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Chapter 5: Problem 19

Let \(V\) be a finite dimensional vector space over the field \(K\), and let \(S\) be the set of all linear maps of \(V\) into itself. Show that \(V\) is a simple \(S\) -space.

Short Answer

Expert verified
The vector space V is a simple S-space, as it has no nontrivial S-subspaces (invariant subspaces) other than V itself and {0}. We showed this by considering any nonzero vector v in V and proving that its S-orbit spans the entire vector space V. We also showed that the subspace generated by the S-orbit is an S-invariant subspace. Concluding, any finite-dimensional vector space V over the field K is a simple S-space when S is the set of all linear maps from V to itself.

Step by step solution

01

Definition of simple S-space and invariant subspace

: A simple space V is called a simple S-space if there are no nontrivial S-subspaces, where S-subspace is an invariant subspace. An invariant subspace W is a subspace of V such that for any w in W and any linear map T in S, T(w) is also in W.
02

The trivial case

: If V = {0}, then V is a simple S-space since it contains no other subspace besides itself.
03

Nontrivial case

: Assume V is nonzero and let S be the set of all linear maps from V to itself. Let v be any nonzero vector in V. We need to show that for any nonzero vector w in V, there exists an S-linear combination of v that equals w i.e., w can be obtained as the result of applying an element T from the set S to the vector v.
04

Define the S-orbit of v

: Consider the set O_v, called the S-orbit of v, defined as {T(v) | T 鈭� S}. This set contains all possible results of applying linear maps T to the vector v.
05

Span of the S-orbit of v

: Now, we want to show that for any nonzero vector w in V, w can be written as a linear combination of vectors in O_v. In other words, we need to prove that the span of O_v is the entire vector space V.
06

Subspace generated by the S-orbit

: Let W be the subspace generated by O_v. Since v 鈭� O_v (using the identity map), we know W is a nontrivial subspace of V. Therefore, W is an S-invariant subspace of V.
07

Show that W = V

: Suppose W 鈮� V, then there exists a nonzero vector w 鈭� V such that w 鈭� W. Since W is an S-invariant subspace of V, we can define the projection map: \( P: V \to V/W \) The projection map P is a linear map, so P is in S, and for the vector w 鈭� V: \( P(w) = w + W \) Since w is not an element of W, \(P(w) 鈮� 0\) in V/W, which contradicts that W is an S-invariant subspace of V. Therefore, W must be equal to V.
08

Final Conclusion

: Since we have shown that for any nonzero vector v in V, its S-orbit O_v spans the entire vector space V, we can conclude that V is a simple S-space.

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