- 51

- 0

[tex]A = U \Sigma V^T => A^T A = V \Sigma^2 V^T[/tex] and [tex]A A^T = U \Sigma^2 U^T[/tex]

So 2 independent eigenvectors of [tex]A^T A[/tex] are a basis for the row space of A and 2 independent eigenvectors of [tex]A A^T[/tex] are a basis for the column space of A.

If I pick

[tex]v_{1} = \left(\begin{array}{cc}1/\sqrt{2}\\1/\sqrt{2}\end{array}\right)[/tex] [tex]v_{2} = \left(\begin{array}{cc}1/\sqrt{2}\\-1/\sqrt{2}\end{array}\right)[/tex]

and

[tex]u_{1} = \left(\begin{array}{cc}1\\0\end{array}\right)[/tex] [tex]u_{2} = \left(\begin{array}{cc}0\\1\end{array}\right)[/tex]

as these bases, respectively, I get into trouble. Matrix A will not equal the singular value decomposition, there will be a sign error even though these vectors truly are correct eigenvectors.

For those who know him, this is an example that prof. Strang gave during one of his video lectures but he didn't quite see where the error came from. And since he never gave a solution, I hope I can find one here.

I started thinking about this and I noticed that these 2 bases don't have the same orientation. So I tried to find a new basis for one of the 2 spaces, with respect to the eigenvalues: just change one vector into the opposite direction. But the problem now is, this only works half the time. So for the 4 new bases I found (with respect to the eigenvalues) only 2 work out the way it's supposed to.

So if I change [tex]u_{2}[/tex] to [tex]\left(\begin{array}{cc}0\\-1\end{array}\right)[/tex] everything works out. But if I change [tex]u_{1}[/tex] to [tex]\left(\begin{array}{cc}-1\\0\end{array}\right)[/tex] things get into trouble again.

Similar for [tex]v_{1}[/tex] and [tex]v_{2}[/tex].

Can anyone tell me what is going on here?

Phew, that was a lot of typing. :tongue2: I hope I didn't make any errors, but my main point is there and that's what counts.

Thanks in advance.