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).
Thank you in advance!
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).
Thank you in advance!
I will prove the forward direction, you try proving the other direction.
Say that $\displaystyle T:V\to W$ is a linear transformation. Then by rank-nullity theorem we know $\displaystyle \text{rank}(T) + \text{nullity}(T) = \text{dim}(V)$. But we are told that $\displaystyle \text{null}(T) = U$ thus $\displaystyle \text{nullity}(T) = \text{dim}(U)$. Putting this together we have that $\displaystyle \text{dim}(U) = \text{dim}(V) - \text{rank}(T)$. But $\displaystyle T[ U ] \subseteq W$ so $\displaystyle \text{rank}(T) = \text{dim}(T[ U ]) \leq \text{dim}(W)$ this means $\displaystyle -\text{rank}(T) \geq - \text{dim}(W)$. Finally, $\displaystyle \text{dim}(U) \geq \text{dim}(V) - \text{dim}(W)$.
$\displaystyle (\Longrightarrow)$ trivial by rank-nullity theorem.
$\displaystyle (\Longleftarrow)$ suppose $\displaystyle \dim W \geq \dim V - \dim U.$ let $\displaystyle B_1=\{u_1, \cdots , u_m \}$ be a basis for $\displaystyle U$ and extend $\displaystyle B_1$ to a basis for $\displaystyle V: \ \ B=\{u_1, \cdots , u_m, v_1, \cdots , v_n \}.$
so we have that $\displaystyle \dim W \geq n.$ thus there exists a set of $\displaystyle n$ linearly independent elements of $\displaystyle W,$ say $\displaystyle w_1, \cdots , w_n.$ now define the map $\displaystyle T:V \longrightarrow W$ by:
$\displaystyle T(a_1u_1 + \cdots a_m u_m + b_1v_1 + \cdots + b_n v_n)=b_1w_1 + \cdots + b_nw_n.$ it's obvious that T is a linear map and $\displaystyle \ker T=U. \ \ \ \square$