TITLE: Write $\left\Vert T\right\Vert _{OP}$ using eigenvalues of orthogonal basis of $\mathbb{R}^k$.
QUESTION [0 upvotes]: I need help with the following question:

If  $T\in Hom(\mathbb{R}^k,\mathbb{R}^k)$ and $v_1,\dots,v_k$ orthogonal basis of  $\mathbb{R}^k$ and $\lambda_1,\dots,\lambda_k \in \mathbb{R}^k$   such as $T(v_i)=\lambda_i\cdot v_i$ for every $i\in {1,\dots,k}$.
Write $\left\Vert T\right\Vert _{OP}$ using $\lambda_1,\dots,\lambda_k$.

My solution: I tried many things
$$
\left\Vert T\right\Vert _{OP}=\sup_{\left\Vert x\right\Vert _{\mathbb{R}^{k}}=1}\left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}=\sup_{x\neq0}\frac{\left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}}{\left\Vert x\right\Vert _{\mathbb{R}^{k}}}=\sup_{x\neq0}\frac{\left\Vert \sum_{i=1}^{k}\alpha_{i}\lambda_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}{\left\Vert \sum_{i=1}^{k}\alpha_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}=\sup_{x\neq0}\frac{\left\Vert \sum_{i=1}^{k}\alpha_{i}\lambda_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}{\left\Vert \sum_{i=1}^{k}\alpha_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}
$$
and then I tried to use the norm and I got
$$
\sup_{x\neq0}\frac{\sqrt{\sum_{i=1}^{k}\left(T\left(x\right)\right)_{i}^{2}}}{\sqrt{\sum_{i=1}^{k}\left(x_{i}\right)^{2}}}=\sup_{x\neq0}\sqrt{\frac{\sum_{i=1}^{n}\left(T\left(x\right)\right)_{i}^{2}}{\sum_{i=1}^{n}\left(x_{i}\right)^{2}}}
$$
I also tried this
$$
\sup_{\left\Vert x\right\Vert _{\mathbb{R}^{k}}=1}\left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}=\sup\left\{ \left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}:\left\Vert x\right\Vert _{\mathbb{R}^{k}}=1\right\} =\sup\left\{ \sqrt{\sum_{i=1}^{n}\left(\lambda_{i}\alpha_{i}\right)^{2}}:\sqrt{\sum_{i=1}^{n}\alpha_{i}^{2}}=1\right\} 
$$
but I didn't know how to continue. I would like to get some help :).
Thank you very much in advance.

REPLY [0 votes]: Hint: Continue with your second approach. If we order the eigenvalues so that $|\lambda_1| \geq \cdots \geq |\lambda_k|$, then we have
$$
\|T(x)\|_{\Bbb R^k}^2 = \sum_{i=1}^n \alpha_i^2\lambda_i^2 \leq \sum_{i=1}^n \alpha_i^2 \color{red}{\lambda_1^2}.
$$
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