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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    Have you tried simple examples? – Friedrich Philipp Mar 15 '16 at 1:18
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    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. – Nick Alger Mar 15 '16 at 1:30

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 $n^2.$

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

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.

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    You are right - thanks. I modified my answer. – user322903 Mar 15 '16 at 1:50

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 $A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$ $\Lambda = \begin{bmatrix}2 & 0\\0 & 3\end{bmatrix}$. 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.

We want to find $\Lambda$ such that $A\Lambda=\Lambda A$. 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.

  • Hello, what is the name of this theorem? Or where is it in a book? I want to reference it. Thanks a lot. – Vladimir Vargas Apr 27 at 22:09
  • I am not aware of a theorem whose result is this since it is just derived from the definition. – TheSimpliFire Apr 28 at 8:20

In general, $A\Lambda\ne\Lambda A$. However, you might be interested in showing that $A^T\Lambda =\Lambda A$, which does hold for all $A,\Lambda\in\mathbb{R}^{N\times N}$ where also $\Lambda$ is diagonal.

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