$AB=BA$ implies $AB^T=B^TA$ when $A$ is normal

Question

I am looking for an elementary proof (if such exists) of the following: $$ AB=BA \quad\Longrightarrow\quad AB^T=B^TA, $$ where $A$ and $B$ are $n\times n$ real matrices, and $A$ is a normal matrix, i.e., $AA^T=A^TA$ - it is true for complex matrices as well, with $A^T$ replaced by $A^*$.

There is a non-elementary proof of this using the exponential of a matrix and properties of entire functions.

Update. In the first version of the question, both $A$ and $B$ were supposed to be normal, but as Shlomi correctly pointed out, this is true even in the case when only one of them is normal.

Answer 1

I'm not sure whether this qualifies as "elementary", but here's one way:

Note that $A$ and $B$ form a commuting family of matrices, so that $A$ and $B$ are simultaneously upper-triangularizable. That is, there is some unitary $U$ such that $U^*AU$ and $U^*BU$ are both upper triangular.

Note that $U^*AU$ and $U^*B U$ are both normal and upper-triangular, which means that they are diagonal. That is, we have $$ A = U\pmatrix{ \lambda_1 & &\\ &\ddots&\\ &&\lambda_n}U^* \quad B = U\pmatrix{ \mu_1 & &\\ &\ddots&\\ &&\mu_n}U^* $$ From here, it is easy to verify that $AB^* = B^*A$.

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Alternatively, there's a nice inductive proof here if we use the following two facts:

• Commuting matrices have a common eigenvector
• A matrix of the form $\pmatrix{A&B\\0&C}$ is only normal if $B = 0$

With that, we may use the fact that $A$ and $B$ have a common eigenvector to write $$ A = U\pmatrix{ \lambda_1 & c^T\\ 0&\tilde A} U^* \quad B = U\pmatrix{ \mu_1 & d^T\\ 0&\tilde B}U^* $$ for a unitary $U$ (whose first column is a common eigenvector to $A$ and $B$). Because $A$ and $B$ are normal, we see that $c = d = 0$. Because $A$ and $B$ are commuting normal matrices, $\tilde A$ and $\tilde B$ are (smaller) commuting normal matrices.

By induction, we may conclude (as before) that for some unitary $U$, we have $$ A = U\pmatrix{ \lambda_1 & &\\ &\ddots&\\ &&\lambda_n}U^* \quad B = U\pmatrix{ \mu_1 & &\\ &\ddots&\\ &&\mu_n}U^* $$ and we may reach the desired conclusion, just like last time.

Answer 2

Proof when one of them is normal:

Assume that only $A$ is normal. So, there is an orthogonal matrix $P$ such that:

$A=P^*DP$, where $D$ is a diagonal matrix. We shall prove that there is a polynomial $q$, such that: $q(D)=D^*$. The fact, it is true from the interpolation polynomial. Hence, we conclude that: $q(A)=A^*$. Since $AB=BA$ we can prove that $q(A)B=Bq(A)$. Thus, $A^TB=BA^T$ and we are done.