1
$\begingroup$

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 $|\lambda_0|+|\lambda_1|$.

1 Answers 1

1

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.

  • 0
    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.2012-11-10
  • 0
    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.2012-11-10
  • 0
    Would it help to say that $A$ is positive definite?2012-11-13
  • 0
    Nope, the bound is still sharp for positive definite, take $$A=D=\begin{bmatrix}2&0\\0&1\end{bmatrix}.$$2012-11-13