Hi,here is the question that I am having trouble with.thanks.

Show that the matrix(0 1;0 0) is not diagonalizable..

semicolon represents row

Printable View

- Jul 9th 2013, 08:00 PMn22diagonalization-help
Hi,here is the question that I am having trouble with.thanks.

Show that the matrix(0 1;0 0) is not diagonalizable..

semicolon represents row - Jul 10th 2013, 02:41 AMchiroRe: diagonalization-help
Hey n22.

Hint: Consider the determinant of the matrix. How does this relate to the conditional of being diagonalizable? - Jul 10th 2013, 06:42 AMjohngRe: diagonalization-help
Hi n22,

Perhaps I'm just being dense, but I don't see how Chiro's hint helps. Here's a solution:

Attachment 28774 - Jul 10th 2013, 08:36 AMHallsofIvyRe: diagonalization-help
I need to learn to read

**all**posts in a thread! I spent some time writing out a proof then realized that Johng had already given exactly that proof!

I am not sure what Chiro's point is. Obviously the determinant is 0 but that alone does not prevent the matrix being diagonalizable. It**does**mean that 0 is an eigenvalue. If the other eigenvalue is NOT 0, the matrix**is**diagonalizable. If**both**eigenvalues are 0, the matrix still might be "diagonalizable" but, in fact, the matrix in that case would have to be the 0 matrix. - Jul 10th 2013, 05:14 PMtopsquarkRe: diagonalization-help
Correct me if I'm wrong but the inverse of a matrix A contains a factor $\displaystyle \frac{1}{|A|}$. So if |A| = 0 that means that the matrix does not have an inverse. I believe that is what chiro was referring to.

-Dan - Jul 10th 2013, 05:38 PMchiroRe: diagonalization-help
I was actually thinking about the rank of the diagonal matrix in the diagonalization decomposition with regards to its eigen-values, but I don't think this was actually a useful hint after all.

- Jul 10th 2013, 05:49 PMtopsquarkRe: diagonalization-help
- Jul 10th 2013, 05:57 PMchiroRe: diagonalization-help
We can use |A| to assess the eigen-values (since the product of the eigen-values is equal to |A|) and from this, I was going to look at the results of a non full-rank (i.e. has at least one row of zeroes since the matrix is diagonal) and its effect on being able to diagonalize the matrix.

Also note that the diagonalization is a form of a rotation so if you aren't able to do the rotation, then you can also prove results regarding whether the diagonalization exists (with regards to whether the rotation provided by the eigen-vectors and eigen-values actually works).