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Some related problems:

  1. Relationship between inner product and norm
  2. What norm Induced inner product?

My problem comes from one step of a certain proof:

$\|Av\|^2=(Av)^T(Av)=v^TA^TAv=v^TIv = v^Tv =\|v\|^2 \Rightarrow \|Av\| = \|v\|$

Note: $A$ is an orthogonal matrix.

More general one:

$\|Av\|^2=\langle Av,Av \rangle=\langle v,A^TAv \rangle=\langle v,Iv \rangle= \langle v,v\rangle=\|v\|^2 \Rightarrow \|Av\| = \|v\|$

They look similar.

It seems the above equality only holds on $l_2$ norm(the first case) and inner product induced norm (the second case).

So such equality cannot hold for general norm, right?

I just want to make my proof more strict.

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sleeve chen
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  • I'm not sure what the question is, since you don't state any assumptions about, e.g., $A$. What the calculations show is that those operators which satisfy $A^TA = I$, where $(.)^T$ is the adjoint operation of the scalar product, are isometries of the norm induced by the scalar product. – Thomas Jul 01 '16 at 06:40
  • What property on $A$ is given? Is it just an orthogonal matrix (linear operator) acting on a finite dimensional linear space? – b00n heT Jul 01 '16 at 06:40
  • @Thomas I fix that. – sleeve chen Jul 01 '16 at 06:41
  • @b00nheT Yes $A$ is an orthogonal matrix and finite dimensional linear space. – sleeve chen Jul 01 '16 at 06:42
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    Then no. Just consider any other induced inner product induced by a positive definite symmetric matrix and its induced norm: i.e choose a PDS matrix $B$ and the inner product $(x,y):=x^\top B y$Then your orthogonality of $A$ (which is wrt the standard inner product) won't hold anymore. – b00n heT Jul 01 '16 at 06:46

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Assuming $A$ is an orthogonal linear map (i.e. $A^T A = I$), you are correct that $\|Av\| = \|v\|$ does not hold in general for norms other than the $2$-norm. For example, take $$A = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}$$ which you can check satisfies $A^T A = I$, and $$v = \begin{pmatrix}1 \\ 0 \end{pmatrix}$$ and compute the $1$-norm of $Av = \begin{pmatrix}1\sqrt{2} \\ -1/\sqrt{2}\end{pmatrix}$ and $v$: we see that $\|Av\|_1 = 2/\sqrt{2}$ whereas $\|v\|_1 = 1$.