Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Use Lagrange's method to find the maximum value of $\langle A\mathbf{x},\mathbf{x}\rangle$ subject to condition $\langle \mathbf{x},\mathbf{x}\rangle=1$ and $\langle \mathbf{u}_1,\mathbf{x}\rangle =0$ where $\mathbf{u}_1$ is a non zero vector in $N(s_1^2I_n-A^TA)$, $s_1$ is the largest singular value of A and $A=A^T \in \mathbb{R}^{n\times n}$.

Can anybody please explain me how to proceed with this question?

share|improve this question
    
Maybe a stupid question, but what is $N(s_1^2I_n-A^TA)$ ? –  sonystarmap Jan 31 '13 at 13:31
1  
Just a guess: the null space of the matrix $s_1^2 I_n - A^T A$? –  Siminore Jan 31 '13 at 15:00
    
N represents the null space –  aneps Jan 31 '13 at 18:50

1 Answer 1

This is connected to something proved in various places in MS (for instance Question related to Lagrange multipliers): the maximum of a quadratic form $Q$ on the unit sphere of a real euclidean space $E$ is the biggest eigenvalue of the matrix of $Q$ wrt any ortonormal basis of $E$. So we take $E\subset\mathbb R^n$ defined by $\langle \mathbf{u}_1,\mathbf{x}\rangle =0$ and $Q$ is the restriction to $E$ of $\langle A\mathbf{x},\mathbf{x}\rangle$. Thus one has to describe the eigenvalues of $Q$ looking carefully at $\mathbf{u}_1$. This can be done using the Spectral Theorem: there is always an orthonormal basis consisting of eigenvalues (an spectral basis). Many assertions that follows are more or less direct consequences of this fact.

First, as $s_1$ is an eigenvalue of $A$, $s_1^2$ is one of $A^TA=A^2$ ($A$ is symmetric). In fact, the eigenspace of $s_1^2$ is that $N$ containing $\mathbf u_1$, and there are two possibilities:

Case 1: $-s_1$ is not an eigenvalue of $A$. Then $N=N(s_1I-A)$ and $\mathbb R^n$ splits into an orthogonal sum $N\oplus F$, were $F$ is the sum of the other eigenspaces of $A$. This can be seen using an spectral basis for $A$. Now:

$\quad$ (i) If $\dim(N)=1$, that is, $s$ has multiplicity $1$, $E=F$ and the matrix of $Q$ wrt an orthonormal basis of $F$ consisting of eigenvectors of $A$ has the eigenvalues of $A$ except $s$. Thus the maximum we look for is the maximum eigenvalue of $A$ smaller than $s_1$.

$\quad$ (ii) If $\dim(N)>1$, we find ${\mathbf u}_2\in N$ with norm $1$ and orthogonal to $\mathbf u_1$, so that $Q(\mathbf u_2)=s_1$, being maximum on $\|\mathbb x\|=1$, is also the maximum we seek.

Case 2: $-s_1$ is also an eigenvalue of $A$. Then $N$ is the sum of the eigenspace of $s_1$ and that of $-s_1$, which can give any maximum $\le s_1$. For instance, for $n=2\ $ let $A$ be the diagonal matrix $D(1,-1)$. Then $\mathbf u_1$ can be any vector in $\mathbb R^2$ so that the maximum of $Q$ we are interested in can be any number between $s_1=1$ and $-1$. Still, one can at least say as in (ii) above (by the same argument):

$\quad$ If $\dim(N)>1$, $s_1$ is the maximum we seek.

share|improve this answer

Your Answer

 
discard

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.