Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: $$\left ( DJ,DJ \right )=...=\left ( D,D \right )=\left \| D \right \|^{2}$$?

share|cite|improve this question
What norm are you using? – Calle Feb 25 '12 at 21:08
I do not think it is not true for all matrix norms. Moreover, since you are using an inner product in your proof-to-be, the parallelogram law has to hold for the norm. – Calle Feb 25 '12 at 21:34
up vote 5 down vote accepted

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 $2$.

share|cite|improve this answer

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 $$ \|A\|_p=\left(\sum_{j=1}^n s_j(A)^p\right)^{1/p},\ \ \ \ \ p\geq1 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.