Let $T$ be a self-adjoint ($T^*=T$), compact operator over a hilbert space $H$. Assume that $T$ has at least one strictly positive eignvalue. Let $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq...$ is a series of strictly positive eignvalues of $T$, when we repeat each eignvalue according to its multiplicity. If the series is finite we complete it to infinite serie with zeros.
1.
A. Prove that $\lambda_1=max_{x\in H,\|x\|=1} (Tx,x)$.
B. Prove that if $x_1\neq 0 \in H$ is an eignvector for $\lambda_1$, then
(*) $\lambda_2=max_{x\in H,\|x\|=1, (x,x_1)=0} (Tx,x)$.
C. Prove that # $\lambda_2= min_{y\in H} max_{x\in H,\|x\|=1, (x,y)=0} (Tx,x)$, in the following way:
(1). Let $y\in H$ a random vector and $x_1$ same as in part B. Show that there are $a_1,a_2 \in F$ that at least one of them not zero, and $x_2\neq 0 \in H$ an eignvector for $\lambda_2$ and orthogonal to $x_1$ such that $a_1x_1+a_2x_2$ orthogonalc which it orthogonal to y. Deduce $\lambda_2\leq (Tx,x)$.
(2). If we choose $y=x_1$ in # we get equality, by (*).Then # is accomplished.
D. Explain shortly, why for all k,
$\lambda_{k+1} = min_{y_1,...,y_k \in H} max_{x\in H, \|x\|=1, (x,y_i)=0 \forall 1\leq i \leq k} (Tx,x)$.
Note: when I mentioned strictly positive eignvalue I meant $\lambda > 0$.
2. Let $S,T$ be self-adjoint, compact operators over a hilbert space H. We count their strictly positive eignvalues, including multiplicity (similar to part 1), by $\lambda_1\geq \lambda_2\geq....$ and $\mu_1\geq \mu_2\geq....$ respectively. Assume, $(Tx,x)\leq (Sx,x)$ for all $x\in H$. Show that, $\lambda_n \leq \mu_n$, $\forall n$
What I think:
In A, I tried to use the claims:
- a normal operator $T\in B(H)$ satisfies $r(T)=\|T\|$ where $r(T)$ is the spectral radius of $T$.
- If $T\in B(H)$ is self adjoint then its spectrom $\sigma(T) \subset R$.
Part B, we know that "if $T\in B(H)$ is normal and $\lambda, \mu$ are different eignvalues of $T$ then $Ker(T- \lambda I)$ is orthogonal to $Ker(T-\mu I)$". However, I don't know how to show these parts formally using these facts.
For 2, I see that the idea to use part D, we can substitut $k=n-1$ so, by the assumption $(Tx,x)\leq (Sx,x)$ we get:
By def of Maximum on sets : $Max_{x\in H, \|x\|=1, \forall 1\leq i\leq n-1 (x,y_i)=0} (Tx,x) \leq Max_{....} (Sx,x)$, then by definition of minimum over sets, we'll get the desired result.
I will appreciate your help/guide so much. It will surely add to my understanding in this topic!
All of the answers to question 1 will use the following:
Let $\mathbf{e}_i$ be a set of unit vectors with the property that $T \mathbf{e}_i = \lambda_i \mathbf{e}_i$. Then any vector $\mathbf{x}$ can be written as $\mathbf{x} = \sum_i x_i\mathbf{e}_i$, and we also have $$ \left<Tx,x\right> = \sum_i \lambda_i|x_i|^2. $$ You will need the spectral theorem to prove this. It's not too hard.
Part A follows immediately from $\lambda_i \le \lambda_1$ for all $i$: $$ \left<T\mathbf{x},\mathbf{x}\right> = \sum_i \lambda_i|x_i|^2 \le \sum_i \lambda_1|x_i|^2 = \lambda_1 \sum_i |x_i|^2 = \lambda_1. $$ And since $\left<T\mathbf{e}_1,\mathbf{e}_1\right> = \lambda_1$, this is indeed the maximum value.
For part B, note that $\left<\mathbf{x},\mathbf{e}_1\right> = 0$ implies $x_1 = 0$. Then you can use a similar argument from $\lambda_i \le \lambda_2$ for all $i \ne 1$.
The strategy for part C is saying for any vector $\mathbf{y}$, show there is a unit vector $\mathbf{x}$ perpendicular to it that only has components in $\mathbf{e}_1$ and $\mathbf{e}_2$. Clearly such a vector has $\left<T\mathbf{x},\mathbf{x}\right> = |x_1|^2\lambda_1 + |x_2|^2\lambda_2$, so $\lambda_2 \le \left<T\mathbf{x},\mathbf{x}\right> \le \lambda_1$, so the max over all such $\mathbf{x}$ is always at least as large as $\lambda_2$. Then show $\lambda_2$ is a possible value, so that the minimum over $\mathbf{y}$ is indeed $\lambda_2$.
For part D, show that for a set of $k$ vectors $\mathbf{y}$, you can construct a vector perpendicular to them using the first $k+1$ $\mathbf{e}_i$. Then use a similar argument to part C.
For question 2, let $T_n$ be the space spanned by the first $n$ eigenvectors of $T$, and $S_n$ be similarly defined. Show there is a unit vector $\mathbf{x} \in T_{n+1}$ that is not in $S_n$. Then show that $\lambda_{n+1} \le \left<T\mathbf{x},\mathbf{x}\right>\le\left<S\mathbf{x},\mathbf{x}\right> \le \mu_{n+1}$.