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Is these representations completely reducible?

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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?

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  • $\begingroup$ Have you first found the invariant subspaces? $\endgroup$ – Max Jan 24 '19 at 15:11
  • $\begingroup$ math.stackexchange.com/questions/3083310/… @Max $\endgroup$ – Idonotknow Jan 24 '19 at 15:12
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    $\begingroup$ 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) $\endgroup$ – Maxime Ramzi Jan 24 '19 at 17:08
  • $\begingroup$ 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 $\endgroup$ – Secretly Jan 26 '19 at 14:12
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    $\begingroup$ @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 $F$ ? $\endgroup$ – Maxime Ramzi Jan 26 '19 at 22:52
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  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.

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  • $\begingroup$ what is the invariant complement in 3? $\endgroup$ – Idonotknow Feb 1 '19 at 16:34
  • $\begingroup$ you mean that the complement of $V_{1}$ in $V_{0}$ is the constant term? $\endgroup$ – Idonotknow Feb 1 '19 at 16:36
  • $\begingroup$ Is F in 2. L and t is 1-a and f is x +a I am confused about the notation .... could you clarify it please? $\endgroup$ – Idonotknow Feb 1 '19 at 16:39
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    $\begingroup$ 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 $V_1$. $\endgroup$ – Alec B-G Feb 2 '19 at 9:31
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    $\begingroup$ $V_1$ 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 $a,\lambda\in\mathbb C$. $\endgroup$ – Alec B-G Feb 3 '19 at 9:20

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