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Here is my question:

-How do we prove that $\left \| A \right \|=\left \| D \right \|$ where $A$ is a square complex $n$ by $n$ matrix that satisfies: $A=J^{-1}DJ$ where $J$ is unitary (i.e $A$ and $D$ are similar)?

-Can Anyone show me how to prove the above statement?

I started like this: $\left \| A \right \|^{2}=\left ( J^{*}DJ,J^{*}DJ \right )=\left ( DJ,DJ \right )$

and I need to prove that: $\left ( DJ,DJ \right )=...=\left ( D,D \right )=\left \| D \right \|^{2}$?

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    I do $n$ot think it is not true for all matri$x$ norms. Moreover, since you are using an inner product in your proof-to-be, the parallelogram law has to hold for the norm.2012-02-25

2 Answers 2

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The result is true when we work with the subordinated norm to the euclidian norm. Indeed, we have $||A||^2=\sup_{x\neq 0}\frac{||Ax||^2}{||x||^2}=\sup_{x\neq 0}\frac{||J^*DJx||^2}{||x||^2}=\sup_{x\neq 0}\frac{||DJx||^2}{||Jx||^2}\frac{||Jx||^2}{||x||^2}=\sup_{x\neq 0}\frac{||Dx||^2}{||x||^2},$ using the fact that $J$ and $J^*$ conserve the euclidian norm.

But it we take an arbitrary norm the result may not be true. For example, consider the $2\times 2$ matrices $\pmatrix{a&b\\\ c& d}$ with the norm $\max\{|a|,|b|,|c|,|d|\}$. Take $A:=\pmatrix{1&1\\\ 1&1}$, then $||A||=1$ but a is unitary diagonalizable and its eigenvalues are $0$ and $2$ so the norm of the corresponding $D$ is $2$.

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What you are trying to do, without saying so, is to show that your norm (whatever it is, you haven't told us) is unitarily invariant.

Some norms are not unitarily invariant, as Davide's example shows.

A good number of examples of unitarily invariant norms can be produced by defining them in terms of the singular values. For example, if we denote $s_1(A),\ldots,s_n(A)$ the singular values of $A$ (i.e. the square roots of the eigenvalues of $A^*A$) in non-increasing order, then the "subordinated norm to the euclidean norm" of Davide's (that I would call the "operator norm") is given by $ \|A\|_\infty=s_1(A). $ It is an example of a Ky-Fan norm: these are the norms $ \|A\|_{(k)}=\sum_{j=1}^k s_j(A),\ \ \ k=1,\ldots,n-1 $ and the $p$-norms $ \|A\|_p=\left(\sum_{j=1}^n s_j(A)^p\right)^{1/p},\ \ \ \ \ p\geq1 $