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Diagonalizing a symmetric matrix with non-distinct eigenvalues

  • Thread starter Fractal20
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Homework Statement


Given the matrix A:

4 2 2
2 4 2
2 2 4

Find the matrix P such that P-1AP is diagonal


Homework Equations





The Attempt at a Solution


So I had this question today on a placement exam and it threw me for a loop. I found the eigenvalues to be 2,2, and 8. The eigenvectors are (-1,1,0), (-1, 0, 1), and (1,1,1) respectively. So my understanding was that with any real symmetric matrix it should be possible to write it in this form. However, my understanding was also that it had to have n linear independent eigenvectors which meant no repeated eigenvalues. Which way is it?
 

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  • #2
Ray Vickson
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Homework Statement


Given the matrix A:

4 2 2
2 4 2
2 2 4

Find the matrix P such that P-1AP is diagonal


Homework Equations





The Attempt at a Solution


So I had this question today on a placement exam and it threw me for a loop. I found the eigenvalues to be 2,2, and 8. The eigenvectors are (-1,1,0), (-1, 0, 1), and (1,1,1) respectively. So my understanding was that with any real symmetric matrix it should be possible to write it in this form. However, my understanding was also that it had to have n linear independent eigenvectors which meant no repeated eigenvalues. Which way is it?
With repeated eigenvalues, say of multiplicity m (m = 2 in your case) you can have: (i) m linearly independent eigenvectors for that eigenvalue; or (ii) fewer than m linearly independent eigenvectors for that eigenvalue. In your case you have a 3x3 matrix with 3 linearly independent eigenvectors, so your matrix is diagonalizable.

An nxn matrix with fewer than n linearly eigenvectors is termed *defective*. Of course, if all n eigenvalues are different the matrix is not defective; defectiveness is possible only for a matrix with repeated eigenvalues, and even then the issue of whether or not it IS defective depends on other details about the matrix.

RGV
 
  • #3
vela
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However, my understanding was also that it had to have n linear independent eigenvectors which meant no repeated eigenvalues. Which way is it?
You can see this idea is wrong by considering the nxn identity matrix. Every non-zero vector is an eigenvector, so you can obviously find n linearly independent vectors. Yet all the eigenvalues are equal to 1.
 
  • #4
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Okay, I see my error in that regard. The book I have been studying (Linear Algebra and its app. Strang) repeatedly states that if a matrix has no repeated eigenvalues then its eigenvectors are independent. I was stupidly reading this as iff.

But the real problem was that the eigenvectors weren't orthogonal, (-1,0,1) and (-1, 1, 0) specifically. My understanding here was that any symmetric matrix can be diagonalized using orthogonal matrices. I'm aware that the development in my text assumes unique eigenvalues, but the theorem explicitly states any symmetric matrix...?
 
  • #5
vela
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If you have two vectors corresponding to the same eigenvalue, any linear combination of them will also be an eigenvector with the same eigenvalue, so you can find linear combinations so that you end up with two orthogonal eigenvectors.
 

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