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!
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 $6$.