Timeline for How to compute the induced $||\cdot||_{2} $ matrix norm of an SPD matrix
Current License: CC BY-SA 2.5
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| Jan 25, 2011 at 22:48 | comment | added | Suvrit | I think Zenon, you will benefit from also visiting the sister site: math.stackexchange.com Let me add my comments as an answer so that your question gets an "answered" status. Also, it is good to leave all these comments; might prove useful to someone else searching google for proofs. However, I did want to stress that since your question requested textbook knowledge, I pointed you to matlab and numerics book. | |
| Jan 25, 2011 at 22:32 | comment | added | Zenon | Thanks Suvrit, finally this is what I wanted. This is the first time I use the forum, should I delete this question and ask the more precise one? Or could you create an answer and I could give you the points? | |
| Jan 25, 2011 at 21:53 | comment | added | Suvrit | to show that $\lambda_{\max}$ is the norm, we need to then only show that there is a vector $x$ for which actually the above inequality holds with equality. Based on the above, just pick $x=v_1$, the first eigenvector, and you have equality, concluding the proof. (note that proof also used $\lambda_i \ge 0$ | |
| Jan 25, 2011 at 21:50 | comment | added | Suvrit | the proof that you want, can be found in the cited textbooks. Here is a short version of it (not writing as an answer, because this is really a math.SE question). Let us look at the matrix $A$. We want $\max \|Ax\| / \|x\|$, where $x \neq 0$. Since $\|\alpha x\| = |\alpha| \|x\|$, wlog, take $\|x\|=1$. Now, consider $\|Ax\|^2=x^TA^TAx$. We wish to maximize this quadratic form over $x^Tx=1$. Use eigendecomposition of $A^TA=V\Lambda V^T$; let $y=V^Tx$. Then, $x^TA^TAx=\sum_i \lambda_i y_i^2 \le \lambda_{\max}y^Ty = \lambda_{\max}$, since $y^Ty=x^TV^TVx=x^Tx=1$ | |
| Jan 25, 2011 at 20:59 | history | answered | Zenon | CC BY-SA 2.5 |