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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}$

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  • $\begingroup$ Do you know Jordan normal form? $\endgroup$ Jun 12, 2020 at 14:04
  • $\begingroup$ @FilippoGiovagnini Yes, I know $\endgroup$
    – cirmish
    Jun 12, 2020 at 14:11
  • $\begingroup$ Every linear operator on $\mathbb{R}^3$ fixes $\{0\}$ and fixes $\mathbb{R}^3$ so these are always invariant subspaces. So your two lists of the invariant subspaces of $A$ and of $B$ are both missing these (rather trivial) ones. $\endgroup$ Jun 12, 2020 at 14:14
  • $\begingroup$ @ancientmathematician You're right. I forgot to write them. But I know they should be there $\endgroup$
    – cirmish
    Jun 12, 2020 at 14:18
  • $\begingroup$ Your eigenvectors for $A$ are wrong: try $(1,-1,0)$ and $(1,0,-1)$. $\endgroup$ Jun 12, 2020 at 14:19

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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.

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