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$\mathcal{H}$ stands for an Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle{\cdot}| {\cdot}\rangle$ and the norm $\|{\cdot}\|$ and let $\mathcal{B}(\mathcal{H})$ the algebra of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}$. I look for an example of maximal vector $x\in \mathcal{H}$ for $T\in\mathcal{B}(\mathcal{H})$ i.e. a vector $x$ satisfying $\|x\|=1$ and $\|Tx\|=\|T\|$.

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  • $\begingroup$ Can you find such a vector on a finite-dimensional space? Then you can do it on a finite-dimensional subspace of $\mathcal H$. $\endgroup$ – Yurii Savchuk Jul 24 '17 at 9:11
  • $\begingroup$ What about $0$ operator? $\endgroup$ – Yurii Savchuk Jul 24 '17 at 9:12
  • $\begingroup$ we assume that $T$ is non zero operator $\endgroup$ – Student Jul 24 '17 at 9:13
  • $\begingroup$ unfortunately I don't find such a vector on a finite-dimensional space. $\endgroup$ – Student Jul 24 '17 at 9:16
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    $\begingroup$ A non-trivial question is of an operator, where such vector does not exist. $\endgroup$ – Yurii Savchuk Jul 24 '17 at 14:15
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Take the shift operator $S:l^2 \rightarrow l^2$ such that $S(x_1,x_2....)=(0,x_1,x_2...).$

Then $||S||_2=1$ and $||Se_1||_2=1$ a where $e_1=(1,0..0...)$ and $||e_1||_2=1$

Also you can take shift the operator $T:l^2 \rightarrow l^2$ such that $T(x_1,x_2...)=(x_2,x_3....)$ and the vector $e_2=(0,1,0..0..)$

Now in the finite dimensional case take the hilbert space $\mathbb{R}^2$ with the usual inner product and the bounded linear operator $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $T(a,b)=(a,0)$

You can easily prove that $||T||_2=1$ and $||Te_1||_2=1$ where $e_1=(1,0)$ and $||e_1||_2=1$

Hope this helps.

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  • $\begingroup$ Thank you. This is example where $H$ is infinite-dimensional Hilbert space. Could you please give me an example where $H$ is finite-dimensional Hilbert space?? Thank you very much $\endgroup$ – Student Jul 24 '17 at 12:36
  • $\begingroup$ ok i will edit my answer providing also such an example $\endgroup$ – Marios Gretsas Jul 24 '17 at 12:40
  • $\begingroup$ Thank you for your help $\endgroup$ – Student Jul 24 '17 at 12:48
  • $\begingroup$ You are welcome. $\endgroup$ – Marios Gretsas Jul 24 '17 at 12:49
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consider the simple example where $T$ is the identity operator.

Then every vector that satisfies $\| x\|=1$ is a maximal vector, because the operator norm of the identity operator is $1$.

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  • $\begingroup$ Thank you for your answer but I want an example when T is not equal to identity $\endgroup$ – Student Jul 24 '17 at 11:56

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