For a commutative exact sequence show that $V_1$ and $V_3$ are finite dimensional iff $V_2$ and $V_4$ also are.

Question

For the following commutative exact sequence: \begin{array}\\ &V_1 & \stackrel{A_1}{\longrightarrow} & V_2\\ & \uparrow_{A_4} &&\downarrow _{A_2}\\ &V_4 & \stackrel{A_3}{\longleftarrow} & V_3 \end{array} Let $V_1$, $V_2$, $V_3$, $V_4$ be vector spaces and $A_1$, $A_2$, $A_3$, $A_4$ be homomorphism. Show that $V_1$ and $V_3$ are finite dimensional if and only if $V_2$ and $V_4$ also are.

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As the diagram is exact and commutative, it can be said hat: \begin{array}\\ \ker(A_1) = \operatorname{Im}(A_4) & \ker(A_2) = \operatorname{Im}(A_1) \\ \ker(A_3) = \operatorname{Im}(A_2) & \ker(A_4) = \operatorname{Im}(A_3) \end{array} Also, if $V_1$ and $V_3$ are finite-dimensional, then: \begin{array}\\ \dim(V_1)&=\dim(\ker(A_1))+ \dim(\operatorname{Im}(A_1))\\ \dim(V_3)&=\dim(\ker(A_3))+ \dim(\operatorname{Im}(A_3)) \end{array}

I don't know what should I do next. I would really appreciate your answers.

Answer 1

Hint

The short exact sequence $\;0\longrightarrow \ker A_2\longrightarrow V_2\longrightarrow \operatorname{Im}A_2\longrightarrow 0$ is split since we have vector spaces. Now $\ker A_2\simeq \operatorname{Im}A_1$, which is finite dimensional, and similarly $ \operatorname{Im}A_2\simeq\ker A_3$, which is also finite dimensional.

Can you take it from there?