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.
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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*}
(AD)_{ij}=\sum_{k=1}^n a_{ik}d_{kj}=a_{ij}d_j.
\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.
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