How do I find the eigenvalues and eigenvectors of a given matrix?

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The discussion revolves around finding the eigenvalues and eigenvectors of a specified matrix. The eigenvalues identified are -5, 5/2 + √(5/2), and 5/2 - √(5/2). The participant initially struggles with deriving the eigenvectors, mistakenly concluding that all components equal zero. Clarifications reveal that the equations for the eigenvectors must account for free variables, leading to valid eigenvector solutions. Additionally, there is a discussion about transforming the strain rate tensor using the eigenvalues and eigenvectors, emphasizing the need for the inverse of the eigenvector matrix for proper diagonalization.
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Homework Statement



Find the eigenvalues and eigenvectors of this matrix.

[4 0 1/2 ]
[0 -5 0 ]
[1/2 0 1 ]

3. The Attempt at a Solution

I have found the eigenvalues = -5, 5/2 + sqrt(5/2), 5/2 - sqrt(5/2)

I know to get the eigenvectors you subtract the eigenvalues from thediagonal and set each row equal to zero and solve for the 3 components.

for -5

[9 0 1/2]
[0 0 0]
[1/2 0 -5]

so 9x +1/2 z = 0
1/2x -5z=0

and y =0

but i get zero for all components even when i do the other eigenvalues. can anyone show me what i am doing wrong?
 
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Hi LoopQG! :smile:
LoopQG said:
[9 0 1/2]
[0 0 0]
[1/2 0 -5]

so 9x +1/2 z = 0
1/2x -5z=0

and y =0

nooo :redface:

your equations don't mention y, so y can be anything!

(and isn't it obvious that the eigenvectors must be [0,y,0]? :wink:)
 
Yes but what about the vectors for the other values. I still get all the components equal zero. I have plugged it into MATLAB and i know what the other eigenvectors are but i don't see how to get them analytically
 
For the first and third rows, for eigenvalue -5, you got two different equations for x and z, so they could only be satisfied by x = z = 0.

But for either of the other eigenvalues, the equations for the first and third rows should be the same (hint: (3 + √10)(3 - √10) = -1 :wink:).
 
ok so after figuring out i had to normalize i first chose x=1 so z=-3 +sqrt(10) for 5/2 + sqrt(5/2)

and for 5/2 - sqrt(5/2) i set z=1 and x=3-sqrt(10)

and y=0 for both

can you verify if this is correct? thank you for all the guidance.
 
Hi LoopQG! :smile:

Yes, (1, 0, -3+√10) and (3-√10, 0, 1) look fine.

But there are other ways which you might think are neater …

eg √10 > 3, so you could write them (1, 0, √10 - 3) and (√10 - 3, 0, -1)

or you could even use (√10 - 3)(√10 + 3) = 1, and start both of them with x = ±1. :wink:
 
excellent. hanks so much for your help. I do have one more question though.

The idea behind this was to find the principal axes of the strain rate tensor, which is the matrix in post 1. And find the principal rates of strain. For the principal rates of strain i am getting from my text that I should get a diagonal 3x3 with the eigenvalues on the diagonal.

the strain tensor should trans form like E(principal rate of strain) = C E(original strain tensor) C (transpose)

where C is the matrix of all 3 eigenvectors.

When I use the eigenvectors i have calculated i do not get a diagonal matrix, is there a certain method i should use for normalizing the eigenvectors to get the principal axes or am I just doing the algebra wrong.

Thanks again for all the clarification this has been a good refresher in linear algebra.
 
Hi LoopQG! :smile:
LoopQG said:
… For the principal rates of strain i am getting from my text that I should get a diagonal 3x3 with the eigenvalues on the diagonal.

the strain tensor should trans form like E(principal rate of strain) = C E(original strain tensor) C (transpose)

where C is the matrix of all 3 eigenvectors.

ah, no, it's not CT, it's C-1

see http://en.wikipedia.org/wiki/Jordan_normal_form#Example :wink:
 
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