How to prove that $\shortparallel T \shortparallel \leq \|T\|^{n}$? My question. How to prove that  $\shortparallel T \shortparallel \leq \|T\|^{n}$ and, if $T$ is iniective, prove that $\|T^{-1}\|^{n}\leq \shortparallel T \shortparallel \leq \|T\|^{n}$?
First I need to state what I am using: I have two Hilbert spaces $V,W$ and $L:V\to W$ is a linear application between them. Then $\DeclareMathOperator{\dimn}{\operatorname{dim}} \DeclareMathOperator{\Sym}{\operatorname{Sym}}$

*

*if $\dimn V\leq \dimn W$ then exist $S\in \Sym (n)$ and $O\in O(V,W)$ such that $L=O\circ S$,


*If $\dimn W\leq \dimn V$  then exist $S\in \Sym (n)$ and $O\in O(V,W)$ such that $L=S\circ O^{*}$,
where $O(V,W)$ is an orthogonal injective linear transformation and $\Sym(n)$ is a linear symmetric transformation.
In this case, when $L=O\circ S$ or $L=S\circ O^{*}$ then $\DeclareMathOperator{\Det}{\operatorname{Det}}$
$$ 
\shortparallel L \shortparallel\, \triangleq |\Det S|.
$$
What I have tried. Let $T\in L(\mathbb{R}^{n},\mathbb{R}^{m})$ with $n\leq m$ and consider
$$
\|T\|=\sup \frac{\|T(x)\|}{\|x\|},
$$
My idea is to use the following inequality
$$
\begin{split}
\|T\|^{n} & = \sup \frac{(\|T(x)\|)^{n}}{\|x\|^{n}} = \sup \frac{(\|O\circ S(x)\|)^{n}}{\|x\|^{n}} \\
& = \sup \frac{(\| S(x)\|)^{n}}{\|x\|^{n}}\geq \frac{\|S(x)\|^{n}}{\|x\|^{n}} = \frac{\|T^{*}\circ T (x)\|^{n}}{\|x\|^{n}},
\end{split}
$$
but how to continue? Maybe could I use the fact that
$$
(\det(T^{*}\circ T))^{\frac{1}{2}}= |\det S| =  \shortparallel T \shortparallel\;? $$
Please I need some help or any hint, thank you so much.
 A: There is a way to do it without explicitly using SVD, though, SVD is super useful, and you should definitely learn about it. Anyways, let $A$ be the matrix associated with $L$. By definition:
$$ \|L\|_2 = \|A\|_2 
= \sup_{x:x\neq 0} \frac{\|Ax\|_2}{\|x\|_2}
= \max_{x: \|x\| = 1} \|Ax\|_2
= \max_{x: \|x\|^2 = 1} \|Ax\|_2^2
$$
The Lagrangian of this constrained optimization problem is $\mathcal L(x, \lambda) = \|Ax\|_2^2 - \lambda (\|x\|_2^2-1)$ which yields the first order condition
$$ 0=\frac{\partial}{\partial x} \mathcal L(x, \lambda) = 2A^* Ax-2\lambda x \iff A^* A x=\lambda x $$
Thus the maximum is achieved for an eigenvector; in fact $\|A\|_2^2=\lambda_{\max}(A^* A)$. On the other hand you have already seen that $\shortparallel T \shortparallel^2 = \det(T^* T) = \prod_k \lambda_k(T^* T)$,  then$^{(*)}$, since all eigenvalues of $T^* T$ are positive it follows that
$$
\shortparallel T \shortparallel^2 = \prod_k \lambda_{k=1}^n(T^* T) \le \prod_{k=1}^n \lambda_{\max}(T^* T) =\lambda_{\max}(T^* T)^n =\|T\|_2^{2n}
$$
Note that by definition $\lambda_k (T*T) = \sigma_k(T)$ are exactly the singular values of $T$
$(*)$ if $\dim V <\dim W$, instead we have $\shortparallel T \shortparallel^2 = \det(TT^*)$ and the same argument holds
