Intersection of two invariant subspaces I need ro find all subspaces of $\mathbb{R}^3$ that are invariant simultaneously with respect to two linear transformations defined by matrices
$$ A=\begin{pmatrix}
  5& -1& -1\\
  -1& 5& -1\\           
-1&   -1& 5
\end{pmatrix}$$
and
$$B=\begin{pmatrix}
  -6& 2& 3\\
  2& -3& 6&\\           
3&   6& 2
\end{pmatrix}$$
I found it eigenvalues and eigenvectors:
for A $$\lambda_1=3, v_1=(1,1,1)^T\\ \lambda_2=\lambda_3=6, v_2=(1,-1,0)^T+(1,0,-1)^T$$
Invariant subspace:
$${R}^3, \begin{Bmatrix}{0}\end{Bmatrix},<(1,1,1)^T> and <(1,-1,0)^T+(1,0,-1)^T>$$
for B:
f$$\lambda_1=7, v_1=(1,2,3)^T\\ \lambda_2=\lambda_3=-7, v_2=(-2,1,0)^T+(-3,0,1)^T$$
Invariant subspace:
$${R}^3, \begin{Bmatrix}{0}\end{Bmatrix}<(1,2,3)^T> and <(-2,1,0)^T,(-3,0,1)^T>$$
I know, that the answer is $$  \mathbb {R}^3, \begin{Bmatrix}{0}\end{Bmatrix},<(1,-2,1)^T>, <(1,1,1)^T,(1,2,3)^T>$$
But I can't get it. I understand why there is $ \mathbb {R}^3 and \begin{Bmatrix}{0}\end{Bmatrix}$
 A: I assume that all your calculations are now correct.
To find the invariant subspaces let us look for those of each possible dimension.
We have agreed that the dimension $0$ case gives $\{0\}$ and nothing else.
We have agreed that the dimension $3$ case gives $\mathbb{R}^3$ and nothing else.
For the dimension $1$ case we are looking for a common eigenvector of $A$ and $B$: clearly it is neither a multiple of $(1,1,1)$ nor of $(1,2,3)$. So if such exists it lies in the intersection of the two $2$-dimensional eigenspaces. That is, it will be some $(x,y,z)$ such that $x-y-z=0$ and $x-2y-3z=0$: that is it is a multiple of $(1,-2,1)$.
For the dimension $2$ case let $U$ be a $2$ dimensional subspace which is invariant under both $A$ and $B$. Note that the $2$-dimensional eigenspace of $A$ is not $B$-invariant, and that the $2$-dimensional eigenspace of $B$ is not $A$-invariant.
Suppose that $U$ does not contain $(1,1,1)$; then $\mathbb{R}^3=\langle(1,1,1)\rangle\oplus U$. Then $A$ restricted to $U$ must have characteristic polynomial dividing $(X-3)(X-6)^2$, and as the eigenvalue $3$ is already accounted for on $\langle(1,1,1)\rangle$ the characteristic polynomial of $A$ restricted to $U$ is $(X-6)^2$, and so $U$ must be the $2$-dimensional eigenspace of $A$: which we know is not so. So $U$ contains $(1,1,1)$.
A similar argument shows that $U$ contains $(1,2,3)$.
All that remains now is to check that the subspace $U=\langle(1,1,1),(1,2,3)\rangle$ actually is both $A$- and $B$-invariant. This is an easy calculation.
