# Make a matrix defective if possible

1. Apr 24, 2010

### Dustinsfl

$$\begin{bmatrix} 1 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & \alpha \end{bmatrix}$$ and $det(A-\lambda I)=\lambda(\alpha-\lambda)(\lambda-2)=0$

Therefore, $\lambda_1=0$, $\lambda_2=\alpha$, and $\lambda_3=2$.

In order to make this matrix defective, I need to make the left two column vectors dependent$$\begin{bmatrix} 1-\alpha & 1 & 0\\ 1 & 1-\alpha & 0\\ 0 & 0 & 0 \end{bmatrix}$$.

Is there an easy way to find or determine if this can be done for a given matrix?

2. Apr 24, 2010

### Dustinsfl

I am thinking the only way this will work if alpha=0 or 2? Then I need to check those values to see if the eigenspace has only one eigenvector. However, is that necessarily true that it will only work at those values or is it possible that another value besides 0 or 2 will work?

3. Apr 24, 2010

### lanedance

well to be defective it must have an eigenvalue with >1 algebraic multiplicity, but looking at
http://en.wikipedia.org/wiki/Defective_matrix
says normal matricies are not defective, and your matrix is real & symmetric as far as I can tell...

Last edited: Apr 25, 2010
4. Apr 24, 2010

### Dick

If a matrix has an eigenvalue, then it has at least one eigenvector corresponding to that eigenvalue. So, yes, the only way it can be defective is if alpha=0 or alpha=2.

5. Apr 24, 2010

### Dustinsfl

Thanks

6. Apr 25, 2010

### gabbagabbahey

Careful, what Dick said was that the matrix can be defective if $\alpha=0$ or $\alpha=2$...In order to determine if it actually is defective for either of those two values, you need to check whether or not they yield an incomplete basis of eigenvectors.

7. Apr 25, 2010

### Dick

Right. I should have emphasized the 'CAN'. In fact, for those two values, if you think about lanedance's general observation, you don't actually have to compute the eigenvectors.