# Thread: T normal operator has splitting characteristic polynomial - eigenvectors form basis?

1. ## T normal operator has splitting characteristic polynomial - eigenvectors form basis?

Question: $T$ is a normal operator on a finite-dimensional real inner product space $V$ with a characteristic polynomial that splits. Prove that $V$ has an orthonormal basis of eigenvectors of $T$.

By Schur's theorem, because the characteristic polynomial of $T$ splits, there exists an orthonormal basis $\beta$ for $V$ such that $[T]_{\beta}$ is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of $T$? Thanks!

2. Originally Posted by Last_Singularity
Question: $T$ is a normal operator on a finite-dimensional real inner product space $V$ with a characteristic polynomial that splits. Prove that $V$ has an orthonormal basis of eigenvectors of $T$.

By Schur's theorem, because the characteristic polynomial of $T$ splits, there exists an orthonormal basis $\beta$ for $V$ such that $[T]_{\beta}$ is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of $T$? Thanks!

Schur's theorem tells us that $U^{*}TU=A$ , where $A$ is upper triangular and $U$ is unitary.
As all the eigenvalues of T are real (otherwise its char. pol. wouldn't split over the reals!), T is self-adjoint or Hermitian (i.e., $T=T^{*}$), and then:

$A^{*}=\left(U^{*}TU\right)^{*}=U^{*}T^{*}U=U^{*}TU =A\Longrightarrow\,A=A^{*}$ , and since $A$ is upper triangular then $A^{*}$ is lower triangular, thus $A$ is in fact diagonal and we're done.

Tonio