Dtyjlui

Let $A$ be an $n times n$ matrix and let $Lambda$ be an $n times n$ diagonal matrix. Is it always the case that $ALambda = Lambda A$? If not, when is it the case that $A Lambda = Lambda A$?

If we restrict the diagonal entries of $Lambda$ to being the equal (i.e. $Lambda = text{drag}(a, a, dots, a)$), then it is clear that $ALambda = AaI = aIA = Lambda A$. However, I can't seem to come up with an argument for the general case.

If all the diagonal entries of$Lambda$ are distinct, it commutes only with diagonal matrices.

In contrast, for each $k$ consecutive equal diagonal entries in $Lambda,$ we may allow $A$ to have anything at all in the corresponding $k$ by $k$ square block with both corners on the main diagonal.

This means that the set of matrices that commute with $Lambda$ has a minimum dimension $n$ and a maximum dimension $n^2.$ Suppose we have $r$ different diagonal entries, and there are $k_i$ copies of diagonal entry $lambda_i.$ Each $k_i geq 1,$ and we have
$$ k_1 + k_2 + cdots + k_r = n. $$
Then by the block construction I mentioned above, the dimension of the space of matrices that commute with $Lambda$ is
$$ k_1^2 + k_2^2 + cdots + k_r^2. $$
The minimum is when $r=n,$ so all $k_i = 1,$ and the dimension is $n$

The maximum is when $r=1,$ and $k_1=n,$ the matrix is a scalar multiple of the identity matrix, and the dimension is $n^2.$

It is possible that a diagonal matrix $Lambda$ commutes with a matrix $A$ when $A$ is symmetric and $A Lambda$ is also symmetric. We have

$$
Lambda A = (A^{top}Lambda^top)^{top} = (ALambda)^top = ALambda
$$

The above trivially holds when $A$ and $Lambda$ are both diagonal.

A diagonal matrix will not commute with every matrix.

$$
begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}*begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}=begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$$

But:

$$begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix} * begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix} = begin{pmatrix} 0 & 2 \ 0 & 0 end{pmatrix}.$$

The claim is false in general. Take $A = begin{bmatrix}1 & 2\3 & 4end{bmatrix}$
$Lambda = begin{bmatrix}2 & 0\0 & 3end{bmatrix}$. Then

$A Lambda = begin{bmatrix}2 & 6\6 & 12end{bmatrix}$

$Lambda A = begin{bmatrix}2 & 4\9 & 12end{bmatrix}$

On a more useful note, you can look up commuting matrices on Wikipedia.

We want to find $Lambda$ such that $ALambda=Lambda A$. Then we have $$left( begin{array}{cc}
sum a_{1n}lambda_{n1} & sum a_{1n}lambda_{n2} & sum a_{1n}lambda_{n3} & ... \
sum a_{2n}lambda_{n1} & sum a_{2n}lambda_{n2} & sum a_{2n}lambda_{n3} & ...\
sum a_{3n}lambda_{n1} & sum a_{3n}lambda_{n2} & sum a_{3n}lambda_{n3} & ...\
... & ... & ... & ...end{array} right)=left( begin{array}{cc}
sum a_{n1}lambda_{1n} & sum a_{n1}lambda_{2n} & sum a_{n1}lambda_{3n} & ... \
sum a_{n2}lambda_{1n} & sum a_{n2}lambda_{2n} & sum a_{n2}lambda_{3n} & ...\
sum a_{n3}lambda_{1n} & sum a_{n3}lambda_{2n} & sum a_{n3}lambda_{3n} & ...\
... & ... & ... & ...end{array} right)$$ Hence for the equality to hold, both $A$ and $Lambda$ must be symmetric since $a_{in}=a_{ni}$ and $lambda_{in}=lambda_{ni}$ for $i=1,2,3,...$ A diagonal matrix is a special case of this.

The answer from @AOK is not generally true. Obviously for diagonal identical matrix $Lambda$ is valid only when $A$ is symmetric matrix.