# A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal

## Problem 492

Let
$D=\begin{bmatrix} d_1 & 0 & \dots & 0 \\ 0 &d_2 & \dots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \dots & d_n \end{bmatrix}$ be a diagonal matrix with distinct diagonal entries: $d_i\neq d_j$ if $i\neq j$.
Let $A=(a_{ij})$ be an $n\times n$ matrix such that $A$ commutes with $D$, that is,
$AD=DA.$ Then prove that $A$ is a diagonal matrix.

## Proof.

We prove that the $(i,j)$-entry of $A$ is $a_{ij}=0$ for $i\neq j$.

We compare the $(i,j)$-entries of both sides of $AD=DA$.
Let $D=(d_{ij})$. That is, $d_{ii}=d_i$ and $d_{ij}=0$ if $i\neq j$.
The $(i,j)$-entry of $AD$ is
\begin{align*}
\end{align*}

The $(i,j)$-entry of $DA$ is
$(DA)_{ij}=\sum_{k=1}^n d_{ik}a_{kj}=d_ia_{ij}.$

Hence we have
$a_{ij}d_j=a_{ij}d_i.$ Or equivalently we have
$a_{ij}(d_j-d_i)=0.$

Since $d_i\neq d_j$, we have $d_j-d_i\neq 0$.
Thus, we must have $a_{ij}=0$ for $i\neq j$.

Hence $A=(a_{ij})$ is a diagonal matrix.

### 2 Responses

1. Lalitha says:

How did value “aijdj” derived from the summation (The (i,j)-entry of AD is – sentence). Please explain.

• Yu says:

This follows from the definition of the matrix product $AD$. Please review how $AD$ is defined.

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