# Prove $\exists T\in\mathfrak{L}(V,W)$ s.t. $\text{null}(T)=U$ iff $\text{dim}(U)\geq\text{dim}(V)-\text{dim}(W)$.

The entire problem statement is,

Suppose that $V$ and $W$ are finite dimensional and that $U$ is a subspace of $V$. Prove that there exists $T\in\mathfrak{L}(V,W)$ such that $\text{null}(T)=U$ if and only if $\text{dim}(U)\geq\text{dim}(V)-\text{dim}(W)$.

My attempt at the proof is:

Suppose first there exists $T\in\mathfrak{T}(V,W)$ such that $U=\text{null}(T)$. We know that $$\text{dim}(V)=\text{dim null}(T)+\text{dim}(\text{range}(T))$$ and so $$\text{dim}(U)=\text{dim}(V)-\text{dim}(\text{range}(T))\geq\text{dim}(V)-\text{dim}(W).$$ For the other direction, let $(u_1,\dots,u_m)$ be a basis for $U$, which can then be extended to a basis in $V$, namely, $(u_1,\dots,u_m,v_1\dots,v_n)$. Also, let $(w_1,\dots,w_p)$ be a basis for $W$. So then by the hypothesis, $m\geq(m+n)-p$ which means $n\leq p$.

That’s as far as I got for the other direction. I’m not sure how else to continue, but I do know I have to figure out a linear map such that $U=\text{null}(T)$. Is there an obvious construction?

Thanks for help or hints!

#### Solutions Collecting From Web of "Prove $\exists T\in\mathfrak{L}(V,W)$ s.t. $\text{null}(T)=U$ iff $\text{dim}(U)\geq\text{dim}(V)-\text{dim}(W)$."

You have basically done it : You know that $n\leq p$, so just define $T: V\to W$ by
$$T(u_i) = 0, 1\leq i\leq m, \text{ and } T(v_j) = w_j, 1\leq j\leq n$$
Then $T$ extends to a linear map $T \in \mathcal{L}(V,W)$. By this construction
$$U \subset \text{null}(T)$$
Furthermore, if $T(v) = 0$, then write
$$v = \sum_{i=1}^m \alpha_i u_i + \sum_{j=1}^n \beta_j v_j$$
Then
$$T(v) = \sum_{j=1}^n \beta_j w_j = 0$$
By linear independence of $\{w_j\}_{j=1}^n$, it follows that $\beta_j = 0$ for all $j$, and hence
$$v = \sum_{i=1}^m \alpha_i u_i \in U$$
Hence, $U = \text{null}(T)$.