If $T^\ast=T^{2017}$ then $T$ is diagonalizable

Question

Let be $T:V\rightarrow V,k=\mathbb{C}$ so that $T^*=T^{2017}$ $\implies$ $T$ is diagonalizable.

Well, the statement is from a T or F and I already know it's true. I am 99% sure it has to do with properties but I'm really confused by this subject. How can I deduce that the statement is true?

$T^*$ denotes the adjoint operator of $T$ (In case that is not the standar notation).

score 4 · Accepted Answer · answered Jan 29 '20 at 13:40

4

Since $T$ conmutes with its adjoint operator it implies $T$ is normal. Then, given dimension $V$ be finite, it will be diagonalizable.

answered Jan 29 '20 at 13:40

astro

649

How do I know that T conmutes with its adjoint? – xtreyreader Jan 29 '20 at 18:39
@xtreymer: $T^T=T^{2018}=TT^$. – Martin Argerami Jan 29 '20 at 20:47

score 2 · Answer 2 · answered Jan 29 '20 at 13:36

2

By Schur's Triangularization there is a unitary $U$ such that $T=UAU^*$, with $A$ upper triangular. It follows that $A^*=A^{2017}$, so $A^*$ is both lower triangular and upper triangular; thus $A^*$ is diagonal, and so $A$ is diagonal.

answered Jan 29 '20 at 13:36

Martin Argerami

205,756

score 1 · Answer 3 · answered Jan 29 '20 at 19:53

since we're in $\mathbb C$, using the fact that conjugate transposition is an involutive operation, you have

$m:= 2017^2$

$T^* = T^{2017}\longrightarrow T = (T^*)^{2017} = (T^{2017})^{2017}= T^m$

so
$\mathbf 0 = T^m - T = T(T^{m-1} - I) = (T-0I)\prod_{j=1}^{m-1}(T-\omega_j I\big)$
where $\omega_j$ are the (m-1)th roots of unity. So $T$ is annhilated by a polynomial with no repeated roots which implies $T$ has a minimal polynomial with no repeated roots which implies $T$ is diagonalizable.

If $T^\ast=T^{2017}$ then $T$ is diagonalizable

3 Answers3