# Existence of T-invariant complement of T-invariant subspace when T is diagonalisable

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?

• 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? Commented Sep 6, 2012 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. Commented Sep 6, 2012 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. Commented Feb 8, 2014 at 13:08
• See math.stackexchange.com/q/383970/18880 for a converse statement. Commented Jan 12, 2015 at 9:43

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 $V_{\lambda} = \{v \in V\mid Av = \lambda v\}$).

• I am unable to see where has the property that "T" is diagonalizable has been used in the solution? Commented Aug 14, 2018 at 21:18

The minimal polynomial is of the form $$p=(x-c_1)(x-c_2)\cdots (x-c_k),$$ where $c_1,c_2,\ldots,c_k$ are the distinct eigenvalues of $T$.

By primary decomposition $$V=W_1 \oplus W_2 \oplus \cdots \oplus W_k,$$ where $W_i$ is the eigenspace corresponding to $c_i$, $1\leq i \leq k$.

From Hoffman & Kunze, Page 226, Exercise 10, one should be able to see that $$W=(W\cap W_1) \oplus (W\cap W_2) \oplus \cdots \oplus (W\cap W_k).$$ Clearly, $W\cap W_i$ is $T$-invariant, $1\leq i \leq k$.

Let $\{\alpha_1,\alpha_2,\ldots,\alpha_{r_i} \}$ be an ordered basis for $W\cap W_i$. Since $W\cap W_i$ is a subspace of the eigenspace $W_i$, $\{\alpha_1,\alpha_2,\ldots,\alpha_{r_i} \}$ can be extended to $\{\alpha_1,\alpha_2,\ldots,\alpha_{r_i},\alpha_{r_i+1},\ldots,\alpha_{s_i} \}$, a basis for $W_i$. Let $V_i$ be the subspace spanned by $\{\alpha_{r_i+1},\ldots,\alpha_{s_i} \}$. Then $W_i=(W\cap W_i)\oplus V_i$.

Hence $$V=(W\cap W_1)\oplus V_1 \oplus (W\cap W_2)\oplus V_2 \oplus \cdots \oplus (W\cap W_k)\oplus V_k,$$ i.e., $W$ has $T$-invariant complementary subspace of $V$, $V_1\oplus V_2 \oplus \cdots \oplus V_k$.

Let $$\{e_1,\dots,e_n\}$$ be the set of eigenvectors and define $$W'$$ as the complement of $$W$$ with respect to $$V$$. Without losing generality, let $$W\ne\{0\}$$ and $$W\ne V$$ (these cases are trivial). For each $$e_i$$, it should happen that $$e_i\in W$$ or $$e_i\in W'$$. Since we assume that $$W\ne\{0\}$$ and $$W\ne V$$, this implies that $$W'\ne\{0\}$$ and $$W'\ne V$$. If $$w\in W$$, then $$w=a_1e_1+\ldots+a_ne_n$$, and this implies $$a_{k+1}e_{k+1}+\ldots+a_ne_n=w-a_1e_1-\ldots-a_ke_k\in W$$, then $$a_{k+1}=\ldots=a_n=0$$, so $$span\{e_1,...e_k\}=W$$ and $$span\{e_{k+1},...e_n\}=W'$$, for some $$k\ge1$$. This proves that $$V= W \oplus W'$$ since $$\operatorname{span}\{e_1,\dots,e_k\} \cap \operatorname{span}\{e_{k+1},\dots,e_n\}=\varnothing$$. They are invariant under $$T$$ since if $$e_i\in W$$ (or $$e_i\in W^{\mathrm c}$$) then $$T(e_i)=\lambda_ie_i\in W$$ (or $$\lambda_ie_i\in W^{\mathrm c}$$).

• I don't see why $e_i \in W$ or $e_i \in W'$ should happen? Commented May 11, 2022 at 3:45
• And where do we use that $W$ is invariant? Commented May 11, 2022 at 3:52
• Given an $T$-inv subspace $W$ since $W$ is a subspace we have that $x\in W$ or $x\in W^c$ for any $x\in V$. We use that $W$ is inv. in the last part where we apply $T$ to get $T(e_i)=\lambda_ie_i\in W$ (or $\lambda_ie_i\in W^{\mathrm c}$). Commented May 14, 2022 at 15:28
• So you mean $W'=V\setminus W$? But then we cannot say $W'$ is also a subspace, can we? Because $0$ is not in $W'$. And the following argument doesn't make much sense.. Commented May 14, 2022 at 16:51

I will suppose there are $k$ distinct eigenvalues $\lambda_1,\ldots,\lambda_k$ (where $k$ may be less than the dimension$~n$).

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 eigenspace (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.