Proof verification: If $A \in \mathcal{M}_n$ commutes with every $B$ then $A = \lambda I$

Question

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:

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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.

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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!

Answer 1

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:

1. You consider the equality $A\left[ce_l(e_l+e_k)^T\right]=\left[ce_l(e_l+e_k)^T\right]A$ for some $c\ne0$ and look at the $(l,l)$-th entry of both sides. In other words, you are considering $e_l^TA\left[ce_l(e_l+e_k)^T\right]e_l=e_l^T\left[ce_l(e_l+e_k)^T\right]Ae_l$. From this you obtain $e_k^TAe_l=0$, i.e. all off-diagonal entries of $A$ are zero.
2. Then, you consider the equality $A\left[c_{l,k}e_le_k^T\right]=\left[c_{l,k}e_le_k^T\right]A$ for some $c_{l,k}\ne0$ and look at the $(l,k)$-th entries of both sides. This amounts to considering the equality $e_l^TA\left[c_{l,k}e_le_k^T\right]e_k=e_l^T\left[c_{l,k}e_le_k^T\right]Ae_k$. The conclusion is that $e_l^TAe_l=e_k^TAe_k$, i.e. all diagonal entries of $A$ are the same.

If you look at your proof closely, you will find that there is much room of improvement:

• There is no need to carry around the useless constants $c$ and $c_{l,k}$. You may just put replace them by $1$.
• In the first part of your proof, instead of considering the commutation between $A$ and $e_l(e_l+e_k)^T$, you may simply consider the commutation between $A$ and $e_le_k^T$.
• Since you also consider $e_le_k^T$ in the second part of your proof, you can actually merge the two parts.

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.