# Given $T^2=\frac{T+T^{\ast \:}}{2}$. Find all eigenvalues of $T$ and show that $T^2 = T$

I need help to show that $T^2 = T$

$T:V\to V\:$ is linear transformation over $F=\mathbb{R}\:or\:F=\mathbb{C}$

$V$ is a finite dimensional vector space.

First I think that it will help to show that $T$ is normal:

$T^2=\frac{T+T^{\ast \:\:}}{2}\:\Rightarrow \:T^{\ast }=2T^2-T\Rightarrow \:T^{\ast }T=2T^3-T^2=T\left(2T^2-T\right)=\:TT^{\ast }$

Now I manages to show that $\left(T^{\ast \:}\right)^2=T^2\:\:and\:\left(T^{\ast \:\:}\right)^3=T^3$

this helped me to observe that

$T^3 - T^4 = 0 \implies T^3(T-I)=0$

Now, $if\:q\left(t\right)=t^3\left(t-1\right)\:then\:q\left(T\right)=0\:$ Since it is known that the minimal polynomial of T divides $q(t)$ therefore the possible eigenvalues of $T$ exists and equals to zero or one.

How I managed to show what are the eigenvalues of $T$:

let $v\in V$ be an eigenvector of $T$, assuming $\lambda$ is the corresponding eigenvalue.

Therefore

$T^2v=\lambda \:^2v\:=\:\frac{1}{2}\left(T+T^{\ast }\right)v=\:\frac{1}{2}\left(\lambda \:v\:+\:\bar{\lambda \:}v\right)$

the last equality is because if $T$ is normal then $\lambda \:eigenvalue\:of\:T\:\Rightarrow \:\bar{\lambda }\:eigenvalue\:of\:T^{\ast }$

Now $\lambda \:\in \:\left\{0,1\right\}\:or\:\left\{1\right\}\:or\:\left\{0\right\}\:\Rightarrow \:\lambda \:\:\in \:\mathbb{R}\:\Rightarrow \lambda =\bar{\lambda }$

So we get that:

$\lambda ^2v=\lambda v\:\Rightarrow \lambda \left(\lambda -1\right)=0\:\Rightarrow \lambda \in \left\{0,1\right\}$

Now I struggle to show that $T^2 = T$ and need some help with that.

• From $T^2=\frac{T+T^*}{2}$ you get that $\frac{T^2+TT^*}{2}=T^3=\frac{T^2+T^*T}{2}$. So $TT^*=T^*T$, is normal. Therefore, there is $U$ unitary and $D$ diagonal such that $T=U^*DU$. It follows that $U^*D^2U=T^2=\frac{T+T^*}{2}=U^*Re(D)U$, where $Re(D)$ is the diagonal matrix with the real parts of the diagonal elements of $D$. Therefore, $D^2=Re(D)$, from where it follows that in the diagonal there are only $0$'s and $1$'s. Thus, $D^2=D$, and then $T^2=U^*D^2U=U^*DU=T$.
– user580373
Aug 6 '18 at 13:12
• But if $V$ is over the field $\mathbb{R}$ then if $T$ is normal then it is not always the case that what you wrote exists. for example $Tv\to \begin{pmatrix}0&1\\ 1&0\end{pmatrix}v$ Aug 6 '18 at 13:22
• It doesn't matter, because you can extend the coefficients and prove the same equation. It doesn't matter that $U^*DU$ has to be written in $\mathbb{C}$ if it is going to be multiplied back into $\mathbb{R}$.
– user580373
Aug 6 '18 at 13:24

$$T^2 = \frac{T+T^*}{2}.$$ Therefore $T^2=(T^*)^2$ follows from taking adjoints of both sides. Using this, $$2T^2-T=T^* \\ 4T^4-4T^3+T^2=T^2 \\ T^4-T^3=0 \\ T^3(T-1)=0.$$
$T$ is normal because $T$ commutes with $T^*=2T^2-T$. Therefore, $$\mathcal{N}(T)=\mathcal{N}(T^2).$$ Therefore, $T^3(T-I)=0$ implies $T(T-I)=0$, which is the desired result that $T^2=T$.