Exercise 3.B.19

Question

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

Arden Shi · Accepted Answer

To prove if and only if, we need to directions.

Direction 1:

Let $	ext{null }T = U$.

By 3.22, 
\begin{equation}
\begin{split}
	ext{dim }U &= 	ext{dim null }T \
&= 	ext{dim } V - 	ext{dim range }T \
&\geq 	ext{dim } V - 	ext{dim }W
\end{split}
\end{equation}
Direction 2:

Let $	ext{dim } U \geq 	ext{dim } V - 	ext{dim }W$
Let $v_1, ..., v_n$ be a basis of $V$ and $w_1, ..., w_m$ a basis of $W$.
Obviously, $m \geq n - 	ext{dim }U$
Now, define $T \in \mathcal{L} (V, W)$ as
$$
T(a_1v_1+ \dotsb +a_nv_n) = a_1w_1+ \dotsb + a_{n - 	ext{dim }U}w_{n - 	ext{dim }U}
$$
Where $v_{n + 1 - 	ext{dim }U}, ..., v_n \in U$
Thus, $	ext{null} T = U$.
\qed

Exercise 3.B.19

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Exercise 3.B.19

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