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 $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 &2 & 0\\ 0 & 0 & 3\\ \end{pmatrix}\ \in M_{3} (\mathbb{R}).$$


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 $\pmatrix{0&0&1}^T$.

  • $\begingroup$ If my matrix were not diagonal. Would that make it harder (impossible) to find the invariant subspaces. Because now we can no longer user the argument about the eigenvectors as there may be repeats. I am not aware of any other argument if this were the case $\endgroup$ – user104221 Mar 9 '14 at 3:34
  • $\begingroup$ Do you mean diagonal or diagonalizable? As long as the matrix is diagonalizable, the same argument holds. $\endgroup$ – Ben Grossmann Mar 9 '14 at 3:34
  • $\begingroup$ Oh I mean if it were not diagonalizable $\endgroup$ – user104221 Mar 9 '14 at 3:35
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    $\begingroup$ 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. $\endgroup$ – Ben Grossmann Mar 9 '14 at 3:38
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    $\begingroup$ 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. $\endgroup$ – Ben Grossmann Mar 9 '14 at 3:41

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