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Let $R^3$ denote the Euclidean space ,$|.|$ denote the usual norm. For a matrix $$T=begin{bmatrix}
a_1 & a_2 & a_3 \
a_4 & a_5 & a_6 \
a_7 & a_8 & a_9 \
end{bmatrix}.$$ How to find the desired $M$ such that for every $alpha in R^3$ $$|T(alpha) | le M| alpha |.$$

(a) Find $e_1 ,e_2 ,e_3$ which form a basis of $R^3$ then we can have $| T(e_i) | le A_i |e_i | $ , let $M_1 = max {A_1,A_2,A_3 }.$

(b) Find $e_1 ,e_2 ,e_3$ which form a orthogonormal basis of $R^3$ then we can have $| T(e_i) | le A_i |e_i | $ , let $M_2 = max {A_1,A_2,A_3 }.$

(c) Let $t_1 , t_2 ,t_3$ denote the eigenvalues of $T$ , and let $M_3 = max {t_1,t_2,t_3 }$

Can we prove that $M_1$ , $M_2$ or $M_3$ are the desired $M$ we are looking for?

Then in general conditions . Let $H_1 , H_2 $ denote two normed space , $T$ is an operator from $H_1$ to $H_2$ with a slight modification , can we prove that $M_1$ , $M_2$ or $M_3$ are the desired $M$ we are looking for?

If you are looking for the smallest such $M$ (you don't say what "desired" means), then none of your candidates is.

The most common characterization of the smallest $M$ is that it is the square root of the largest eigenvalue of $M^tM$.

If you consider $$T=begin{bmatrix} 0&1&1\0&0&0\0&0&0end{bmatrix}$$ and ${e_1,e_2,e_3}$ is the canonical basis, then $$|Te_1|=0, |Te_2|=|e_1|=1, |Te_3|=|e_1|=1,$$ while $|T|=sqrt2>1$.