# Invertibility and Diagonalizability

1. Nov 26, 2011

### 0pt618

Does there exist a matrix which is both not invertible and not diagonalizable? If so, please provide an example.

Thanks,
David

2. Nov 26, 2011

### micromass

Staff Emeritus
Is this a homework question??

What did you try alrready??

3. Nov 26, 2011

### Deveno

hint: see if you can find a 2x2 matrix using only 0's and 1's with 0 determinant. i can think of 2 such matrices right off the bat that fulfil both your criteria.

4. Nov 26, 2011

### HallsofIvy

Staff Emeritus
One that comes to mind immediately is
$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

A matrix is not invertible if and only if it has 0 as an eigenvalue. A matrix is diagonalizable if the exist a basis for the space consisting of eigenvectors. Those are not contradictory.

5. Nov 26, 2011

### micromass

Staff Emeritus
Uuh, that is already a diagonal matrix...

6. Nov 27, 2011

### HallsofIvy

Staff Emeritus
Yes. So? A diagonal matrix is trivially "diagonalizable".

If you want a non-diagonal, diagonalizable, matrix that is not invertible, start with a diagonal matrix, say
$$D= \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
and multiply by A and $A^{-1}$ where A is some invertible matrix.

For example, if
$$A= \begin{bmatrix}1 & 0 & 0 \\ 2 & -1 & 0 \\ 5 & -2 & 1\end{bmatrix}$$
then
$$A^{-1}= \begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 2 & 1\end{bmatrix}$$

and then
$$ADA^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ -2 & -2 & 0 \\ 13 & -4 & 0\end{bmatrix}$$
which is a non-invertible matrix which can be "diagonalized" to the original matrix, D.

7. Nov 27, 2011

### micromass

Staff Emeritus
It's fine. Except that the OP want a matrix that is not diagonalizable. :tongue:

8. Nov 27, 2011

### pol92

You can simply choose a nilpotent non-null matrix, as it's easy to see that the only diagonalizable AND nilpotent matrix is the null one.
for example:
01
00

9. Nov 27, 2011

### nucl34rgg

How about [0 0 0]? Definitely a matrix, and definitely not invertible since it isn't even square. Also, certainly not diagonalizable.

10. Nov 27, 2011

### HallsofIvy

Staff Emeritus
dang! Don't you just hate those little words like "not"?

Okay, lets start with a Jordan Normal form, non-diagonal matrix:
$$P= \begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
which also has 0 as an eigenvalue and so is not invertible. Using the same "A" as before,
$$APA^{-1}= \begin{bmatrix}0 & 1 & 0 \\ 4 & 0 & 0 \\ 8 & 1 & 0\end{bmatrix}$$

That is neither invertible nor diagonalizable.