# Singular values

• Mar 9th 2010, 01:23 PM
Susaluda
Singular values
Let A be an orthogonal 2x2 matrix. Use the image of the unit circle to find the singular values of A.

Seems like it should be an easy question but I don't understand how to do it...help please!
• Mar 9th 2010, 01:27 PM
tonio
Quote:

Originally Posted by Susaluda
Let A be an orthogonal 2x2 matrix. Use the image of the unit circle to find the singular values of A.

Seems like it should be an easy question but I don't understand how to do it...help please!

What do you mean by "singular values of a matrix", anyway?

Tonio
• Mar 9th 2010, 01:29 PM
Susaluda
Singular values are what the chapter is on. It's a topic. I don't really know much about them though...

Edit: The singular values are the square roots of the eigenvalues of the transpose of matrix A times A.
• Mar 9th 2010, 06:25 PM
tonio
Quote:

Originally Posted by Susaluda
Singular values are what the chapter is on. It's a topic. I don't really know much about them though...

Edit: The singular values are the square roots of the eigenvalues of the transpose of matrix A times A.

Ok, so now we know what they are. Now, an orthogonal (real, of course...otherwise you'd call it unitary, right?) matrix is

of the form $\displaystyle A=\begin{pmatrix}\cos x&-\sin x\\\sin x&\cos x\end{pmatrix}$ , or $\displaystyle A=\begin{pmatrix}\cos x&\sin x\\\sin x&-\cos x\end{pmatrix}$ ,where we can (not must) take $\displaystyle 0\leq x\leq 2\pi$

Take for example the first kind, then $\displaystyle A^tA=I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ , so the singular values are ...$\displaystyle \pm 1$ . (Happy)

Of course, this follows at once from the fact that $\displaystyle A^t=A^{-1}$ for orthogonal (real) matrices....

Tonio