# Invariant subspaces using matrix of linear operator

I am attempting the following problem but stuck at some parts:

How does one find the (2 dimensional) subspaces that are invariant under $A$ for $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 &2 & 0\\ 0 & 0 & 3\\ \end{pmatrix}\ \in M_{3} (\mathbb{R}).$$

Solution:

I found the 1-dimensional subspaces: They are just the span of individual eigenvectors

2-d subspaces: I know we need to satisfy $\mathbf{W}=\{\alpha\mathbf{w}_1 +\beta\mathbf{w}_2\mid\alpha,\beta\in\mathbb{R}\}$. Such a subspace is invariant if and only if $A\mathbf{w}_1\in\mathbf{W}$ and and $A\mathbf{w}_2\in\mathbf{W}_2$. So, does that mean it is the span of 2 eigenvectors?

3-d subspaces: $\mathbb{R^3}$

Also, out of curiosity if I had a $4\times4$ diagonal matrix would it 3-d invariant subspace be the span of three eigenvectors?

You have the right idea. The $k$-dimensional invariant subspaces of a diagonalizable linear operator can be found by taking the span of any $k$ eigenvectors.
Note that in this particular case, the eigenvectors are $\pmatrix{1&0&0}^T,\pmatrix{0&1&0}^T$, and $\pmatrix{0&0&1}^T$.
• If $A$ were not diagonalizable, then we would not have a basis of eigenvectors, which means we can no longer find all invariant subspaces in the same way. However, the span of any of $A$'s eigenvectors still forms an invariant subspace. – Ben Grossmann Mar 9 '14 at 3:38
• If you include complex eigenvectors, then $A$ can only fail to be diagonalizable if we have generalized eigenvectors, in which case the generalized eigenspaces are the invariant subspaces. $$\,$$ If you don't include complex eigenvectors, then you could end up with something like $$A = \pmatrix{1&-1\\1&1}$$ Where we don't necessarily have $1$-dimensional subspaces, since we don't have any eigenvectors. – Ben Grossmann Mar 9 '14 at 3:41