0
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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 $1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .