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

Let $A$ be real symmetric and $D$ shall contain the eigenvalues of $A$. I've learned that $\|A\|_{\text{max}}< \|D\|_{\text{max}}$, where $\|A\|_{\text{max}}$ means the Max norm.

I want to get a sharper bound $\|A\|_{\text{max}}$ by using the knwoledge about more than one eigenvalue, let say two. Ky-Fan norms seem appropriate, so I'm looking for something like $$ \|A\|_{\text{max}}\not <\frac12\|D\|_2, $$ where $\|D\|_2=\lambda_0+\lambda_1$ sums up the largest eigenvalues. Numerics showed that it doesn't hold, ven if I use absoulute values $|\lambda_0|+|\lambda_1|$.

share|cite|improve this question
up vote 1 down vote accepted

What you want is impossible, unless you put further restrictions on which $A$ are allowable. To see why, start with $A$ diagonal, so that $D=A$. In this example, your original inequality is sharp. Then any linear combination of the diagonal of $D$ with coefficients less than $1$ in absolute value will fail your desired inequality.

share|cite|improve this answer
Sounds a little like Luke Skywalker: 'You want the impossible.' Anyway, so are you saying that the knowledge about more than one eigenvalue doesn't help at all when I want to know something about $\|A\|_\max$? Help me Obi-Wan Kenobi, you are my only hope. – draks ... Nov 10 '12 at 12:55
I gave you an example where $\|A\|_{\max}$ is exactly the biggest eigenvalue, and all the information about the remaining eigenvalues is irrelevant. You cannot expect to improve on that. – Martin Argerami Nov 10 '12 at 14:23
Would it help to say that $A$ is positive definite? – draks ... Nov 13 '12 at 6:36
Nope, the bound is still sharp for positive definite, take $$A=D=\begin{bmatrix}2&0\\0&1\end{bmatrix}.$$ – Martin Argerami Nov 13 '12 at 10:59

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.