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Thread: Eigenvalue, norm

  1. #1
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    Eigenvalue, norm

    Hello!

    I've got a question according the solution of this exercise

    $\displaystyle A \in M(n, \mathbb{K})$

    Show that $\displaystyle \|A \| := \sqrt{E_{max}}$, where

    $\displaystyle E_{max}$ is the biggest eigenvalue $\displaystyle of A^*A$

    and

    $\displaystyle \| A \| := \sup_{\| x \|_2 = 1} \| Ax\|^2 = \sup_{\| x\| \le 1} \langle Ax, Ax \rangle$

    $\displaystyle = \sup \langle A^* Ax,x \rangle$

    SOLUTION

    $\displaystyle A^*A$ is positive definit, so there is a unitary $\displaystyle U \in GL(n, \mathbb{K}) : $


    $\displaystyle UA^*AU^{-1} = diag (E_1, ... , E_n )=$ and $\displaystyle E_i \ge 0$ denote the eigenvalues of $\displaystyle A^*A$


    So


    $\displaystyle \| A \|^2 = \sup \langle A^*Ax, x \rangle$

    $\displaystyle = \sup \langle U^{-1}DUx,x \rangle$

    $\displaystyle = \sup \langle DUx , Ux \rangle , \ Ux =: y$

    $\displaystyle =\sup \langle Dy, y \rangle$

    $\displaystyle = \sup \sum^n_{i=1} E_i |y_i |^2$

    $\displaystyle = E_{max}$

    So why is the last line? Why is it not $\displaystyle \ge E_{max}$?


    Rapha
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  2. #2
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    That is because $\displaystyle y = Ux$ is supposed to be a unit vector, so $\displaystyle \sum^n_{i=1} |y_i |^2 = 1$. Therefore $\displaystyle \sum^n_{i=1} E_i |y_i |^2\leqslant \sum^n_{i=1} E_{\max} |y_i |^2\leqslant E_{\max}$.
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  3. #3
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    Hello Opalg!

    Thanks alot.
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