aalma said:
Yes, I agree. But it is still hard for me to connect all things tohether. Can you provide some details of how this works (Starting with ##g\in H## how to find ##\lambda, v##?
What is ##H##? I assume it is a subgroup of the center by what you wrote, maybe the entire center of ##G## However, if ##G## is commutative per given condition, then ##H## is any subgroup, or ##H=G##?
We have a representation ##\rho\, : \,G\longrightarrow GL(V).##
I assume that ##V## is a finite-dimensional, complex vector space. It means that ##\rho(g)## is a square matrix of the size of the dimension of ##V.## We are looking for values ##\lambda \in \mathbb{C}## such that
$$
\rho(g)(v)=\lambda \cdot v \, \Longleftrightarrow \, \rho(g)(v)-\lambda v=(\rho(g)-\lambda \cdot \operatorname{id}_V)(v)=0
$$
The notation ##g(v)## is sloppy since ##g## has nothing to do with ##v##, ##\rho(g)## has. You can write ##g(v)## or better ##g.v## but you must be aware that ##(\rho(g))(v)## is meant. This is especially important if ##G## itself consists of matrices. Then we can use matrix multiplication as ##\rho.##
You really need to tell us such things. Anyway.
We are looking for a complex number ##\lambda ,## a vector ##v\in V## and have an equation, namely
$$
(\rho(g)-\lambda \cdot \operatorname{id}_V)(v)=0 \text{ or } v\in \operatorname{ker}(\rho(g)-\lambda \cdot \operatorname{id}_V)
$$
I cannot hold a lecture on linear algebra here, the more as I'm basically already gone, and knowledge of linear algebra is definitely what you need. The equation above that defines eigenvalues and eigenvectors is solved by finding the zeros of the characteristic polynomial. We have in particular
$$
\det(\rho(g)-\lambda \cdot \operatorname{id}_V) = 0
$$
This is where we need the complex numbers. Only there is guaranteed that all zeros of this polynomial of degree dimension of ##V## always exist. Any polynomial over the complex numbers splits into linear factors, i.e. (I write ##x## for the complex indeterminate because we want to find values ##x##)
\begin{align*}
0&=\det(\rho(g) - x \cdot \operatorname{id}_V)=(a_1-x)\cdot(a_2-x)\cdot(a_3-x) \cdot\ldots\cdot (a_n-x)\\
\end{align*}
This polynomial is zero, if and only if ##x=a_k## for some ##k.## Note that the ##\{a_k\}## can have repetitions! So we actually have something like ##0=(b_1-x)^{m_1} \cdot\ldots \cdot (b_k -x)^{m_k}## and the polynomial is zero if and only if ##x=b_r## for some ## r .###
Hence ##\lambda \in \{b_1,\ldots,b_m\}.## To find a corresponding vector, you have to solve
$$
\rho(g)(v)=\lambda \cdot v
$$
Note, that there is, in general, more than one vector that fulfills the equation!
aalma said:
I tried to show that the given set is a subrep but I got stuck:
Given ##g\in G, w\in V: g(w)=\lambda w## then why is ##\rho(g)(w)## in this set, i.e.
##g(\rho(g)(w))=\lambda(\rho(g)(w)##
How then this leads to what they ask).
Next, we have the eigenspace
$$
W(\lambda )=\left\{v\in V\,|\,\rho(g)(v)=\lambda \cdot v\;\textbf{ for all }\;g\in G\right\}.
$$
The forall quantifier is essential here as we will soon see, at (*)!
We need to show that ##W(\lambda )## is a vector subspace in ##V,## i.e. that
\begin{align*}
\alpha v+\beta w &\in W(\lambda )\text{ for all }\alpha,\beta \in \mathbb{C} \text{ and } v,w\in W(\lambda )\\
&\Longleftrightarrow \\
\rho(g)(\alpha v+\beta w)&=\ldots =\lambda \cdot (\alpha v+\beta w)
\end{align*}
This proves that ##W(\lambda ) \subseteq V## is a subspace.
At last, we want to show that ##\left. \rho\right|_{W(\lambda )}\, : \,G\longrightarrow GL(W(\lambda ))## is a representation again. This implies to show that ##\left. \rho\right|_{W(\lambda )}## is a group homomorphism, i.e. that
$$
\left. \rho\right|_{W(\lambda )}(g\cdot h)(v)=(\left. \rho\right|_{W(\lambda )}(g)\circ \left. \rho\right|_{W(\lambda )}(h))(v)
$$
or since ##\left. \rho\right|_{W(\lambda )}(g)## and ##\left. \rho\right|_{W(\lambda )}(h)## are matrices
$$
\left. \rho\right|_{W(\lambda )}(g\cdot h)=\left. \rho\right|_{W(\lambda )}(g) \, \cdot \,\left. \rho\right|_{W(\lambda )}(h)
$$
Fortunately, there is nothing to do here because ##\left. \rho\right|_{W(\lambda )}## inherits this property from ##\rho .## The restriction on ##W(\lambda )## does not change this. However, and this is essential,
we do not know whether
$$
\left. \rho\right|_{W(\lambda )}(W(\lambda )) \subseteq W(\lambda )
$$
That means we have to show it. Let ##w\in W(\lambda ).## Then by definition of ##W(\lambda )##
\begin{align*}
(\,\left. \rho\right|_{W(\lambda )}(g)\,)\,(\left. \rho\right|_{W(\lambda )}(h))\,(w)&=(\rho(g))(\rho(h))(v)\\
&=\rho(g\cdot h)(w)\stackrel{(*)}{=}\lambda \cdot w
\end{align*}
So ##\left. \rho\right|_{W(\lambda )}(h)(w)\in W(\lambda )## or generally
$$
\left. \rho\right|_{W(\lambda )}(g)(W(\lambda )\subseteq W(\lambda )\;\text{ for all }\;g\in G
$$
and the restriction of ##\rho## on the subspace ##W(\lambda )\subseteq V## is a subrepresentation.
Hint: Don't be so sloppy. Use your definitions, separate both sides of ##A\Rightarrow B## clearly, and trace your dependencies. Maybe your book or professor writes ##W_\lambda ## instead of ##W(\lambda )## but it should be noted: different eigenvalues ##\lambda ## give different eigenspaces ##W(\lambda ).##
Maybe you want to have a look at:
https://www.physicsforums.com/insights/10-math-tips-save-time-avoid-mistakes/
or
https://www.physicsforums.com/insights/how-most-proofs-are-structured-and-how-to-write-them/