Diagonalizable Matrices & Eigenvalues!

  • Thread starter Este
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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.
 

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  • #2
Hurkyl
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You should empirically test your hypothesis; try it out on the simplest matrices you can imagine.
 
  • #3
HallsofIvy
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For example, how many distinct eigenvalues does

[tex]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
have?
 
  • #4
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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
Hurkyl
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The most direct way to tell if a matrix isn't diagonalizable is to try and diagonalize it.
 
  • #6
HallsofIvy
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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
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Great! Things are pretty clear now, thanks HallsofIvy!
 
Last edited:

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