Is $u^TAu \geq 0$ true for all symmetric matrices $A$?

Question

we know from the definition of inner product and norm, that $u^Tu$ is always larger than zero, except the case where $u=0$ at which case it is zero.

I came across a question that infers that $u^TAu \geq 0$ for all symmetric matrices $A$, I'd like to know why this is true, it seems odd to me.

The question asks us to prove an equality about some function: $f(p) =f(x_0)+\frac{1}{2}(p-x_0)^TA(p-x_0$) where $A$ is a symmetric matrix and $p$ is some point, and infer from this result that $f(x_0)$ is a global minimum of $f$.

I won't go into details, it doesn't matter, I showed that this equality indeed holds. My problem now is showing that $f(x_0)$ is a global minimum. If it is, then $f(p) \geq f(x_0)$. which would imply that $(p-x_0)^TA(p-x_0) \geq 0$ which is unclear to me why this is true for all cases of $A \in Sym(n)$.

Answer 1

No, for example $A=-I$, where $I$ is the identity matrix.

$A$ being symmetric just means that it has real eigenvalues.

$u^T Au\ge 0$ means for a symmetric matrix that $A$ is positive (semi)definite, i.e. all the eigenvalues are non-negative.

Answer 2

This is true for the non-negative symmetric matrix i.e. for the symmetric matrix with non-negative eigenvalues. In fact let $(e_1,\ldots,e_n)$ an orthonormal basis of eigenvectors of $A$ associated to the eigenvalues $\lambda_k\ge0$ and let $$u=\sum_{k=1}^n x_k e_k$$ then $$\langle Au,u\rangle=\sum_{k=1}^nx_k^2\lambda_k\ge0$$