# Is this representation completely reducible?

Is these representations completely reducible? Definition:

A linear representation is said to be completely reducible if every invariant subspace has an invariant complement.

But I have no idea how to apply the definition, and I know that the answer for 2 is no while that for 3 is yes, could you please clarify for me how to check the definition in each case?

• Have you first found the invariant subspaces? – Max Jan 24 '19 at 15:11
• – Idonotknow Jan 24 '19 at 15:12
• Knowing the invariant subspaces, all you have to do is look at the list and check whether every element on that list has a complement that's also on the list (I'm a different Max by the way) – Maxime Ramzi Jan 24 '19 at 17:08
• For number 3 the invariant subspaces are "Any subspace spanned by some set of eigenvectors of the operator $\alpha$" ...... This is the answer written at the back of the book @Max – Secretly Jan 26 '19 at 14:12
• @hopefully : ok well then ? Take a space $F$ in the set of invariant subspaces, and look at all other invariant subspaces : is there one of them that is a complement of $F$ ? – Maxime Ramzi Jan 26 '19 at 22:52

1. Consider polynomials of degree less than or equal to 1 $$V_1=\mathbb C\cdot x\oplus\mathbb C\cdot 1,$$ this is clearly invariant under $$F(t)$$ as $$F(t)$$ does not change degree. Similarly consider $$V_0=\mathbb C\cdot 1,$$ the space of constant functions. It is clear that $$V_0\subset V_1$$ and that they are both invariant under $$F$$. However, if we consider the complement of $$V_1$$ in $$V_0$$, $$W$$, then there exists some $$a\in\mathbb C$$ such that $$x+a\in W$$. If $$W$$ is invariant, then we have that $$L(1-a)(x+a)=x+1\in W$$. $$W$$ is a vector space so $$x+a-L(1-a)(x+a)=1\in W$$. This means $$V_0\subset W$$, a contradiction. Hence the space is not completely reducible.
2. $$\alpha\in$$ End$$(V)$$, with $$V\cong\mathbb C^n$$, so $$\alpha$$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $$\alpha=\text{diag}(\lambda_1,\dots,\lambda_n)$$. So $$F(t)=\text{diag}(e^{t\lambda_1},\dots,e^{t\lambda_n}),$$ and $$V=\mathbb C_{\lambda_1}\oplus\dots\oplus\mathbb C_{\lambda_n},$$ where $$\mathbb C_{\lambda_i}$$ is the one dimensional representation $$v\mapsto e^{t\lambda_i}v$$. So $$V$$ is completely reducible.
• you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term? – Idonotknow Feb 1 '19 at 16:36
• If $V$ is a subspace of $W$, both with an action by a group $G$, an invariant complement is a vector space $U$ such that $G\cdot U=U$ and $W=V\oplus U$. A complement of $V_1$ in $V_0$ would be any one dimensional space spanned by $x-a$ for $a\in\mathbb C$. I showed that such a space cannot be invariant under $F(t)$ without also containing all of $V_1$. – Alec B-G Feb 2 '19 at 9:31
• $V_1$ is two dimensional and $V_0$ is one dimensional, so the complement, $W$, is a subspace of $V_1$ such that $V_0\oplus W=V_1$, and so has to be one dimensional. Write $\mathbf 1$ for the constant function 1, a vector in both $V_0$ and $V_1$, and $\mathbf x$ for the function $x$, a vector in $V_1$. $W$ is a one dimensional vector space that sits inside $(\mathbb C\mathbf 1\oplus \mathbb C\mathbf x)\backslash\mathbb C\mathbf 1$. This means that arbitrary elements of $W$ are of the form $\lambda\cdot(a\mathbf 1+\mathbf x)$ for $a,\lambda\in\mathbb C$. – Alec B-G Feb 3 '19 at 9:20