Question

A =
[ 3 2 4 ]
[ 2 0 2 ]
[ 4 2 3 ]

show that A is diagonalizable even though one eigenvector has algebraic multiplicity 2. Do this by brute force computation. write A = Q/\Q^T where Q's columns are orthogonal unit vectors of A.

Answer 1

I'm embarrassed to say I did this the hard way. I solved \[|A-\lambda I| = 0\] which involved factoring a cubic function.
\[\Lambda _{1} = 8, \Lambda _{2} = -1 , \Lambda _{3} = -1.\] Then once I found the repeated eigenvalue, I substituted it for lambda. The matrix\[A - \lambda I\] was only rank 1, so there were two different eigenvectors for the repeated lambda, which made three linearly independent eigenvectors in total! Which means A is diagonalizable. I would however appreciate someone showing how to factor
\[A =Q \Lambda Q^{T}\] as I am only really familiar with the \[A = S \Lambda S ^{-1}\] factorization, and I feel there must be an easier way.

Answer 2

Thank you!

Answer 3

@datanewb FYI
http://openstudy.com/study#/updates/50037f89e4b0848ddd694f98

Answer 4

ah, thanks. That is a good thread. I now see that S is basically Q, almost. :) I think Strang talks about Q in the next section I'm about to read anyhow. I thought there might be an easier way to calculate lambda with a symmetric matrix.

One correction in the way the question in this thread is stated: It says "one eigenvector has algebraic multiplicity 2," I believe it should be eigenvalue instead of eigenvector.