Existence of linear operator for given subspaces of vector space Let $V$ be a finite-dimensional vector space and $L_1$ and $L_2$ are subspaces of $V$ such that $\dim L_1+\dim L_2=\dim V$. Show that exists linear operator $f$ such that, $$\ker(f)=L_1,\quad \text{Im}(f)=L_2.$$
My approach: Since $L_1$ is a subspace of $V$ then one can define quotient space $V/L_1$ which has dimension $\dim V-\dim L_1=\dim L_2$ and hence $V/L_1\cong L_2$. Suppose $p:V/L_1\to L_2$ is a desired isomorphism. Then one can define the map $f:V\to V$ by equation $f(x)=p(x+L_1)$.
Then it is trivial to check that $f$ is operator with $\ker(f)=L_1$ and $\text{Im}(f)=L_2$.
Is the reasoning correct?
Remark: I was thinking that if $\dim L_1+\dim L_2=\dim V$ then $V=L_1\oplus L_2$ but I've realized that this is false. Indeed, if we take  $V=\mathbb{R}^2$ with standard basis and $L_1=L_2=\langle e_1\rangle $. Btw I was not able to come with non trivial example such that $\dim L_1+\dim L_2=\dim V$ but $V$ is not direct sum of $L_1$ and $L_2$. Can anyone give some example?
 A: The reasoning of your argument looks ok. 
While your counterexample might "feel" trivial, in that case it's the only thing you can do not to get a direct sum splitting. Indeed, if $L_1, L_2 \subset V$ are two subspaces of a finite-dimensional vector space $V$, then $V \cong L_1 \oplus L_2$ if and only if both $L_1 + L_2 = V$ and $L_1 \cap L_2 = \{0\}$. The first condition means for all $v \in V$ there exists some $w_1 \in L_1, w_2 \in L_2$ such that $v = w_1 + w_2$, while the second condition guarantees that this representation is unique. 
So if you have two subspaces intersecting only at the identity that are of complementary dimensions, in the sense that $\dim(L_1) = n - \dim(L_2)$, then they must "fill up" all of $V$ in that $L_1 + L_2 = V$. By the aforementioned result, $V$ splits as a direct sum of $L_1$ and $L_2$. Applying this to the situation where $V = \mathbb{R}^2$, we see that any two distinct lines satisfy these conditions and give a splitting of $V$.     
A: A different approach you might already be aware of but still:
Let $\{b_1, \ldots, b_k\}$ be a basis for $L_1$ and extend it to a basis $B = \{b_1, \ldots, b_k, b_{k+1}, \ldots, b_n\}$ for $V$.
Let $\{c_1, \ldots, c_{n-k}\}$ be a basis for $L_2$ and define a linear map $f : V \to V$ on the basis $B$ as
$$b_1 \mapsto 0$$
$$\vdots$$
$$b_k \mapsto 0$$
$$b_{k+1} \mapsto c_1$$
$$\vdots$$
$$b_{n} \mapsto c_{n-k}$$
Then it is obvious that $\ker(f) = L_1$ and $\operatorname{Im}(f) = L_2$.
