$\|T^*T\|=\max\{\lambda: \lambda \text{ is an eigen value of } T^*T\}$

Question

I want to prove this $\|T^*T\|=\max\{\lambda: \lambda \text{ is an eigen value of } T^*T\}$\

Suppose $\lambda_0>\lambda_1>\dots>\lambda_{n-1}\ge 0$ are distinct eigen values of $T^*T$ and $\{v_0,\dots,v_{n-1}\}$ be the orthonormal set of eigen vectors of $T^*T$ corresponding to the above taken eigen values.

so any $v\in \mathbb{C}^n,\|v\|=1$ can be written as $$v=\sum_{i=0}^{n-1} v_i$$ $$T^*Tv=\sum_{i=0}^{n-1} T^*T v_i= \sum_{i=0}^{n-1} \lambda_i v_i$$

$$\|T^*Tv\|^2=\langle T^*Tv,T^*Tv\rangle=\sum_{i=0}^{n-1}\lambda_i^2$$ If I want to proceed in this way, how can I conclude the result, please help

Answer 1

If $\|v\| = 1$, you should have $$v = \sum_{i=1}^{n-1} a_i v_i$$ with $\sum_{i=1}^{n-1}a_i^2 = 1$.

Thus $$T^*Tv = \sum_{i=0}^{n-1}a_iT^*Tv_i = \sum_{i=0}^{n-1} a_i \lambda_i v_i$$

and $$\|T^*Tv\|^2 = \sum_{i=1}^{n-1}a_i^2\lambda_i^2 \leq (\sum_{i=1}^{n-1}a_i^2)\max\lambda_i^2 = \max \lambda_i^2$$

And take $i_0 = \arg\max \lambda_i^2$ and $v = v_{i_0}$, the equality holds