I'm given a matrix a 3x3 matrix $A $ , where I need to find a vector x such that $x'Ax $ it gives me a negative number (or x'Ax < 0). I was thinking I have a vector x set to x = [a, b, c] and perform gaussian elimination but not sure where to go from there.
Any tips on how to approach this would be great!
First of all, find the eigenvalues $\lambda_k$ and eigenvectors $V_k$ associated with them. Two cases :
(a) either all $\lambda_k$s are $\geq 0$ , in such a case, you will never find an $X$ such that $X^TAX <0$.
(b) or one at least of the eigenvalues, say $\lambda_1$ is $<0$, in this case, take $X=V_1$. Indeed, $X^TAX=V_1^T(AV_1)=V_1^T(\lambda_1 V_1)$ (by definition of an eigenvector), which gives $\lambda_1 \|V_1\|^2 < 0$. That's all.
Considering the alternative you give in your title (find a way to make $A$ "no longer Symmetric Positive Definite"), you just have to "shift left" the spectrum of $A$. For example, if Spec$(A)=\{2,3,7\}$, replace $A$ by, say, $A-2.001 I_3$, the spectrum of the new matrix will be shifted by $-2$ , i.e., will become$\{-0.001,0.999,4.999\}$, and we are now in case (b)