1. ## linear mapping

Hi,
I really need your help with this:
Suppose that V and W are finite-dimensional and that U is a subspace of V. Prove that there exists a T – linear map from V to W such that null(T)=U if and only if dim(U)>=dim(V)-dim(W).

2. I will prove the forward direction, you try proving the other direction.

Say that $T:V\to W$ is a linear transformation. Then by rank-nullity theorem we know $\text{rank}(T) + \text{nullity}(T) = \text{dim}(V)$. But we are told that $\text{null}(T) = U$ thus $\text{nullity}(T) = \text{dim}(U)$. Putting this together we have that $\text{dim}(U) = \text{dim}(V) - \text{rank}(T)$. But $T[ U ] \subseteq W$ so $\text{rank}(T) = \text{dim}(T[ U ]) \leq \text{dim}(W)$ this means $-\text{rank}(T) \geq - \text{dim}(W)$. Finally, $\text{dim}(U) \geq \text{dim}(V) - \text{dim}(W)$.

3. Originally Posted by bamby
Hi,
I really need your help with this:
Suppose that V and W are finite-dimensional and that U is a subspace of V. Prove that there exists a T – linear map from V to W such that null(T)=U if and only if dim(U)>=dim(V)-dim(W).
$(\Longrightarrow)$ trivial by rank-nullity theorem.
$(\Longleftarrow)$ suppose $\dim W \geq \dim V - \dim U.$ let $B_1=\{u_1, \cdots , u_m \}$ be a basis for $U$ and extend $B_1$ to a basis for $V: \ \ B=\{u_1, \cdots , u_m, v_1, \cdots , v_n \}.$
so we have that $\dim W \geq n.$ thus there exists a set of $n$ linearly independent elements of $W,$ say $w_1, \cdots , w_n.$ now define the map $T:V \longrightarrow W$ by:
$T(a_1u_1 + \cdots a_m u_m + b_1v_1 + \cdots + b_n v_n)=b_1w_1 + \cdots + b_nw_n.$ it's obvious that T is a linear map and $\ker T=U. \ \ \ \square$