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I have two questions about linear algebra. I was doing a exercise that says: Let $A$ be a real symmetric matrix $3 \times 3$ and $\det A = 6 $. Suppose that $u =(4,8,-1)$ and $v=(1,0,4)$ are eigenvectors of $A$ and $1$ and $2$ the eigenvalues associated respectively.

My first conclusion about $\det A=6\neq 0$ is the matrix $A$ is full rank so its rows are L.I (Linear independent), I was trying to figure out a way to link this result with the number of eigenvalues of matrix $A$, because the first question says:

1) The eigenvalues of $A$ are only $1$ and $2$?

If my first conclusion is wrong(i'm not sure), how to use the information from the statement to conclude that is false.

2) The cross product $u \times v$ is necessarily a eigenvector ?

For this one, I've calculated the cross product that result $(32, -17, -8)$, but I don't how to follow from this and prove that.

Thanks for help!

Widawensen
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Kutz
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    Also remember that the determinant is equal to the product of the eigenvalues – Triatticus Sep 27 '17 at 00:46
  • I didn't know it! But i've read a little in: https://math.stackexchange.com/questions/507641/show-that-the-determinant-of-a-is-equal-to-the-product-of-its-eigenvalues

    So the first question is answered by this fact!

     – Kutz Sep 27 '17 at 01:04
  • Yes, since the matrix is of full rank, there is only one possibility for the last eigenvalue. Of course the eigenvectors need not be completely orthogonal unless the matrix is symmetric – Triatticus Sep 27 '17 at 01:08
  • if matrix A wasn't full rank, is there other possibility? – Kutz Sep 27 '17 at 01:13
  • Well it might mean a zero eigenvalue or a repeated one – Triatticus Sep 27 '17 at 01:45
  • You could try going back to basics for part b: multiply $u\times v$ by $A$ and see what you get. – amd Sep 27 '17 at 02:42
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    On the other hand, it’s a basic fact about real symmetric matrices that eigenvectors corresponding to different eigenvalues are orthogonal. – amd Sep 27 '17 at 07:18
  • I can't multiply $u \times v$ by $A$ because I don't know $A$, I just know that is symmetric. Is a orthogonal vector in relation the eigenvectors a eigenvector too – Kutz Sep 27 '17 at 18:10

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Since $A$ is a real symmetric matrix, it is diagonalizable and there is an orthogonal basis of $\Bbb R^3$ which consists of eigenvalues of $A$. So, automatically you have that $(4,8,1)\times(1,0,-4)$ is an eigenvector of $A$. And the corresponding eigenvalue has to be $3$, since the product of the eigenvalues has to be $\det A$, which is $6$.

José Carlos Santos
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