Projection identity

Question

Let A be a $n\times n$ real symmetric matrix. Let $\lambda_1$ and $\lambda_2$ have multiplicity 2 and 3, respectively, with eigenvectors $u_1,u_2$ and $v_1,v_2,v_3$, respectively. Define the subspace $$V=\operatorname{span} \left\{ \alpha_1 u_1+\alpha_2 u_2, \beta_1 v_1+ \beta_2 v_2+ \beta_3 v_3 \right\}$$ for some fixed real $\alpha$'s and $\beta$'s. Then for the orthogonal projection matrix $P$ onto $V$ and any $x\in V$ it holds $PAx=Ax$.

It seems to be the case, tough I never saw it anywhere else. Can anyone prove this property?

Answer 1

Indeed, let $x=t(\alpha_1 u_1+\alpha_2 u_2)+s(\beta_1 v_1+ \beta_2 v_2+ \beta_3 v_3) $.

$Ax=t(\alpha_1 Au_1+\alpha_2 Au_2)+s(\beta_1 Av_1+ \beta_2 Av_2+ \beta_3 Av_3) = \lambda_1 t(\alpha_1 u_1+\alpha_2 u_2)+\lambda_2 s(\beta_1 v_1+ \beta_2 v_2+ \beta_3 v_3) \in V$.

Thus, $PAx\in V$, which implies $PAx=Ax$.

Answer 2

This is true. Hints to prove it:

1. Show or recall that for $\mathbf{v}\in\Bbb{R}^{n}$ and $W$ any subspace of $\Bbb{R}^{n}$, if $P$ is the matrix for orthogonal projection onto $W$, we have $P\mathbf{v}=\mathbf{v}$ if and only if $\mathbf{v}\in W$.

2. Show that if $\mathbf{x}\in V$, then $A\mathbf{x}\in V$.