# Show that $A$ is normal if it commutes with some normal matrix with distinct eigenvalues.

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

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

and

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

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

$$\mathbf{T}\mathbf{D}=\mathbf{D}\mathbf{T}$$

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.

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

## 1 Answer

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 $$i\neq j$$.