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Let A be an m$x$n matrix.

(a)If $Q$ is an orthogonal matrix, prove: $||QA||_2=||A|_2$

(b)Using what you proved (a); If $A$ has a singular value decomposition $U\Sigma V^T$,

Prove: $||A||_2 = ||\Sigma ||_2 = \sigma_1$

where $\sigma_1$ is the first(i.e.,largest) singular value of $A$.

Note that $|| ||_2$ is the usual 2-norm for matrices. That is,

$||A||_2 = \max_{\vec{x}\neq 0}\frac{||A\vec{x}||_2}{||\vec{x}||_2}$

Is my answer acceptable, or is there more to it for these proofs?

PF

(a) $||QA||_2^2 = (QAx)^T(QAx)= (Ax)^T(Ax) = ||A||_2^2$

(b)$||A||_2 = \max_{\vec{x}\neq 0}\frac{||U\Sigma Vx||_2}{||x||_2} = \frac{||\Sigma x||_2}{||x||_2} = \sigma_1$

  • 2
    For (a), the idea is correct but I think you need maxima over $x$ with unit norm. – 2017-02-17
  • 2
    MathJax: typing `\|` gives you a better norm sign, $\|$. – 2017-02-17

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