Representation theory question

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>>8185669
if the matrix A has distinct EigValues x1,...,xn and v1,...,vn is a basis of V consisting of EigVectors of A, the the induced endomorphism on Sym(d,V) has EigVectors and values
[math]x_{i_1}, \hdots ,x_{i_d}[/math]
[math]v_{i_1}, \hdots,v_{i_d}[/math]
for all sequences
[math]i_1 \leq \hdots \leq i_d[/math]

>>8185731
and if W is a subrepresentation of Sym(d,V), then it must be spanned by a subset of these eigenvectors.

>>8185731
of course eigenvals are
[math]x_{i_1} \cdot ... \cdot x_{i_d}[/math]
eigenvectors
[math]v_{i_1} \cdot ... \cdot v_{i_d}[/math]
for all sequences
[math]i_1 \leq ... \leq i_d[/math]

>>8185669
assume its not irreducible

get a contradiction

>>8185669
I don't think Young's Lattice represents partitioning integers well.

>>8185765
I think it does, and it's very relevant to the irreducible representations of GL(V).

back to the proof
>>8185737
cont.

now look at the elementary matrices [math]E_{ij}[/math] defined such that
[math]E_{ij} v_k = \delta_{jk] v_i[/math]

since W is invariant under the endomorphisms [math]1 + E_{ij}[/math] it follows.. hm.. something is supposed to follow here, not sure how to finish

>>8185807
[math]E_{ij} v_k = \delta_{jk} v_i[/math]
/sci/ really needs a preview button.

ANYWHO, not sure how W's invariance under the endomorphisms [math]1+E_{ij}[/math] implies that all eigenvectors of the _induced_ endomorphism must already lie in W, ie that W=Sym(d,V).

Say W was spanned by [math]v_1^d[/math] then with the above construction I get that all [math]v_j^d[/math] would also have to be in W, but those arent nearly all the eigenvectors

>>8185839
>/sci/ really needs a preview button.
>what is the TeX button in the reply window
Learn how to use a website newfag.

Is it really irreducible unless dim V = 1?

Because for any basis vector v \in V, v \otimes v \otimes v ... \otimes v should span a proper submodule.

>>8185968
it is always irreducible, yes.
also a single vector [math]v^d=v \otimes ... \otimes v[/math] can never span an invariant subspace, because it would have to be an eigenvector of every single element of GL(V)