Let $V$ be a complex inner product space and $T$ a self-adjoint linear operator on $V$. Show that $I + iT$ is non-singular.

Question

The question is:

Let $V$ be a complex inner product space and $T$ a self-adjoint linear operator on $V$. Show that $I + iT$ is non-singular.

Taking $\Vert(I+iT)v\Vert^2$, I'm able to obtain $$\left\langle v|v\right\rangle + \left\langle TT^*v|v\right\rangle$$ Chegg says that this should get me to $\left\langle v|v\right\rangle + \left\langle v|v\right\rangle = 2\left\langle v|v\right\rangle$.

But this would imply that $T$ is also unitary, which I don't see why it necessarily would be.

Answer 1

We assume $V$ is finite dimensional.

Suppose that $(I+iT)(x)=0$ we have, $<x+iT(x),x+iT(x)>=<x,x>-i<x,T(x)>+i<T(x),x>+<T(x),T(x)>$ since $T$ is self-adjoint, $<x,T(x)>=<T(x),x>$ we deduce that $<x+iT(x),x+iT(x)>=<x,x>+<T(x),T(x)>$=0. This implies that $<x,x>=0$ and henceforth $Ker T=0$ thus $T$ is invertible.

Answer 2

I assume $V$ is finite dimensional.

Then $I+iT$ is singular if and only if $-i(I+iT)=T-iI$ is singular, that is $i$ is an eigenvalue of $T$. But as $T$ is self-adjoint, its eigenvalues are real. Indeed, if $Tv=\lambda v$, $v\ne0$, you have $$ \lambda\langle v,v\rangle= \langle v,\lambda v\rangle= \langle v,Tv\rangle= \langle Tv,v\rangle= \langle \lambda v,v\rangle= \bar\lambda\langle v,v\rangle $$ and, since $\langle v,v\rangle\ne0$ we get $\lambda=\bar\lambda$.

Alternatively, from $(I+iT)v=0$, you get $$ 0=\langle v+iTv,v+iTV\rangle= \langle v,v\rangle-i\langle Tv,v\rangle+ i\langle v,Tv\rangle+\langle Tv,Tv\rangle= \langle v,v\rangle+\langle Tv,Tv\rangle $$ Since $\langle v,v\rangle\ge0$ and $\langle Tv,Tv\rangle\ge0$, we get both are zero, in particular $v=0$.

(Note: I use semilinearity in the first variable, but it would be the same for semilinearity in the second variable.)