Suppose that $\mathbf{A} \in \mathbb{C}^{n \times n}$ and there is a normal matrix $\mathbf{X} \in \mathbb{C}^{n \times n}$ such that $\mathbf{X}$ has distinct eigenvalues (none of them repeat) and $\mathbf{A}\mathbf{X} = \mathbf{X}\mathbf{A}$.

I want to show that $\mathbf{A}$ is normal.

$\textbf{My attempt}$

Since both matrices commute and $\mathbf{X}$ has distinct eigenvalues, they share the same eigenvectors. Taking the Schur decompositions of both matrices, we obtain

$$\mathbf{X}= \mathbf{U}\mathbf{D}\mathbf{U}^{*}$$


$$\mathbf{A}= \mathbf{U}\mathbf{T}\mathbf{U}^{*}$$

Where $\mathbf{U}$ has the eigenvectors of both matrices as columns, $\mathbf{D}$ is diagonal and contains the eigenvalues of $\mathbf{X}$, and $\mathbf{T}$ is upper triangular and contains the eigenvalues of $\mathbf{A}$.

This implies


Now, I want to show $\mathbf{T}$ is diagonal. But how exactly do I do this? I've seen this answer, but it is hard to follow.

  • $\begingroup$ If I commute with a diagonal matrix that has distinct entries, then I must be diagonal, as s.harp’s answer shows. $\endgroup$ – Santana Afton Mar 2 at 17:01

Since $D$ is diagonal you have $$(DT)_{ij}=D_{ii}T_{ij}, \qquad(TD)_{ij}=T_{ij}D_{jj}$$ so $$(DT-TD)_{ij}=T_{ij}(D_{ii}-D_{jj})\overset!= 0.$$ If $T_{ij}\neq0$ you get $D_{ii}=D_{jj}$, since you are supposing that $D$ only takes on distinct eigenvalues you find $T_{ij}=0$ if $i\neq j$.


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