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Eigenvectors of 3x3 matrices watch

1. Hey

I was wondering about the following question(sorry i don't know how to do the brackets)

Given matrice
A= -2 2 -3
2 1 -6
-1 -2 0
Given that A squared- 5A = 0 find an eigen vector of A.

I know how to find an eigenvector but how does the Asquared - 5A come into it-

Thanks.
2. I'm not too sure, but maybe find your eigenvector as normal, so it would be of the form of

lambda*(a,b,c)T (i.e. a column vector with elements a, b and c)

And work out A2 - 5A, equate to 0 to get another expression?

Remember that eigenvectors aren't unique since they can be multiplied by any scalar lambda. So the question probably wants you to find one specific eigenvector that works with the given equation too.
3. (Original post by roxy potter)
Hey

I was wondering about the following question(sorry i don't know how to do the brackets)

Given matrice
A= -2 2 -3
2 1 -6
-1 -2 0
Given that A squared- 5A = 0 find an eigen vector of A.
But A^2 - 5A doesn't equal 0!

If it did equal 0, you could say that for any vector x,

(A^2 - 5A)x = 0
A^2 x = 5Ax
A(Ax) = 5(Ax)
Ax is an eigenvector of A with eigenvalue 5

Please check that you have typed A correctly.
4. That was part b of a question and the matrix A was given in part a of the question so maybe it was my fault and they didn't want you to use A from the earlier bit( I assumed they would)

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