I'm doing an exercise where I have a similar statement concerning linear transformations. More specifically: $\forall \sigma \in \mathcal{L} (V) (\tau \sigma = \sigma \tau) \implies \tau = a \iota$, where $\iota$ is the identity operator. Since every linear transformation with respect to a basis is equivalent to a matrix, I thought I could instead prove the statement in the title, that is, a square matrix which commutes with every other square matrix of same size must be a scalar multiple of $I$. I tried constructing matrices to assist me in the proof, so I wanted to be sure if the proof is actually valid.
The proof starts below:
Let $M_c (l, k) \in \mathcal{M}_n (F)$, where $n \geq 2$ be a matrix where $m_{l, l} = c$, $m_{l, k} = -c$, and for any other index $m_{i, j} = 0$. Also, $l \neq k$ and $c \neq 0$. Summarizing, the $n$th entry in the main diagonal has value $c$, and another entry in the same row has value $-c$, and all other entries have value $0$.
We intend to prove that any matrix $A \in \mathcal{M}_n$ which commutes with $M_a (l, k)$ for every $1 \leq l, k \leq n$ is a diagonal matrix. For the sake of readability, the matrix will be written only as $M$. Since both commute, for any $l$ and $k$, we have:
\begin{equation}
\begin{split}
[B M]_{l, l} = \sum^n_{i = 1} b_{l, i} m_{i, l} = b_{l, l} m_{l, l} = c b_{l, l} \\
= [M B]_{l, l} = \sum^n_{i = 1} m_{l, i} b_{i, l} = b_{l, l} m_{l, l} + b_{k, l} m_{l, k} = c b_{l, l} + (- c b_{k, l}) \; .
\end{split}
\end{equation}
We have that $- c b_{k, l} = 0$. Since $a \neq 0$, the same holds for its additive inverse. This implies that $b_{k, l} = 0$ for arbitrary $k$ and $l$ where $k \neq l$. We can choose every $k$ and $l$ between $1$ and $n$ and arrive at the same conclusion for these indexes, so we have that $b_{i, j} = 0$ for $i \neq j$, therefore $B$ must be a diagonal matrix, with all entries outside the main diagonal equal to $0$.
Now, we pick some matrix $C \in \mathcal{M}_n$ where $c_{l, k} \neq 0$ for $l \neq k$. Since $A$ commutes with $C$:
\begin{equation}
\begin{split}
[A C]_{l, k} = \sum^n_{i = 1} a_{l, i} c_{i, k} = a_{l, l} c_{l, k} \\
= [C A]_{k, j} = \sum^n_{i = 1} c_{l, i} a_{i, k} = a_{k, k} c_{l, k} \; .
\end{split}
\end{equation}
Since $c_{l, k} \neq 0$, this implies any two entries in the diagonal are equal, and since $l$ and $k$ are arbitrary, this applies for all entries. Since $A$ is also diagonal, we conclude that $A$ must be the multiple of an identity matrix.
I just wanted to be sure that I didn't make any mistakes and that these conclusions are enough to prove the statement. Is the proof correct?
Thanks in advance!
Best Answer
Apart from some minor typos, your proof looks correct to me, but it can be simplified quite a bit.
Let $e_i$ be the $i$-th vector in the standard basis of $F^n$. That is, $e_1=(1,0,0,\ldots,0)^T,\,e_2=(0,1,0,\ldots,0)^T$, etc.. In matrix language, your proof is composed of two parts:
If you look at your proof closely, you will find that there is much room of improvement:
This gives rise to the following proof:
Proof 1. Let $l\ne k$. Since $Ae_le_k^T=e_le_k^TA$, we have $$ \begin{cases} e_l^T(Ae_le_k^T)e_l=e_l^T(e_le_k^TA)e_l\\ e_l^T(Ae_le_k^T)e_k=e_l^T(e_le_k^TA)e_k \end{cases} \Rightarrow \begin{cases} 0=e_k^TAe_l\\ e_l^TAe_l=e_k^TAe_k \end{cases} \Rightarrow \begin{cases} a_{kl}=0,\\ a_{ll}=a_{kk}. \end{cases} $$ Hence $A$ is a scalar multiple of the identity matrix.
The above proof can be refined further. Note that the matrix $e_le_k^T$ above is a rank-one matrix. This motivates us to consider the situation where $A$ commutes with all rank-one matrices:
Proof 2. For any two vectors $u,v\in F^n$, we have $Auv^T=uv^TA$. In turn, we have $Auv^Tx=uv^TAx$ for any three vectors $u,v$ and $x$. Since $v^Tx$ and $v^TAx$ are scalars, the previous equality means that $(v^Tx)Au=(v^TAx)u$. Pick a pair of vectors $v$ and $x$ such that $v^Tx\ne0$. Then $Au=\frac{v^TAx}{v^Tx}u$ for every vector $u$. Since the constant $c=\frac{v^TAx}{v^Tx}$ does not depend on $u$, we conclude that $A=cI$.
This proof, although written using matrix and vectors, is essentially coordinate-free. If you replace $v^T$ by a linear functional $f$ and $uv^T$ by the linear transformation $x\mapsto f(x)u$ in the above, you can get rid of all matrices and rewrite the proof in terms of linear transformations. This is actually Robert Israel's proof in another thread.