Let $A$ be an $n \times n$ matrix and let $\Lambda$ be an $n \times n$ diagonal matrix. Is it always the case that $A\Lambda = \Lambda A$? 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 $A\Lambda = AaI = aIA = \Lambda A$. However, I can't seem to come up with an argument for the general case.

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    $\begingroup$ When the diagonal matrix is on the right, it scales the columns of the matrix it is multiplying. when the diagonal matrix is on the left, it scales the rows. Since column-scaling and row scaling are different operations, there are only very limited circumstances that the matrices will commute. $\endgroup$
    – Nick Alger
    Mar 15, 2016 at 1:30

5 Answers 5


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

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

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


A diagonal matrix will not commute with every matrix.

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


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


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 $n^2.$

  • $\begingroup$ Do the copies have to be right next to each other? e.g. diag(1,1,1,2,2,7,8,8)? $\endgroup$
    – Not Euler
    Feb 8, 2018 at 5:25
  • $\begingroup$ @HritRoy, no. You can change your basis to express the copies next to each other if you want to. But it is not necessary $\endgroup$ Apr 27, 2018 at 22:22

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

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

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

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


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


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