0
$\begingroup$

On page 134 of the 3rd edition of Linear Algebra Done Right, Professor Axler makes the following statement (where "The comments above" refers to a brief and trivial proof of the assertion, and $T$ is a linear operator):

The comments above show that $T$ has a 1-dimensional invariant subspace if and only if $T$ has an eigenvalue.

Certainly, the statement as written is correct since, even if the eigenvalue corresponds to an invariant subspace of dimension greater than 1, we can always select one basis vector thereof and consider its span as the one-dimensional subspace in question. I'm wondering whether the following adaptation 1) is correct and 2) might be more illuminating (obviously this is somewhat a matter of opinion):

The comments above show that $T$ has an invariant subspace of at least dimension 1 if and only if $T$ has an eigenvalue.

Maybe it's just me, but I think the latter is more useful because the former seems to hint that there is only an invariant subspace of dimension precisely equal to 1 iff $T$ has an eigenvalue.

$\endgroup$
1
$\begingroup$

The first statement while confusing (I mean not technically, but I had to read a few times, because I definitely misinterpreted it on my first reading) is true, but the second is definitely false (the only if part). For example, take any linear operator on a vector space $V$ of dimension $\geq 2$. Then $V$ will be an invariant subspace, even though $T$ need not have any eigenvalues (think of a rotation in $\Bbb{R}^2$)

$\endgroup$
1
$\begingroup$

Consider $$ \pmatrix{0& 1 & 0 & 0\\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0& 1 \\ 0 & 0 & -1 & 0} $$ This has an invariant subspace of at least dimension $1$ (namely the $xy$-plane), but has no real eigenvalue. (Indeed, that's true for the upper left 2x2 block as well).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.