# Invertibility and Diagonalizability

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

Thanks,
David

## Answers and Replies

Is this a homework question??

What did you try alrready??

Deveno
Science Advisor
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.

HallsofIvy
Science Advisor
Homework Helper
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.

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.

Uuh, that is already a diagonal matrix...

HallsofIvy
Science Advisor
Homework Helper
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.

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

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

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

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

HallsofIvy
Science Advisor
Homework Helper
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.

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