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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.

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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.

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Makes sense, thanks Sanchez. – Hailie Mathieson Dec 14 '12 at 7:16

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