Hermit transformation.

Also sprach Zarathustra

Show that if T is Hermit transformation on inner product space V, so there are exist self values k and t of T so for all v!=0 in V :

k <= <Tv,v> / <v,v> <= t

HallsofIvy

MHF Helper
Do you mean Hermite?

Isomorphism

MHF Hall of Honor
I have not heard of a Hermite transformation. If you mean an operator $$\displaystyle T$$ such that $$\displaystyle T = T^{H}$$ then i know it as a Hermitian operator .

We will find the minimum value of the function. Firstly, we can apply spectral theorem to $$\displaystyle T$$ to get $$\displaystyle T = U\Lambda U^{H}$$ where $$\displaystyle U^{H}$$ stands for Hermitian of $$\displaystyle U$$ and $$\displaystyle \Lambda$$ is diagonal matrix of self values.

Now $$\displaystyle \min_{v \neq 0} \frac{<Tv,v>}{<v,v>} = \min_{v \neq 0} \frac{<U\Lambda U^{H}v,v>}{<v,v>} = \min_{v \neq 0} \frac{<\Lambda U^{H}v,U^{H}v>}{<v,v>}$$

Since U is unitary and hence full rank, as v varies over all non-zero vectors, $$\displaystyle U^{H}v$$ varies over all non-zero vectors. Also, since U is unitary, it preserves inner products, i.e. $$\displaystyle <v,v> = <U^{H}v,U^{H}v>$$.

$$\displaystyle \min_{v \neq 0} \frac{<\Lambda U^{H}v,U^{H}v>}{<v,v>} = \min_{U^{H}v \neq 0} \frac{<\Lambda U^{H}v,U^{H}v>}{<U^{H}v,U^{H}v>} = \min_{x \neq 0} \frac{<\Lambda x,x>}{<x,x>}$$

But,
$$\displaystyle \min_{x \neq 0} \frac{<\Lambda x,x>}{<x,x>} = \min_{||y|| = 1} <\Lambda y,y>$$

Thus we need to minimize $$\displaystyle \sum_{l=1}^{l=t} \lambda_i |y_i|^2$$ subject to $$\displaystyle \sum_{l=1}^{l=t}|y_i|^2 = 1$$ where $$\displaystyle \{\lambda_i\}$$ is the set of self values of $$\displaystyle T$$

Clearly,
$$\displaystyle \min_{||y|| = 1} <\Lambda y,y> = \min_{i \in \{1,2,3, \cdots, t\}} {\lambda_i} = k$$

Note that k is a self value and thus we have established:

$$\displaystyle k \leq \frac{<Tv,v>}{<v,v>}$$

Can you prove the other inequality?

Hint: $$\displaystyle l = \max_{i \in \{1,2,3, \cdots, t\}} {\lambda_i}$$