Show that $C=\{x \in \mathbb{R}^4 \mid x^TAx \geq 1\}$ is not empty?

Question

Consider the set $C=\{x \in \mathbb{R}^4 \mid x^TAx \geq 1\}$ where $A \in \mathbb{R}^{4 \times 4}$ is a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.

Show $C$ is not empty?

My try:

$A$ is symmetric, so it can be written as $A=u \Lambda u^T$. Hence,

$$ x^TAx=x^Tu \Lambda u^Tx \geq 1 $$

So

$$ z^T \Lambda z \geq 1 $$ where $u^Tx=z \in \mathbb{R}^4$. Let $a_1,a_2,a_3,a_4$ be eigenvalues of $A$ so $a_1z_1^2+a_2z_2^2+a_3z_3^2+a_4z_4^2 -1 \geq 0$. Since two of the eigenvalues are nonpositive, assume they are zero. Hence, $a_1z_1^2+a_2z_2^2 -1 \geq 0$ which is the region outside of an ellipsoid in a 2D plain. Am I write, or there is a more beautiful way to show it?

Answer 1

You can't simply assume that the non-positive eigenvalues of $A$ are zero.

Note that to show that $C$ is non-empty, you only need to show that there is at least one vector in $C$.

Let $\lambda$ be the one of the positive eigenvalues of $A$, and let $v$ be a corresponding eigenvector with unit norm.

If $x = \alpha v$ for some constant $\alpha$, then how big is $x^TAx$? Can you find a large enough $\alpha$ such that $x = \alpha v \in C?$

Answer 2

Let $x \in R^4$ be the eigenvector corresponding to one of the positive eigenvalues $\lambda_1$. Then $x^TAx = \lambda_1 x^Tx > 0$. Now all you have to do is multiply $x$ by a normalizing scalar to get some $y \in R^4$ and you have $y^TAy = 1 \Rightarrow y \in C$.