g:M_n(\mathbb{R})\to\mathbb{R}$. The idea is that if $B$ is a basis of $M_n$, also $g$ and $\text{tr}$ agree on $B$ up to some constant multiple $\alpha$, then $$g(A)=\alpha\cdot \text{tr}(A) $$ for all $A\in M_n$.

Let $E_{i,j}$ be the matrix where all the entries are zero except it is 1 at the position $(i,j)$. Let $T_k=\text{diag}(\underbrace{1,1,\ldots,1}_k,0,\ldots,0)$. Notice that for $i\neq j$ we have $E_{i,j}T_j=E_{i,j}$ while $T_jE_{i,j}=0$. Therefore for all $i\neq j$ we have $g(E_{i,j})=0$.

Now let $J_k$ be a $k\times n$ matrix where all the entries are 1. Define an $n\times n$ matrix $S_k$ as a block matrix $$S_k:=\begin{pmatrix}J_k\\0\end{pmatrix}$$.

Note that since $g(E_{i,j})=0$ for $i\neq j$, we have $g(S_k)=g(T_k)$ for all $k$.

Now \begin{align*} S_kS_1&=S_k \\ S_1S_k&=kS_1 \end{align*}

Hence $g(T_k)=g(S_k)=g(kS_1)=kg(S_1)=kg(T_1)$. Notice that the trace map also satisfies $\text{tr}(T_k)=k$. By taking $\alpha=g(T_1)$ then $g$ and $\alpha \cdot \text{tr}$ are two linear maps that agree on $B=\{E_{i,j}\mid i\neq j\}\cup \{T_k\mid k=1,..,n\}$ which is a basis of $M_n$. Hence $g=\alpha\cdot \text{tr}.$

user9077
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