By J. Ikramov

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**Sample text**

4: The characters χR and χ0 on conjugacy classes is mapped into itself by the action of R, and the restriction to E0 gives an irreducible representation R0 of S4 . In fact, C4 = E0 ⊕ C(1, 1, 1, 1) decomposes the space C4 into complementary invariant, irreducible subspaces. The subspace C(1, 1, 1, 1) carries the trivial representation (all elements act through the identity map). Examining the effect of the group elements on the standard basis vectors, we can work out the character of R. For instance, R((12)) interchanges e1 and e2 , and leaves e3 and e4 fixed, and so its matrix is 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 and the trace is χR ((12)) = 2.

Vσ−1 (n) ) for all (v1 , . . , vn ) ∈ Fn and σ ∈ Sn . , v) : v ∈ F} ⊂ Fn Representing Finite Groups 10/2011 39 and E0 = {(v1 , . . , vn ) ∈ Fn : v1 + · · · + vn = 0}. ); (ii) each vector e1 − ej lies in the span of {R(σ)w : σ ∈ Sn }, for any w ∈ E0 ; (iii) the restriction R0 of R to the subspace E0 is an irreducible representation of Sn . 3. Let Pn be the set of all partitions of [n] = {1, . . , n} into k disjoint nonempty subsets, where k ∈ [n]. For σ ∈ Sn and p ∈ Pk , let σ · p = {σ(B) : B ∈ p}.

Nk−1 Nk−2 Nk−3 Nk−4 Nk Nk−1 Nk−2 Nk−3 ... ... ... 0 0 0 .. Ek , . . k − 1 . . 69) by expressing the first column as a linear combination of the other columns and a column vector whose entries are all 0 except for the 46 Ambar N. Sengupta last entry which is (−1)k−1 kEk . This expresses the k-th elementary symmetric sum Ek in terms of the power sums as N1 1 0 0 ... 0 N2 N1 2 0 ... 0 N3 N N 3 . . 0 1 2 1 Ek = det .. 70) .. .. .. , . k! . . . Nk−1 Nk−2 Nk−3 Nk−4 .

### Problemas de Algebra Lineal by J. Ikramov

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