# Linear Algebra / Group Theory

1. Sep 14, 2009

### phagist_

1. The problem statement, all variables and given/known data

Let $$T : V \rightarrow V$$ be a linear operator on a complex inner product space $$V$$ , and let
$$S = I + T^{*}T$$, where $$I : V \rightarrow V$$ is the identity.
(a) Write $$<Sx,x>$$ in terms of $$x$$ and $$Tx$$.
(b) Prove that every eigenvalue $$\lambda$$ of $$S$$ is real and satisfies $$\lambda$$$$\geq 1$$.
(c) Prove that the nullspace of $$S$$ is $$\{0\}$$.

2. Relevant equations

3. The attempt at a solution
Ok I really hardly have any idea on how to tackle this problem.

Am I right in thinking that; in terms of $$Tx$$ and $$x$$
$$<Sx,x>$$ becomes $$<[I+T^{*}T]x,x>$$?
If so, I'm thinking of using the properties of inner products, but so far I've only dealt with scalars inside the <,> brackets, and not linear operators.
I'm really struggling with it and thus haven't attempted parts b or c yet.

Any help/comments would be greatly appreciated,
cheers!

Last edited: Sep 14, 2009
2. Sep 14, 2009

### HallsofIvy

Staff Emeritus
Well, what are the "properties of inner products" that you could use? Stop "thinking" about using them and use them! Mainly what you need about linear operators here is the definition of $T^*$

3. Sep 14, 2009

### phagist_

So I have
$$<[I+T^{*}T]x,x>$$

and the definition of $$T^*$$ is $$<Tx,x>=<x,T^*x>$$

but I'm not really familiar with $$<Sx,x>$$.. this means that S is operating on x, right?
Then another way to say it is $$<[I+T^{*}T]x,x>$$
I'm looking for a way to 'break it up'.. I know the property that $$<ax,x> = a<x,x>$$ and $$<x+y,z> = <x,z> + <y,z>$$but as I said before, I'm only familiar with that when a, x, y and z are scalars from some field.

Sorry if these questions/remarks seem stupid, but I just haven't come a problem of this variety before - and are unfamiliar with how to handle it.