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Question: Find all the linear transformations $f$ from $Mn(R)$ to $R$ such that $f(AB) = f(BA)$ for all matrices $A, B ∈ Mn(R)$.

Doubt: I know what a linear transformation is, but I am not able to use the given linear transformation condition.

Alpha
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2 Answers2

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Hint: First, we get from linearity that $f(AB-BA)=0$ for all $A,B$. $[A,B]=AB-BA$ is known as the commutator of $A$ and $B$.

It is fairly well-known, I gather, that the subspace of commutators is the subspace of traceless matrices.

Let's look at the $2\times2$ case. The space of traceless matrices has dimension $n^2-1$ in general, so $3$ in this case. A basis would be $\{\begin{pmatrix}1&1\\0&-1\end{pmatrix}, \begin{pmatrix}1&0\\1&-1\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix}\}$, for instance.

To complete it to a basis for $\mathcal M_2(\Bbb R)$, we can add $\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

Now a linear transformation is completely determined by its effect on a basis. Thus we would have $f$ of the first three basis vectors being zero. The last one can be taken to any nonzero real number.

Rel this basis, we could represent each such $f$ by a $1\times4$ matrix/row vector $(0,0,0,k)$, where $f(\begin{pmatrix}0&0\\0&1\end{pmatrix})=k$.

To get $f$ rel the standard basis, use the relevant ($4\times4$) change of basis matrix.

All of this generalises to the $n\times n$ case in a rather straight forward way.

  • Perhaps a bit simpler, one could note that $f$ has to send $\begin{pmatrix}1 & 0 \ 0 & 0 \end{pmatrix}$ to $k$, and it has to send both $\begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 0 \ 1 & 0\end{pmatrix}$ to zero. From this it's easy to compute $f(\begin{pmatrix}a & b \ c & d\end{pmatrix})$. –  Oct 22 '20 at 07:36
  • @Bungo Ok. I guess we could say $f(M)=k\cdot\rm{trace}M$ for any $M\in\mathcal M_n(\Bbb R)$. –  Oct 22 '20 at 07:44
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    Yes, looks right to me. The key observation, as you noted, is that $tr(AB) = tr(BA)$, hence $tr(AB-BA) = 0$, hence $f$ must map all zero-trace matrices to zero. The rest follows from linearity of $f$. –  Oct 22 '20 at 07:48
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Here is an approach which I find it more straightforward:

We set $E_{ij}$ to be the matrix whose $(i,j)$-entry is one, while all other entries are zero.

Since $f$ is linear, all that is required to know is how $f$ act on the basis. Clearly, $\{E_{ij}\}_{1\le i, j\le n}$ is a basis of $M_n(\mathbb R)$. Suppose the coefficients are respectively $c_{rs}$, i.e. $$f(E_{rs})=c_{rs}$$

For $1\le r < s \le n$, $$ c_{rs} = f(E_{rs}) = f(E_{r1}E_{1s})= f(E_{1s}E_{r1}) = f(0) = 0$$

For $1\le r = s \le n$, $$ c_{rr} = f(E_{rr}) = f(E_{r1}E_{1r})= f(E_{1r}E_{r1}) = f(E_{11}) = c_{11}$$

Thus, $f$ must has the form of taking the trace and then multiply a constant ($c_{11}$ in the deduction above). We know that $f(AB)=f(BA)$ holds for such linear transformations.

fantasie
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