- #1
0pt618
- 25
- 1
Does there exist a matrix which is both not invertible and not diagonalizable? If so, please provide an example.
Thanks,
David
Thanks,
David
Invertibility and Diagonalizability
- Context: Undergrad
- Thread starter 0pt618
- Start date
Click For Summary
Discussion Overview
The discussion centers on the existence of matrices that are both not invertible and not diagonalizable. Participants explore examples and properties of such matrices, including specific constructions and theoretical implications.
Discussion Character
- Exploratory
- Technical explanation
- Debate/contested
- Mathematical reasoning
Main Points Raised
- David asks if a matrix can be both not invertible and not diagonalizable, seeking an example.
- Some participants suggest looking for a 2x2 matrix with 0's and 1's that has a determinant of 0.
- One participant proposes the zero matrix as an example, noting that it is not invertible but is diagonalizable.
- Another participant argues that a diagonal matrix is trivially diagonalizable, suggesting the need for a non-diagonal matrix that is not invertible.
- A construction is provided using a diagonal matrix multiplied by an invertible matrix to create a non-invertible matrix that can be diagonalized.
- Another participant introduces the concept of nilpotent matrices, stating that the only diagonalizable nilpotent matrix is the zero matrix.
- A participant mentions a Jordan Normal form matrix that is non-diagonal and not invertible, providing a specific example.
Areas of Agreement / Disagreement
Participants express differing views on the definitions and examples of matrices that are not invertible and not diagonalizable. There is no consensus on a single example that satisfies both conditions, and multiple competing views remain.
Contextual Notes
Some participants rely on specific definitions of diagonalizability and invertibility, which may not be universally agreed upon. The discussion includes various assumptions about matrix properties and constructions that are not fully resolved.
- #2
- 22,170
- 3,326
Is this a homework question??
What did you try alrready??
What did you try alrready??
- #3
Deveno
Science Advisor
Gold Member
MHB
- 2,726
- 6
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
HallsofIvy
Science Advisor
Homework Helper
- 42,895
- 983
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.
\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
- 22,170
- 3,326
HallsofIvy said:
Uuh, that is already a diagonal matrix...
- #6
HallsofIvy
Science Advisor
Homework Helper
- 42,895
- 983
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.
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
- 22,170
- 3,326
HallsofIvy said:
It's fine. Except that the OP want a matrix that is not diagonalizable.
- #8
pol92
- 2
- 0
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
for example:
01
00
- #9
nucl34rgg
- 149
- 15
How about [0 0 0]? Definitely a matrix, and definitely not invertible since it isn't even square. Also, certainly not diagonalizable.
- #10
HallsofIvy
Science Advisor
Homework Helper
- 42,895
- 983
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.
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.
Similar threads
- · Replies 3 ·
- · Replies 3 ·
- · Replies 2 ·
- · Replies 4 ·
- · Replies 4 ·
- · Replies 4 ·
- · Replies 1 ·
- · Replies 12 ·
- · Replies 1 ·
Undergrad
Diagonalizability and Invertibility
- · Replies 9 ·