Diagonalizable Matrices & Eigenvalues

In summary, the conversation discusses the sufficiency of determining if a nXn matrix is diagonalizable by showing that it has n distinct eigenvalues. The conversation also explores other methods for checking if a matrix is diagonalizable, such as using eigenvectors. The theorem states that if a matrix has n distinct eigenvalues, then it is diagonalizable, and the conversation also mentions that this is a sufficient but not necessary condition.
  • #1
Este
7
0
Hello,

Is it sufficient to determine that a nXn matrix is not diagonalizable by showing that the number of its distinct eigenvalues is less than n?

Thanks for your time.
 
Physics news on Phys.org
  • #2
You should empirically test your hypothesis; try it out on the simplest matrices you can imagine.
 
  • #3
For example, how many distinct eigenvalues does

[tex]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
have?
 
  • #4
Ok I got your point...I think the question should be now, how to check whether a matrix is diagonalizable or not?

A theorem I've encountered in my textbook:
Theorem 3: Suppose that A is an nXn matrix and that A has n distinct eigenvalues, then A is diagonalizable.

and somewhere else I read this is sufficient to prove a matrix is diagonalizable but not the other way around.. that's why I posted the question..

So now, all I can do is to prove that Matrix x is diagonalizable, but if it's not, I can't tell for sure...what other methods should I be using?

Thanks for your responses.
 
  • #5
The most direct way to tell if a matrix isn't diagonalizable is to try and diagonalize it.
 
  • #6
Este said:
Ok I got your point...I think the question should be now, how to check whether a matrix is diagonalizable or not?

A theorem I've encountered in my textbook:


and somewhere else I read this is sufficient to prove a matrix is diagonalizable but not the other way around.. that's why I posted the question..
Yes- sufficient but not necessary.

So now, all I can do is to prove that Matrix x is diagonalizable, but if it's not, I can't tell for sure...what other methods should I be using?

Thanks for your responses.
An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. Since eigenvectors corresponding to distinct eigenvalues are always independent, if there are n distinct eigenvalues, then there are n independent eigenvectors and so the matrix is diagonalizable. But even if the eigenvalues are not all distinct, there may still be independent eigenvectors.

[tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
as I pointed out before has the single eigenvalue 1 but is already diagonal because <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> are independent eigenvectors.

[tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}[/tex]
also has 1 as its only eigenvalue. Now, <1, 0, 0> and <0, 0, 1> are the only eigenvectors so this matrix cannot be diagonalized.

Finally,
[tex]\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}[/tex]
also has 1 as its only eigenvalue but now only <1, 0, 0> and multiples of that are eigenvectors.
 
  • #7
Great! Things are pretty clear now, thanks HallsofIvy!
 
Last edited:

1. What is a diagonalizable matrix?

A diagonalizable matrix is a square matrix that can be transformed into a diagonal matrix through a similarity transformation. This means that the matrix has a full set of linearly independent eigenvectors.

2. What are eigenvalues and eigenvectors?

Eigenvalues are scalar values that represent the scaling factor of the eigenvectors in a linear transformation. Eigenvectors are non-zero vectors in which the linear transformation only affects the scaling of the vector, not its direction.

3. How do you find the eigenvalues of a matrix?

To find the eigenvalues of a matrix, you first need to find the characteristic polynomial by subtracting the scalar variable from the main diagonal elements of the matrix and then taking the determinant. The roots of the characteristic polynomial will be the eigenvalues of the matrix.

4. Can a matrix have repeated eigenvalues?

Yes, a matrix can have repeated eigenvalues. This means that there is more than one eigenvector associated with the same eigenvalue. In this case, the matrix is still diagonalizable, but it may have fewer linearly independent eigenvectors.

5. What is the significance of diagonalizable matrices and eigenvalues?

Diagonalizable matrices and eigenvalues are important in many areas of mathematics and science. They have applications in solving differential equations, finding the stability of systems, and analyzing complex matrices. They also have practical applications in fields such as computer graphics and quantum mechanics.

Similar threads

  • Linear and Abstract Algebra
Replies
1
Views
765
  • Linear and Abstract Algebra
Replies
8
Views
1K
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
10
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
673
  • Linear and Abstract Algebra
Replies
5
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
488
  • Math Proof Training and Practice
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
7K
Back
Top