I am stuck with two questions that look very similar and I am having a problem solving the, I tried and solved one of them.
Let $V$ be a finite inner product space and let $T:V\to V$ be a linear operator such that
$$(7I-T)T^*=10I$$
- Then $T$ is self adjoint.
- $\lambda$ is an eigenvalue such that $3\le \lambda \le 5$
I could only solve the second part with the assumption that the first part is correct, if $T$ is selfadjoint then
$$(7I-T)T^* = 7T-T^2=10I$$
And by Cayley Hamilton we have that $\lambda =3$ and $\lambda=5$ are the only eigenvalues.
But I couldn't get how to solve the first one.
$(7I-T)T^*=7T-TT^*=10I$ so
$7T=TT^*+10I$
Now if you apply the adjoint operator you have that
$(7T)^*=7T^*=(TT^*+10I)^*=$
$=(TT^*)^*+10I=(T^*)^*T^*+10I=$
$=TT^*+10I=7T$ $\to$
$7T^*=7T$ and so $T=T^*$