I have solved an exercise and would like to know if my solution is correct.

The Exercise

$\kappa_p(A) :=\|A\|_p\|A^{-1}\|_p, 1 \leq p \leq \infty, A \in \mathbb R^{n \times n}$

Show that for a symmetric regular, but not necessarily positive definite matrix $A \in \mathbb R^{n \times n}$ it holds that:

$\kappa_2(A) = \frac{|\lambda|_{max}(A)}{|\lambda|_{min}(A)}$

where $|\lambda|_{max}$ is the absolute value of the eigenvalue with biggest absolute value and $|\lambda|_{min}$ with the smallest.

My solution

If A $\in \mathbb R^{n \times n}$ is a symmetric regular matrix, there is an orthonormal base in $\mathbb R^n$ where A has a diagonal representation A = diag$(\lambda_1,...,\lambda_n)$ with eigenvalues arranged according to

$|\lambda_1| \leq .... \leq |\lambda_n|$

Hence $|\lambda_1| = |\lambda|_{min}(A)$ and $|\lambda_n| = |\lambda|_{max}(A)$. Naturally we obtain $\|A\|_2 = max_{\|x\|_2 = 1} \|Ax\|_2 = |\lambda|_{max}(A)$ Note that since A is regular, then $\lambda_k \neq 0 $ for k=1,..,n. Moreover within the same base $A^{-1}$ has also a diagonal representation: $A^{-1}$ = diag$(\frac{1}{\lambda_1},...,\frac{1}{\lambda_n})$. Hence the statement above is true since $\|A^{-1}\|_2 = \frac{1}{|\lambda|_{min}(A)}$

  • 1
    $\begingroup$ Looks OK to me. $\endgroup$ – Giuseppe Negro Mar 26 '13 at 12:31

It looks good! It's even better if you could elaborate on the reason why $\max_{\|x\|_2 = 1} \|Ax\|_2 = |\lambda|_{\max}(A)$. (The orthogonality of the basis plays an important role here.)


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