General Information about Eigenvalues for an 3x3 symmetric matrix How do we find general information about the eigenvalues of an arbitrary 3x3 symmetric matrix without resorting to explicitly computing the solutions to the cubic characteristic equation?:
\begin{bmatrix}
    a   & b & c  \\
    b    & d & e \\
    c & e &f
\end{bmatrix}
(In this case the sum and product of the eigenvalues, and the sum of the reciprocals of the eigenvalues.) Since diagonalization preserves the trace and determinant of matrices, would it be best to just find the corresponding diagonal matrix and read off the sum and product of the eigenvalues from there? The problem is I can't think of a way to diagonalize a matrix without explicitly computing the eigenvalues.
 A: For a 3x3 real symmetric matrix, a single step of Householder reduction immediately and exactly brings it to tri-diagonal form without changing the eigenvalues.  (See Numerical recipes, chapter 11 for discussion of all these arcane terms.)  That means we can easily reduce the problem to finding the eigenvalues of a matrix of the form
$$\left( \begin{array}{ccc} \alpha & \beta & 0 \\
\beta & \delta & \epsilon \\
0& \epsilon &\phi \end{array} \right)$$
Next, the QR method, which consists of a series of orthogonal transformations found by decomposing the matrix into an orthogonal matrix ($Q)$ times an upper triangular matrix $(R)$ iteratively finds the eigenvalues of the tri-diagonal matrix. The QR algorithm with eigenvalue shifts based on the smaller eigenvalue of the upper left 2x2 submatrix cunverges very rapidly.  This is a very efficient (though approximate) method for finding the eigenvalues.
A: If A is symmetric and has real entries then you know that all roots of its characteristic polynomial are real as well. That is, all eigenvalues will be real.
$\mathbf{Proof:}$  Let $v$ be an eigenvector associated with a eigenvalue of A, $\lambda.$
Then we have $Av=\lambda v$
take the complex conjugate of both sides to obtain 
$ \bar A \bar v= \bar \lambda \bar v $
And we know A is real, so $\bar A= A$
First notice, $  (A \bar v) \bullet v= (\bar \lambda \bar v) \bullet v $
$(A \bar v)^{T} v= (\bar \lambda \bar v)^{T} v$
$$\rightarrow \bar v^{T} A v= \bar \lambda \bar v^{T} v$$ (1)
Now,we also have $ \bar v \bullet A  v = \bar v \bullet \lambda v  $
that is $$\bar v^{T} A v= \bar v^{T} \lambda v$$
(2)
So we can now equate (1) and (2) as $\bar v^{T} \lambda v=\bar \lambda \bar v^{T} v$
$\rightarrow$ $\bar v^{T} \lambda v-\bar \lambda \bar v^{T} v=0$
($\lambda- \bar \lambda$)($\bar v^{T} v)=0$
$\bar v^{T} v \ne 0$  because these are eigenvectors so at least one of its entries will be non zero, so the norm will be non zero as well
thus $\lambda= \bar \lambda$ that is , it is real. $QED$
