Shortcut to finding invariant subspaces

Question

I have looked at the following example:

$M = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$

and found

$ker(M-1Id)=\langle \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\rangle$

and

$ker(M+1Id)=\langle \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\rangle$

and

$ker(M+1Id)^2=\langle \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\rangle$

Using this simple example, I have the following hypothesis:

The invariant subspaces are any combination (i.e. direct sum) of eigenvectors, so any combination of subspaces between $ker(M-1Id)$ and $ker(M+1Id)$. Moreover, the respective "subspaces" from the canonical decomposition of $M$ (i.e. $ker(M-1Id)\oplus ker(M+1Id)^2$) are invariant subspaces, so $ker(M-1Id)$ and $ker(M+1Id)^{2}$ and then of course $0$ and $\mathbb R^{3}$.

This is the image that has emerged from stubborn calculations. Is this correct, any support is greatly appreciated

Answer 1

Indeed, those are invariant subspaces. What you're working is Jordan's theorem (also search Jordan canonical form). On $\mathbb{C}$ you can always find that decomposition in invariant subspaces (in real numbers you can't be sure about Jordan form for every matrix because eigenvalues might be complex)

Reference: Linear Algebra. Hoffman - Kunze (or any other linear algebra textbook)