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Linear Algebra Question

leftrightleftrightleft

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Hey, I've got the following 3x3 matrix:

Row 1: [5, 0.6, 0]
Row 2: [0.6, 4, 0]
Row 3: [0, 0, 1]

I've found the eigenvalues to be: L1 = 1, L2 = 5.28, L3 = 3.72


I know these eigenvalues are correct from checking them on wolfram alpha.

How do you find the eigenvectors?

From wolfram, I know that the the eigenvector for L1=1 is V1 = [0 0 1], but I have no idea how they got it.

Anyone a fan of linear algebra?
 

metherion

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I don't remember much, I took diff equations about 4 years ago instead of linear algebra, but I do remember that eigenvectors remain the same when multiplied by the matrix.

So [0 0 1] vertical would work because, well, anything times 0 is zero, and due to matrix algebra, the 1 stays a 1, and when you multiply it by the eigenvalue, it stays the same.

I don't remember how to FIND other eigenvectors. All I remember is (square matrix) * vector = (eigenvalue) * vector.

So, for two, you'd have to find a vector, where, when you multiply it by the matrix, then divide/multiply (I can't remember which, sorry) by (5.28 for L2, 3.72 for L3), you wind up with the original vector.

Sorry I can't be more help, I hope that was some small help to you!

Metherion
 
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Farinata

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Matrix * eigenvector = eigenvalue * eigenvector

From this form three equations for each eigenvalue. Solve for eigenvector.

Example for eigenvalue 1:

(had to put in three equal signs for sake of formatting, should be read as 3x3 matrix hitting against a 3x1 eigenvector = eigenvalue 1 hitting against the same 3x1 eigenvector)

[5, .6, 0] [x] = [x]
[.6, 4, 0] [y] = 1* [y]
[0, 0, 1] [z] = [z]

5x + .6y +0z = x
.6x + 4y + 0z = y
0x + 0y + 1z = z

solve:

From 1st equation: 4x = -.6y , thus, 20x = -3y
From 2nd equation: .6x = -3y , thus, .6x = 20x
So x must equal 0. Therefore, from first two equations by inspection, y = 0 (5*0 +.6y = 0)
z = z is all that's left. z = 1 makes the final result an eigenvector.
 
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