# Find non-trivial invariant subspaces of a projection onto plane

I have to find the nontrivial invariant subspaces of the projection to the $$x+y+z = 0$$ plane.

These are the lines in the plane (which go through the origin), and the plane itself. Is this right?

But how can I get the answer from the matrix of the transformation.

$$A = \begin{pmatrix} \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \\ \end{pmatrix}\$$

I found the eigenvalules and eigenvectors: $$1,0$$ and $$(-1,0,1), (-1,1,0), (1,1,1)$$.

How can i find the invariant subspaces?

For a linear transformation $$A$$ and its eigenvector $$x$$ and non-zero eigenvalue $$\alpha$$ we have $$Ax=\alpha x$$. The transformation maps $$x$$ onto a multiple of itself, so the space of all multiples of $$x$$ is invariant.
If zero is an eigenvalue, then for its eigenvector $$Ay=0$$. In your example the entire 3d space is mapped onto a 2-d space (a plane), containing the other two eigenvectors. The plane does not include $$(1,1,1)$$, since that has the zero eigenvalue. It does include the other two vectors.
All the vectors $$v$$ in an invariant subspace satisfies $$Av = v$$. As a consequence, a subspace is invariant if and only if it is contained in the eigenspace relative to the eigenvalue $$1$$.