0
$\begingroup$

I have some difficulties with exercises concering the Singular Value Decomposition. I have to find $A = U D_A V^T$ for the singular value decomposition, where $U$ and $V$ are orthogonal matrices. I know the basics of how to compute it, but I always encounter difficulties when I have to find orthogonal eigenvectors that constitute the orthogonal matrices $U$ and $V$. Let me provide an example:

Consider matrix $A = \begin{bmatrix} \sqrt{3} & 2\sqrt{2} \\ \sqrt{3} & 0 \end{bmatrix}$.

$A A^T = \begin{bmatrix} 11 & 3 \\ 3 & 3 \end{bmatrix}$

The eigenvalues are $\lambda_1 = 12$ and $\lambda_2 = 2$

Now I want to find a orthogonal matrix $U \in \mathbb{R}^{2 \times 2}$ be defined by $U = [u_1 \quad u_2]$. Write $u_1 = \begin{bmatrix} x \\ y \end{bmatrix}$ for the orthogonal eigenvector corresponding to eigenvalue $\lambda_1 = 12$ and $u_2 = \begin{bmatrix} p \\ q \end{bmatrix}$ for the orthogonal eigenvector corresponding to eigenvalue $\lambda_2 = 2$. It follows that $[AA^T = \lambda_1 I_2]u_1 = 0 \iff \begin{bmatrix} -1 & 3 \\ 3 & -9 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

I know that the answer is $u_1 = \begin{bmatrix} \frac{\sqrt{3.6}}{2} \\ \frac{\sqrt{0.4}}{2} \end{bmatrix}$, but I don't really see how you can come to such an answer if you have to do it by yourself. It seems to me a bit farfetched. Any idea how you could easily spot such numbers?

$\endgroup$

1 Answer 1

1
$\begingroup$

It's clear that $$x=\left[\begin{matrix} 3 \\ 1 \end{matrix}\right]$$ is in the null space by Gaussian elmination. $\langle x, x\rangle=\sqrt {10}$, so divide both components of $x$ by $\sqrt{10}$ in order to make $x$ orthogonal. The result is equivalent to the $u_1$ you have at the bottom of your answer above.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .