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Find $||T|| $ qnd $||T^2||$

I managed to find $||T^2||$, but I have no idea how to calculate $||T||$, I thought about using inequality $ ||T^2|| \leq ||T||^2$, but I couldn't

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    $\begingroup$ I'd be curious to see by what means you could find $\lVert T^2\rVert$ while remaining completely oblivious of $\lVert T\rVert$. $\endgroup$ – Gae. S. Oct 27 at 14:53
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Since $$\|Tx\|^2 = 0 + 16x_1^2 + x_2^2 + 16 x_3^2 + \cdots \le 16(x_1^2 + x_2^2 + x_3^2 + \cdots) = 16 \|x\|^2$$ you have $\|T\| \le 4$. Equality is attained because in the special case $x = (1,0,0,\ldots)$ you have $$4 = \|Tx\| \le \|T\| \|x\| = \|T\|.$$

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