Let be a finite dimensional vector space over a field and let be a linear transformation. A subspace is called -stable if . Prove that if has a stable subspace , then decomposes as a direct sum of linear transformations. Moreover, show that if each summand is nonsingular, then is nonsingular.
Conversely, prove that if is nonsingular over a finite dimensional vector space, then and are nonsingular. Prove that this statement is not true over an infinite dimensional vector space.
Suppose . Letting denote the natural projection, note that . Moreover, since is -stable, we have . By the generalized first isomorphism theorem, we have an induced linear transformation given by . It is clear that the restriction is a linear transformation. Recall that via the isomorphism . Evidently, . Thus decomposes as a direct sum: , where .
Suppose and are nonsingular linear transformations. That is, . Suppose ; then and , and we have . Thus is nonsingular.
Now suppose is finite dimensional, is -stable for some linear transformation , and let and be the induced maps discussed above. Suppose is nonsingular. Since is finite dimensional, in fact is an isomorphism. This induces the following short exact sequence of vector spaces.
Note that , so that is injective (I.e. nonsingular). Again, since is finite dimensional, is surjective. Using part (a) to this previous exercise, is injective- that is, nonsingular.
Note that this proof strategy depends essentially on the fact that, on finite dimensional vector spaces, injectivity and surjectivity are equivalent. If we are going to find an infinite dimensional counterexample, then the subspace must also be infinite dimensional and the induced mapping must be injective but not surjective.
Consider the vector space . Let be the “right shift operator” given by if and otherwise. Let ; that is, all tuples in whose first coordinate is 0. Certainly is a subspace of , and moreover is stable under . We also have that . Now is injective but not surjective. Suppose ; then . Since , in fact . So is singular.