- #1

- 136

- 0

I'm thinking not ... since even if you have:

[tex]\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right][/tex]

you could just as well say that the eigenvalue(s) are 0 (w/ algebraic multiplicity 2) and the eigenvectors are:

[tex]u_1 = \left[\begin{array}{cc}1 & 0 \end{array}\right]^T, u_2 = \left[\begin{array}{cc}0 & 1\end{array}\right]^T[/tex]