Prove there exist an isomorphism

Question

Let $V$ be a vector space and $U,W,Z$ are subspaces of $V$, where $V=Z \oplus W=Z \oplus U$

Prove there exist linear isomorphism $f:V \to V$ such that for every $\gamma \in Z, \ \ f(\gamma)=\gamma$ and $f(U) \subseteq W$.

Can someone give me a hint ? I know that transfomation is an isomorphism when $\ker f = \{ 0\}$ and $im f=V$

Answer 1

Assuming $V$ is of finite dimension $n$, $V=Z\oplus W=Z\oplus U$ and $Z$ is a subspace of dimension $k\le n$, we can conclude that both $W$ and $U$ must be of dimension $n-k$, since $\dim(A\oplus B)=\dim(A)+\dim(B)$. Since all $n-k$ dimensional $k$-vector spaces are isomorphic, there is an isomorphism $\varphi: U\to W$. Now consider the linear map \begin{align*} \operatorname{id}_Z \oplus \varphi \colon Z\oplus U &\longrightarrow Z\oplus W, \\ z+u &\longmapsto z+\varphi(u). \end{align*}

Edit: Actually, we don't need finite dimensions to make this work. When $V=Z\oplus W$, the quotient map $V\to V/Z$ restricts to an isomorphism on $W$, so $W\cong V/Z \cong U$, and the above argument works out for arbitrary $k$-vector spaces.

Answer 2

$$V=W\oplus Z\iff\;\;\forall\,v\in B\;\;\exists!(\,w_v\in W,\,z_v\in Z)\;\;s.t. v= w+z$$

Now, just define

$$f(v=w_v+z_v):=z_v = \;\text{ the projection on}\;\;Z$$

Here, $\;f(z)=z\;\;\forall\,z\in Z\;,\;\;f(W)={0}\subset W\;$

Now show the above indeed is a linear transformation.

I'm not sure whether the above is what you wanted since I think there's some confusion with your notation.