Eigenvalues and maximum values of the square distance of a vector on a unit sphere multiplied by a matrix

Let $X$ be some matrix, with dimensions $n \times m$, $Y$ be the independent columns of $X$, and $\bf{z}$ be a vector with dimension that is $n \times 1$ that lies on the unit sphere, that is, $||z|| = 1$. I want to find the maximum value of $||Y^T\bf{z}||^2$. I know the answer to the question relies on knowing the eigenvalues of $YY^T$, but I don't see how eigenvalues relate to maximum distances which I (assume) is what this question is about. I'm wondering if anyone can shed some light on how to get started with a question like this?

• you might want to read about rayleigh ritz ratios after reading user copper.hat's answer. – dineshdileep Oct 11 '13 at 7:00

Presumably you are dealing with the Euclidean norm.

You have $\|Y^T z\|^2 = \langle Y^T z, Y^T z \rangle = \langle z, YY^T z \rangle$.

$YY^T$ is symmetric and positive semi-definite, and the maximum value of $\langle z, YY^T z \rangle$ for $\|z\| \le 1$ is given by the maximum eigenvalue of $Y Y^T$ (this can be seen since $YY^T$ can be diagonalized with an orthogonal matrix).

Relevant facts:

If $Q$ is an orthogonal matrix, then $\|Q x\| = \|x\|$.

If $A$ is symmetric and can be diagonalized with an orthogonal matrix, then $\lambda_{\text{min}} \|x\|^2 \le \langle x, Ax \rangle \le \lambda_{\text{max}} \|x\|^2$. To see this, suppose $A = Q \Lambda Q^T$, where $Q$ is an orthogonal and $\Lambda = \operatorname{diag}(\lambda_1,...,\lambda_n)$. Then $\langle x, Ax \rangle = \langle Q^Tx, \Lambda (Q^Tx) \rangle = \sum_k \lambda_k [Q^Tx]_k^2$. Then it is easy to see that $\lambda_{\text{min}} \sum_k [Q^Tx]_k^2 \le \langle x, Ax \rangle \le \lambda_{\text{max}} \sum_k [Q^Tx]_k^2$, and since $\sum_k [Q^Tx]_k^2 = \|Q^T x\|^2 = \|x\|^2$, we have the desired result. (Also, the bounds are achieved for the appropriate corresponding eigenvector.)

• Did you mean to say $\|Y^T z\|^2 = \langle Y^T z, Y^T z \rangle = \langle z, YY^T z \rangle$? – Ken Li Oct 11 '13 at 17:12
• @KenLi: Indeed I did, thanks. – copper.hat Oct 11 '13 at 18:41