Will a 2 by 2 quadratic form be negative definitive if it has repeated eigenvalues which are negative?

Question

Say we have the quadratic form $$ f = x^T Q x \\ Q = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$$ which has repeated eigenvalues $\lambda = -1$. Will the quadratic form be negative definitive? If yes then why?

Answer 1

Let any vector $x \in \mathbb{R}^2$ such that $\|x\|_2> 0$then $$x^TQx = \langle x, Qx\rangle = \langle x, -I_2x\rangle = \langle x,-x\rangle = -\langle x,x\rangle=- \|x\|^2_2<0,$$ where $I_2$ is the $2 \times 2$ identity matrix, $\|\cdot\|_2$ the euclidian norm and $\langle \cdot , \cdot \rangle$ the euclidian inner product. It follows that the matrix $Q$ is negative definite.

Note that from the definition of negative definite matrix proposed by Wolfram, just the fact that the eigenvalues are $-1$ is enough to conclude. However if you use the definition of Wikipedia then you have to do the above computation to show that $Q$ is negative definite.