Find a canonical form through orthogonal transformation I need to find a canonical form through orthogonal transformation, the problem is, that the equation given to me doesn't make sense: $g=2x_1x_2-6x_1x_3-6x_2x_4+2x_3x_4$. The matrix for this form is: $$\left[\begin{matrix}0 & 1 & -3 & 0 \\ 1 & 0 & 0 & -3 \\ -3 & 0 & 0 & 1 \\ 0 & -3 & 1 & 0\end{matrix}\right]$$ So, the way to find eigenvalues would be to solve $\lambda^3-s_1\lambda^2+s_2-s_3=0$ where $s_{1}=$ sum of elements on the diagonal, $s_2=$ the sum of diagonal minors and $s_3$ is a determinant. (I am not sure if there is 4 variables whether there are four $s$?) So accordingly $s_1=0, s_2=3, s_3=64$. So the equation to find eigenvalues looks like this: $\lambda^3+3\lambda-64=0$. How to solve equation like this? Maybe for four variables there is different equation to find eigenvalues?
 A: You solve cubic equations with Cardano's method. In this case you have one irrational real root, plus two non-real complex roots.
However, as it's a $4\times 4$ matrix the characteristic polynomial is actually a quartic, namely $\lambda^4-20\lambda^2+64=(\lambda^2-4)(\lambda^2-16)$, with roots $\pm2,\,\pm4$. Once you know about Cardano's method, you can learn quartic-solving methods too.
A: For a square matrix $M$ of any size, the characteristic polynomial is $\det(\lambda I-M)$, which will be of degree $n$ if the matrix is $n\times n$. So, the polynomial that you’ve come up with for this matrix can’t be correct. As another answer points out, the correct characteristic polynomial is $\lambda^4-20\lambda^2+64$, the roots of which you can find with a few applications of the quadratic formula.  
However, eigenvalues and eigenvectors of this matrix can also be found by inspection. Notice first that all of the row sums are equal. This means that $(1,1,1,1)^T$ is an eigenvector with eigenvalue equal to the row sum, $-2$. The trace of the matrix is equal to $0$, so there must be at least one other distinct eigenvalue and since the matrix is symmetric, its eigenvectors must be orthogonal to this first one, which narrows the search quite a bit. When you have columns that differ only in the signs of their entries, vectors that involve differences of these columns are often eigenvectors, and in fact, we have $M(1,1,-1,-1)^T = (4,4,-4,-4)^T$, so this is an eigenvector with eigenvalue $4$. This might also suggest trying $(1,-1,1,-1)^T$, and sure enough it also turns out to be an eigenvector with eigenvalue $-4$. The last eigenvalue can always be obtained “for free” from the trace, and it is $2$. For a corresponding eigenvector, choose anything that’s orthogonal to all three of the vectors already obtained or compute it using the usual methods. Once you have the four eigenvectors, normalize them for an orthogonal diagonalization.
