Linear operator proof. Let $P$ and $Q$, respectively, be subspaces of the vector spaces $V$ and $W$ over the same field $K$ and let $V$ be a finite dimensional space. If $\dim P + \dim Q = \dim V$ prove that then there exists a linear operator $L:V \rightarrow W$ for which $\ker L=P$ and $\operatorname{Im}L=Q$. Give detailed explanation of your work.
Now, since  $P$ is a subspace of $V$, its dimension is less than the dimension of $V$. Let's say that $\dim V=n$ and that $\dim P=k$, where $k<n$; this means that $\dim Q=n-k$. Now, I know that the kernel is the set of all vectors in the domain that map to zero and that the image is the set of all vectors of the codomain, but how can I prove that there exists such a linear operator that meets these requirements? I've found similar question here and it says that I should extend the basis of $P$ to a basis of $V$. What that actually means, and how am I supposed to do that? Finally, how will that help me to solve this?
 A: Most of the question can be summed up with the following diagram

We are given finite-dimensional subspaces $P$ and $Q$ of vector spaces $V$ and $W$ respectively such that
$$
\dim P+\dim Q=\dim V
$$
We wish to find a linear map $L:V\to W$ with $\ker L=P$ and $\DeclareMathOperator{image}{image}\image L=Q$.
To do so, start with bases $\{p_1,\dotsc,p_m\}$ and $\{q_1,\dotsc,q_n\}$ of $P$ and $Q$ respectively. Note that $\{p_1,\dotsc,p_m\}$ may be extended to a basis $\{p_1,\dotsc,p_m,v_1,\dotsc,v_n\}$ of $V$. We may then define $L:V\to W$ on the basis $\{p_1,\dotsc,p_m,v_1,\dotsc,v_n\}$ of $V$ by
\begin{align*}
L(p_k) &= \mathbf 0 & L(v_k) &= q_k
\end{align*}
and extend linearly.
By "extend linearly" we mean apply the following proposition:
Proposition. Given a basis $\{v_1,\dotsc,v_n\}$ for a vector space $V$ and vectors $\{w_1,\dotsc,w_n\}$ in a vector space $W$, there exists a unique linear map $T:V\to W$ with $T(v_j)=w_j$.
A: If we extend a basis of $P$ to a basis of $V$ we will create (with the other basis) a subspace $P'$ which verify :
$$
P\oplus P'=V
$$
and we have $\dim(P')=\dim(Q)$ so it exist an invertible linear operator $T$ such that $T : \; P'\to Q$ (we can construct it canonically by identifying the basis of $Q$ to our contracted basis)
and let now the projection $S : V \to P'$ to $P'$ defined by :
$$
\begin{array}{l}
S :& V=P\oplus P' & \to & P'\\
&x=x_P+x_{P'} & \mapsto & x_{P'}
\end{array}
$$
this is a linear map who the kernel is clearly $P$ 
and finally we put $L=TS$
then 
$$
L : V  \to Q\\
Ker(L)=ker(S)=P\\
Im(L)=Q
$$
A: The key is precisely the possibility to extend a basis of $P$ to a basis of $V$, together with the possibility of defining a linear map by assigning the images to the elements of a basis of the domain.
Suppose $\{v_1,\dots,v_k\}$ be a basis of $P$; then there exist vectors $u_{1},\dots,u_{n-k}\in V$ such that $\{v_1,\dots,v_k,u_{1},\dots,u_{n-k}\}$ is a basis of $V$.
Now, let $\{w_{1},\dots,w_{n-k}\}$ be a basis for $Q$. Then define $f\colon V\to W$ by setting
\begin{align}
f(v_1)&=0 \\
&\vdots \\
f(v_k)&=0 \\[6px]
f(u_1)&=w_1 \\
&\vdots \\
f(u_{n-k})&=w_{n-k}
\end{align}
Clearly the image of $f$ is contained in $Q$ and the kernel of $f$ contains $P$.
Apply the rank-nullity theorem to show that indeed $\dim\ker f=k$ and $\dim\operatorname{Im} f=n-k$, so we deduce the required equalities.
