The $k$- dimensional invariant subspace of a diagonalizable linear operator can be found by taking the span of any $k$ eigenvectors.
I'm having some difficulties dealing with the following problem: Suppose $f:\Bbb R^{3\times 1} \to \Bbb R^{3\times 1}, f(X)=AX$, where
$$ A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix} $$ And we're asked to find all the $f$-invariant subspaces.
So eigenvalues: $1,1,2$ and eigenvectors: $V_A(1)\setminus \{0\}=\{ae_1+be_2:a,b \in \Bbb R, ab \ne 0\}$ $V_A(2)\setminus \{0\}=\{ce_3:c \in \Bbb R \setminus \{0\}\}$
We know $\{0_{\Bbb R^{3 \times 1}}\}, \Bbb R^{3\times 1}$ are $f$-invariant and have dimensions $0$ and $3$ respectively.
As for the rest dimensions, I gave as an answer:
for $1:$ $span\{λe_1+μe_2\},span\{e_3\},span\{e_1\},span\{e_2\}$
for $2:$ $span\{e_1,e_2\},span\{e_1,e_3\},span\{e_2,e_3\},span\{λe_1+μe_2,e_2\},span\{λe_1+μe_2,e_3\}$
But this answer is false as I include some subspaces that are not invariant. It is not very clear to me wich subspaces are false and why..
Also is the first statement in the box correct? Thanks in advance!
A subspace is said to be invariant if $f(U) = U$. We should note that if $\vec{x}$ is an eigenvector then $f(\vec{x})=\lambda \vec{x} \in \mathrm{span}(x)$ so every eigenspace of $f$ is also $f$ invariant.
Hence, we if you have all the eigenspaces, which are the span of $e_1,e_2,e_3$, then you know that your vector space can be written as $k e_1 \oplus k e_2 \oplus k e_3$, where $k$ is a field, and in particular that eigenspaces are minimal invariant subspaces with respect to inclusion.
In $1$, $\mathrm{span}\{ e_1+ e_2\}$ is invariant, since $A$ applied to any vector with a $0$ in the third co-ordinate lands you in the same subspace, it is just not a great one, because it decomposes further as the direct sum of two eigenspaces.