$A$ be an irreducible matrix, $DA=AD$ then $D$ has to be a scalar multiple of $I$

Question

Let $A$ be an irreducible matrix and $D$ be a diagonal matrix. If $DA=AD$ then $D$ has to be a scalar multiple of $I$, identity matrix.

The result seems to be non trivial to me.

A matrix is irreducible, if it is not reducible.

Answer 1

The result is true for any base field $\mathbb{K}$. Suppose the dimension of the matrices $A$ and $D$ is $n$-by-$n$. Write $A=\big[a_{i,j}\big]_{i,j\in[n]}$ and $D=\text{diag}\big(d_1,d_2,\ldots,d_n\big)$, where $[n]:=\{1,2,\ldots,n\}$ and $a_{i,j},d_k\in\mathbb{K}$ for all $i,j,k\in[n]$.

Let $G(V,E)$ be the simple graph with the vertex set $V:=[n]$ such that, for each unordered pair $\{i,j\}$ of vertices $i,j\in V$ with $i\neq j$, there exists an edge joining $i$ and $j$ if and only if either $a_{i,j}\neq 0$ or $a_{j,i}\neq 0$. We have that $$AD=\begin{bmatrix} d_1a_{1,1}&d_2a_{1,2}&\cdots&d_na_{1,n}\\ d_1a_{2,1}&d_2a_{2,2}&\cdots&d_na_{2,n}\\ \vdots&\vdots&\ddots&\vdots\\ d_1a_{n,1}&d_2a_{n,2}&\cdots&d_na_{n,n} \end{bmatrix}\text{ and }DA=\begin{bmatrix} d_1a_{1,1}&d_1a_{1,2}&\cdots&d_1a_{1,n}\\ d_2a_{2,1}&d_2a_{2,2}&\cdots&d_2a_{2,n}\\ \vdots&\vdots&\ddots&\vdots\\ d_na_{n,1}&d_na_{n,2}&\cdots&d_na_{n,n} \end{bmatrix}\,.$$ Thus, if $\{i,j\}\in E$, then we must have $d_i=d_j$. It follows immediately that, for $i,j\in V$ that lie within the same connected component of $G$, we must have $d_i=d_j$.

Since $A$ is an irreducible matrix, $G$ consists of a single connected component. As a result, $$d_1=d_2=\ldots=d_n\,.$$ Ergo, $D$ is a scalar matrix.