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Let $V$ be a complex linear space of dimension $n$. Let $T \in End(V)$ such that $T$ is diagonalisable. Prove that each $T$-invariant subspace $W$ of $V$ has a complementary $T$-invariant subspace $W'$ such that $V= W \oplus W'$.

Note: Let $\{e_1,...e_n\}$ be the set of eigenvectors together with eigenspaces $V_{\lambda_1},...V_{\lambda_n}$ of $T$. It's sufficient to show that every $T$-invariant subspace $W$ must be a direct sum of eigenspaces, then it'll be trivial to find $W'$ (just take the rest eigenspaces not in the direct sum and glue them to $W$).. But how to prove $W$ is a direct sum of eigenspaces?

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I believe eigenvectors span the whole space because $T$ is diagonalizable. Isn't this enough to show that $V$ is a direct sum of one-dimensional spaces generated by eigenvectors? –  Tunococ Sep 6 '12 at 3:41
    
Decompose $T$ as a $C[x]$ module using the structure theorem for modules over a PID, and notice that $T$-invariant subspaces correspond to $C[x]$-submodules. Since you're working over the complex numbers and $T$ is diagonalizable, the $C[x]$ submodules will be direct sums of the (really nice) elementary divisors. –  jmracek Sep 6 '12 at 3:41
    
Your note is confusing: there is no such thing as the set of eigenvectors. You might means some basis of eigenvectors, but be aware: there might be fewer eigenspaces than vectors in a basis of eigenvectors, so using $n$ to number each of these is not possible. –  Marc van Leeuwen Feb 8 at 13:08
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3 Answers

Suppose $\{ e_i \}$ is an eigenbasis for $T$. Furthermore, suppose $W$ is a $k$-dimensional subspace for which $T(W) \subseteq W$. Consider $e_j$. If $e_j \in W$ then $T(e_j) = \lambda_j e_j$. It follows that $e_j \in W$ places no further conditions on $W$. The fact that it is in $W$ is consistent with the $T$-invariance of $W$. Then continue, sift through values of $j$ until you find $k$ of the eigenvectors inside $W$. Once you have $k$-linearly independent vectors then you know that no more can be found inside $W$. The e-vectors are taken from a basis so you have linear independence.

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I think this is insufficient. You cannot simply claim that $k$ of the eigenbasis vectors lie inside $W$. –  EuYu Sep 6 '12 at 3:53
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That's not what I'm saying from the outset. My argument is that a e-vector is either in $W$ or it isn't. If it is then we can use it for a basis for $W$. Then, go on to the next e-vector and once more find it's either in $W$ or out. If it's in then put it in the basis for $W$... continue. Eventually this terminates because $W$ has finite dimension like $V$. –  James S. Cook Sep 6 '12 at 4:06
    
That still doesn't show that $W$ is a direct sum of eigenspaces. The question essentially asks to show that $T\mid_W$ is diagonalizable. What you show doesn't seem to cover that. –  EuYu Sep 6 '12 at 4:10
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up vote 1 down vote accepted

Based on the hint $W=(W \cap V_{\lambda1}) \oplus...\oplus(W \cap V_{\lambda_s})$ where $\{\lambda_1,...\lambda_s\}$ is the set of eigenvalues one way to show it is as follows:

We can prove the following theorem: If $v_1 + v_2 + \cdots + v_k \in W$ and each of the $v_i$ are eigenvectors of $A$ with distinct eigenvalues, each of the $v_i$ lie in $W$.

Proof: Proceed by induction. If $k = 1$ there is nothing to prove. Otherwise, let $w = v_1 + \cdots + v_k$, and $\lambda_i$ be the eigenvalue corresponding to $v_i$. Then:

$$Aw - \lambda_1w = (\lambda_2 - \lambda_1)v_2 + \cdots + (\lambda_k - \lambda_1)v_k \in W$$

By induction, $(\lambda_i - \lambda_1)v_i \in W$, and since the eigenvalues $\lambda_i$ are distinct, $v_i \in W$ for $2 \leq i \leq k$, then we also have $v_1 \in W \quad \square$

Now each $w \in W$ can be written as a finite sum of nonzero eigenvectors of $A$ with distinct eigenvalues, and by the theorem these eigenvectors lie in $W$.Then we have $W = \bigoplus_{\lambda \in F}(W \cap V_{\lambda})$ as desired (where $V_{\lambda} = \{v \in V\mid Av = \lambda v\}$).

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I will suppose there are $k$ distict eigenvalues $\lambda_1,\ldots,\lambda_k$.

Since $T$ is diagonalisable, it has a minimal polynomial $\mu_T$ that is split with simple roots; indeed one has $\mu=(X-\lambda_1)\ldots(X-\lambda_k)$. Since for the restriction $T|_W$ of$~T$ to$~W$ one certainly has $\mu[T|_W]=0$, this restriction is also diagonalisable, with its eigenvalues among $\{\lambda_1,\ldots,\lambda_k\}$, and each eigenspace of $T|_W$ for some$~\lambda_i$ is a subspace of the eigenspace of$~T$ for$~\lambda_i$. It now suffices to choose in each eigenspace of$~T$ a complementary subspace to the eigenspace of$~T|_W$, or the whole space (a complement of $\{0\}$) in case the eigenvalue does not occur as eigenvalue of$~T|_W$. Now take $W'$ to be the (direct) sum of those complementary subspaces.

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