Orthogonal projection on an eigenspace

1.9k Views Asked by At

For different symmetric matrix $A$ ( eigenspaces are orthogonals ), I've noticed that when I build the matrix of the orthogonal projection of an eigenspace I get identity. Are my calculation wrong or is it always true ? For example with $$A=\begin{pmatrix} 1 &2 \\ 2& 1 \end{pmatrix}$$ We get the two eigenvalues are $3$ and $-1$ for two normalized eigenvector respectively $\displaystyle e_1=\frac{1}{\sqrt{2}}\left(1,1\right)$ and $\displaystyle e_2=\frac{1}{\sqrt{2}}\left(1,-1\right)$ Then the orthogonal projection $p(x)$ of $x=\left(a,b\right) \in \mathbb{R}^2$ is $$ p\left(x\right)=\langle x,e_1\rangle e_1+\langle x,e_2\rangle e_2=\left(a,b\right)=x $$

1

There are 1 best solutions below

1
On

Your $p$ is not the orthogonal projection onto an eigenspace, it is the orthogonal projection onto the span of both eigenspaces. The two eigenspaces span $\Bbb R^2$, so of course that projection is the identity, being the orthogonal projection from $\Bbb R^2$ onto itself.

If $A$ is any symmetric $n\times n$ matrix then the eigenspaces of $A$ span $\Bbb R^n$.