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Let $E$ be a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$. We recall that an operator $T\in \mathcal{L}(E)$ is called normal if $TT^*=T^*T$.

It is true that for a normal operator $T$ we have $$\|Tx\|\leq |\langle Tx\;| \;x\rangle|,\;\forall x\in E\;?$$ Thank you for your help.

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    $\begingroup$ Let $T$ be the $2\times 2$ matrix defined by a -1 in the top right, and a 1 in the bottom left, all other entries zero, and consider the Hilbert space $\mathbb{R}^2$. Then $T^* = -T$ (so it's normal), and for some $(a,b)^t = x\in E$, $Tx=(-b,a)^t$. You can see now that $\langle Tx,x \rangle = 0$ yet $||Tx|| \geq 0$, and will be strictly larger when $x$ is nonzero. $\endgroup$ – DKS Oct 6 '17 at 14:52
  • $\begingroup$ $\|Tx\|$ is order 1 in $x$, while $|\langle Tx,x\rangle|$ is order 2 in $x$. So replace $x$ by $\alpha x$ and allow $\alpha$ to vary in order discover a problem in the stated inequality. $\endgroup$ – DisintegratingByParts Oct 6 '17 at 23:04
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Let $T$ be the $2\times 2$ matrix defined by a -1 in the top right, and a 1 in the bottom left, all other entries zero, and consider the Hilbert space $E=\mathbb{R}^2$. Then $T^∗=−T$ (so it's normal), and for some $(a,b)^t=x\in E$, $Tx=(−b,a)^t$. You can see now that $\langle Tx,x\rangle =0$ yet $||Tx||\geq 0$, and will be strictly larger when $x$ is nonzero.

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  • $\begingroup$ One thing that jumps out to me: remember the Cauchy-Schwarz inequality? This say for two vectors $v,w$ in a Hilbert space, $|\langle v, w \rangle | \leq ||v|| \cdot ||w||$. Think about when this inequality is actually an equality, i.e., is there a $w$ such that $|\langle v, w \rangle | = ||v|| \cdot ||w||$? And what might that $w$ be? $\endgroup$ – DKS Oct 18 '17 at 1:43

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