Invertibility and Diagonalizability

[0pt618]

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

Thanks,
David

 

[micromass]

Is this a homework question??

What did you try alrready??

 

[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.

 

[HallsofIvy]

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.

 

[micromass]

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]

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.

 

[micromass]

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

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

 

[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

 

[nucl34rgg]

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

 

[HallsofIvy]

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

Okay, let's 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.