Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
add comment

1 Answer

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|improve this answer
    
Makes sense, thanks Sanchez. –  hmIII Dec 14 '12 at 7:16
add comment

Your Answer

 
discard

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.