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manw1992's picture

I have problems finding the eigenvectors of a 3 x 3 matrix. Can anyone help?

AA's picture

Say matrix A is 3x3. Let g be the eigenvalue, i.e. a scalar parameter, associated with the matrix. Let v be the 3x1 non-zero eigenvector associated with this eigenvalue. Then A.v=g.v --> (A-g.I)v=0 --> I is the 3x3 unit matrix. g exists if (A-g.I) is non-invertible, which happens only if its determinant is 0 i.e. det(A-g.I)=0. So calculating the determinant of this (that is you subtract g as an unknown scalar from each diagonal entry of A and calculate the determinant of the result), and setting it equal to 0 gives you an equation in one unknown, g, which you can solve (the characteristic equation). It can give multiple solutions for g, all of which are eigenvalues of the matrix A. After you do that, you solve the matrix equation (A-g.I).v=0 for each eigenvalue g you found, which will yield one eigenvector v (with parameters v1,v2,v3 which you can find) defined as a vector times a scalar for each eigenvalue.

For more details and examples see Khan's lectures in Linear Algebra 132-137 and especially lecture 137 http://www.theopenacademy.com/content/lecture-137-linear-algebra-eigenve....

Hope that helped.

manw1992's picture
Thank you

Thanks that makes sense now.

AmySteger's picture
If you still don't have

If you still don't have answer to this problem I suggest you to visit StudyGeek.org they can help you a lot.

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Andrea's picture
This content is written very
htmlcolor's picture
Interesting, I wasn't aware

Interesting, I wasn't aware of that context. I guess it makes some sense, thanks for the history lesson!


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