Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $T$ is a linear operator on some vector space $V$, and suppose $U$ is a $T$-invariant subspace of $V$. Does there necessarily exist a complement (a subspace $U^c$ such that $V=U\oplus U^c$) in $V$ which is also $T$-invariant?

I'm curious because I'm wondering if, given such $U$, it is always possible to decompose the linear operator $T$ into the sum of its restrictions onto $U$ and $U^c$, but I don't know if such a $T$-invariant $U^c$ exists.

share|cite|improve this question
See Proving a diagonal matrix exists for linear operators with complemented invariant subspaces, which shows that (over $\Bbb C$) this can only be expected to be true is $T$ is diagonalisable. – Marc van Leeuwen Jan 12 '15 at 9:47
up vote 5 down vote accepted

No. For example consider $\begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}$ as a linear map from $\mathbb{R}^2$ to itself, and the $T$-invariant subspace generated by $\begin{bmatrix}0 \\ 1 \end{bmatrix}$. If there is a complement, it must have some element of the form $\begin{bmatrix} a \\ b \end{bmatrix}$ with $a \neq 0$, but then apply $T$, you see that $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ also lies in that subspace, which means the subspace is not a complement.

The point is that if a $T$-invariant subspace always has complement, this automatically implies that $T$ is always diagonalizable provided that the eigenvalues lie in the field you are working with - and you can easily find something non-diagonalizable.

share|cite|improve this answer
Makes sense, thanks Sanchez. – Hailie Mathieson Dec 14 '12 at 7:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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