## A linear transformation on a finite dimensional vector space which has a stable subspace decomposes as a direct sum

Let $V$ be a finite dimensional vector space over a field $F$ and let $\varphi : V \rightarrow V$ be a linear transformation. A subspace $W \subseteq V$ is called $\varphi$-stable if $\varphi[W] \subseteq W$. Prove that if $\varphi$ has a stable subspace $W$, then $\varphi$ decomposes as a direct sum of linear transformations. Moreover, show that if each summand is nonsingular, then $\varphi$ is nonsingular.

Conversely, prove that if $\alpha \oplus \beta$ is nonsingular over a finite dimensional vector space, then $\alpha$ and $\beta$ are nonsingular. Prove that this statement is not true over an infinite dimensional vector space.

Suppose $\varphi[W] \subseteq W$. Letting $\pi : V \rightarrow V/W$ denote the natural projection, note that $\pi \circ \varphi : V \rightarrow V/W$. Moreover, since $W$ is $\varphi$-stable, we have $W \subseteq \mathsf{ker}\ \pi \circ \varphi$. By the generalized first isomorphism theorem, we have an induced linear transformation $\overline{\varphi} : V/W \rightarrow V/W$ given by $\overline{\varphi}(v+W) = \varphi(v) + W$. It is clear that the restriction $\varphi|_W : W \rightarrow W$ is a linear transformation. Recall that $V \cong W \oplus V/W$ via the isomorphism $\theta : (w, v+W) = v+w$. Evidently, $\theta \circ (\varphi|_W \oplus \overline{\varphi}) = \varphi \circ \theta$. Thus $\varphi$ decomposes as a direct sum: $\theta = \alpha \oplus \beta$, where $V = A \oplus B$.

Suppose $\alpha : A \rightarrow A$ and $\beta : B \rightarrow B$ are nonsingular linear transformations. That is, $\mathsf{ker}\ \alpha = \mathsf{ker}\ \beta = 0$. Suppose $(a,b) \in \mathsf{ker}\ \alpha \oplus \beta$; then $a \in \mathsf{ker}\ \alpha$ and $b \in \mathsf{ker}\ \beta$, and we have $(a,b) = 0$. Thus $\alpha \oplus \beta$ is nonsingular.

Now suppose $V$ is finite dimensional, $W \subseteq V$ is $\varphi$-stable for some linear transformation $\varphi$, and let $\varphi|_W : W \rightarrow W$ and $\overline{\varphi} : V/W \rightarrow V/W$ be the induced maps discussed above. Suppose $\varphi$ is nonsingular. Since $V$ is finite dimensional, in fact $\varphi$ is an isomorphism. This induces the following short exact sequence of vector spaces.

a diagram of vector spaces

Note that $\mathsf{ker}\ \varphi|_W \subseteq \mathsf{ker}\ \varphi$, so that $\varphi|_W$ is injective (I.e. nonsingular). Again, since $W$ is finite dimensional, $\varphi|_W$ is surjective. Using part (a) to this previous exercise, $\overline{\varphi}$ 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 $W$ must also be infinite dimensional and the induced mapping $\varphi|_W$ must be injective but not surjective.

Consider the vector space $V = \bigoplus_\mathbb{N} F$. Let $\varphi : V \rightarrow V$ be the “right shift operator” given by $\varphi(a)_i = 0$ if $i = 0$ and $a_{i-1}$ otherwise. Let $W = 0 \oplus \bigoplus_{\mathbb{N}^+} F$; that is, all tuples in $V$ whose first coordinate is 0. Certainly $W$ is a subspace of $V$, and moreover is stable under $\varphi$. We also have that $V/W \cong F \cong F \oplus \bigoplus_{\mathbb{N}^+} 0$. Now $\varphi|_W$ is injective but not surjective. Suppose $v+W \in V/W$; then $\overline{\varphi}(v+W) = \varphi(v)+W$. Since $\varphi(v) \in W$, in fact $\overline{\varphi} = 0$. So $\overline{\varphi}$ is singular.