Prove the direct sum $U \oplus \ker f$ is isomorph to $V$

Question

I'm having trouble proving the following:

Let $V$, $U$ be finite-dimensional vector spaces and $f$ the surjective linear map from $V$ to $U$. I am required to prove the direct sum of $U+\ker f$ is isomorph to $V$.

I have gone about it the following way. Since the map is surjective $\dim f(V)=\dim U$, and thus $\dim V = \dim U + \dim\ker f$.

Firstly, How can I prove the sum $\ker f + U$ to be direct if it is not given that $f = f^2$ which is the only way I can prove that the intersection between $\ker f$ and $U$ is the zero vector. And secondly, when the dimensions of the direct sum and $V$ are the same, how would that support my proof?

Thank you for your help.

Answer 1

Let $W$ be a subspace of $V$ such that $\ker(f)\oplus V'=V$. Then $f|_W$ is an isomorphism from $W$ onto $U$:

• $\ker f|_W\subset \ker(f)\cap W=\{0\}$ and therefore $f|_W$ is injective;
• if $u\in U$, then there is a $v\in \ker f$ and a $w\in W$ such that $u=f(v+w)$. Therefore, since $v\in\ker f$, $u=f(w)$ and so $f|_W$ is injective.

So, we can define an isomorphism $\psi\colon U\oplus\ker f\longrightarrow V$ by $\psi(u+v)={f|_W}^{-1}(u)+v$ ($u\in U$ and $v\in\ker f$).