Hermit transformation.

Dec 2009
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


MHF Hall of Honor
Dec 2007
IISc, Bangalore
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. \(\displaystyle <v,v> = <U^{H}v,U^{H}v>\).

\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>}\)

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

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 \(\displaystyle T\)

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

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