For $A$ positive semidefinite and $\theta_1 ^{T} \theta_2 = 0$ is it true that $\theta_1^{T} A \theta_1 + 2 \theta_1^{T} A \theta_2 \geq 0$?

Question

I have been trying to prove or disprove that for $\theta_1, \theta_2 \in \mathbb{R}^P$ such that $\theta_1 ^{T} \theta_2 = 0$ we have the following inequality

$$\theta_1^{T} A \theta_1 + 2 \theta_1^{T} A \theta_2 \geq 0, $$

for some positive semidefinite matrix $A \in \mathbb{R}^{p \times p}$. I have been experimenting with the Cauchy-Schwartz inequality and its variants but I haven't been able to prove the assertion. This makes me believe that the claim is false but I don't have any counterexamples at hand. Could I please then get some help?

Answer 1

This is true if both vectors are orthogonal eigenvectors, then $v_1^TAv_2 = \lambda v_1^Tv_2=0$.

Here is a counterexample: Let $$ A=\pmatrix{1&0\\0&0}, v = \pmatrix{1\\1}, w = \pmatrix{-1\\1} $$ then $$ 0=(v+w)^TA(v+w) = v^TAv + 2 v^TAw + w^Tw, $$ since $w\ne 0$, it follows $v^TAv + 2 v^TAw<0$.