Invertibility and Diagonalizability

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In summary: However, there is another matrix which is both not invertible and not diagonalizable: \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
  • #1
0pt618
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Does there exist a matrix which is both not invertible and not diagonalizable? If so, please provide an example.

Thanks,
David
 
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  • #2
Is this a homework question??

What did you try alrready??
 
  • #3
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
One that comes to mind immediately is
[tex]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

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
HallsofIvy said:
One that comes to mind immediately is
[tex]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

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...
 
  • #6
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
[tex]D= \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]
and multiply by A and [itex]A^{-1}[/itex] where A is some invertible matrix.

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

and then
[tex]ADA^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ -2 & -2 & 0 \\ 13 & -4 & 0\end{bmatrix}[/tex]
which is a non-invertible matrix which can be "diagonalized" to the original matrix, D.
 
  • #7
HallsofIvy said:
Yes. So? A diagonal matrix is trivially "diagonalizable".

It's fine. Except that the OP want a matrix that is not diagonalizable. :tongue:
 
  • #8
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
How about [0 0 0]? Definitely a matrix, and definitely not invertible since it isn't even square. Also, certainly not diagonalizable.
 
  • #10
dang! Don't you just hate those little words like "not"?

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

That is neither invertible nor diagonalizable.
 
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1. What does it mean for a matrix to be invertible?

When a square matrix has a unique solution to the system of linear equations it represents, it is considered invertible. This means that it can be multiplied by another matrix, called its inverse, to give the identity matrix.

2. How do you determine if a matrix is invertible?

A matrix is invertible if its determinant is non-zero. This can be calculated by taking the product of the elements on the main diagonal and subtracting the product of the elements on the opposite diagonal.

3. Can a non-square matrix be invertible?

No, only square matrices can be invertible. This is because non-square matrices do not have an equal number of rows and columns, making it impossible to find an inverse matrix that can multiply with it to give the identity matrix.

4. What is the significance of diagonalizable matrices?

Diagonalizable matrices have a special property where they can be simplified into a diagonal matrix through a change of basis. This makes it easier to perform calculations and solve equations involving the matrix.

5. How can you tell if a matrix is diagonalizable?

A matrix is diagonalizable if it has a full set of linearly independent eigenvectors. This means that the matrix can be decomposed into the product of a diagonal matrix and a matrix of eigenvectors.

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