Exercise.
Suppose $V$ and $W$ are finite-dimensional and $U$ is a subspace of $V$. Let
$$
\mathcal{E} = \{T \in \mathcal{L}(V,W) : U \subseteq \text{null} \ T\}.
$$
(a) Show that $\mathcal{E}$ is a subspace of $\mathcal{L}(V,W)$.
(b) Find a formula for $\text{dim} \ \mathcal{E}$ in terms of $\text{dim} \ V$, $\text{dim} \ W$, and $\text{dim} \ U$.
Hint: Define $\Phi : \mathcal{L}(V,W) \to \mathcal{L}(U,W)$ by $\Phi(T) = T|_{U}$. What is $\text{null} \ \Phi$? What is $\text{range} \ \Phi$?
Notation used.
- $\mathcal{L}(V,W)$ is the set of linear maps from a vector space $V$ to a vector space $W$.
- $T|_{U}$ means the function $T$ restricted to $U$. Thus $T|_{U}$ is the function whose domain is $U$, with $T|_{U}$ defined by $T|_{U}(u) = Tu$ for every $u \in U$.
Source.
Linear Algebra Done Right, 4th edition, Sheldon Axler.
What I want to focus on and what I've tried.
I was able to show (a), so I will omit showing it here. Also, I verified that $\Phi$ is linear for the sake of completeness. What I am struggling with is finalizing part (b); see below for what I was able to do so far:
By definition, $\text{null} \ \Phi = \{T \in \mathcal{L}(V,W) : \Phi(T)=0\}$. The mappings $T$ in $\text{null} \ \Phi$ that satisfy $\Phi(T)=0$ are such that $T|_{U}(u) = Tu = 0$, since $\Phi(T) = T|_{U}$. Notice that $\mathcal{E}$ contains all such $T$'s since if $T \in \mathcal{E}$, then $Tu = 0$ for all $u \in U$. Hence,
$$
\text{null} \ \Phi = \mathcal{E} \tag{1}
$$
By definition, $\text{range} \ \Phi = \{\Phi(T) : T \in \mathcal{L}(V,W)\}$. Since $\Phi(T) = T|_{U}$, we have
$$
\text{range} \ \Phi = \text{range} \ T|_{U} \tag{2}
$$
Now, using the fundamental theorem of linear maps, we have
$$
\text{dim} \ \mathcal{L}(V,W) = \text{dim} \ \text{null} \ \Phi + \text{dim} \ \text{range} \ \Phi \tag{3}
$$
Using $(1)$ and $(2)$, and the fact that $\text{dim} \ \mathcal{L}(V,W) = (\text{dim} \ V)(\text{dim} \ W)$ since $V$ and $W$ are finite-dimensional, $(3)$ becomes
$$
(\text{dim} \ V)(\text{dim} \ W) = \text{dim} \ \mathcal{E} + \text{dim} \ \text{range} \ T|_{U} \tag{4}
$$
$(4)$ is where I got stuck. The only thing I can say about $\text{dim} \ \text{range} \ T|_{U}$ is that it's less than or equal to the dimension of $W$, but that doesn't really help. Am I misinterpreting $\text{range} \ \Phi$ possibly?
Best Answer
My attempt uses the standard basis for $\mathcal L(V,W)$:
Let $\{v_1,\ldots,v_\ell\}$ be a basis for $U$. By basis extension $\{v_1,\ldots,v_\ell,\ldots,v_n\}$ is a basis for $V$. Let $\{w_1,\ldots,w_m\}$ be a basis for $W$. The standard basis for $\mathcal L(V,W)$ is the set $\{T_{pq}\vert p=1,\ldots,m \ \text{and} \ q=1,\ldots,n\}$, which is defined by $$T_{pq}(v_i)=\delta_{iq}w_p,$$ where $\delta_{iq}$ is the Kronecker delta.
If $T\in\mathcal E$, then $$T=\sum_{p=1}^m\sum_{q=1}^na_{pq}T_{pq}, \ \ \ \ \ \ \ \ \ (*)$$ because $T\in\mathcal L(V,W)$, and $$T(v_1)=0 \ \ ; \ \ \ldots \ \ ; \ \ T(v_\ell)=0, \ \ \ \ \ \ \ \ \ (**)$$ because $U\subseteq\operatorname{null}T$.
Substituting $(*)$ into $(**)$, we have \begin{align} &\sum_{p=1}^m\sum_{q=1}^na_{pq}T_{pq}(v_1)=0 \ \ ; \ \ \ldots \ \ ;\ \ \sum_{p=1}^m\sum_{q=1}^na_{pq}T_{pq}(v_\ell)=0\\ &\implies \sum_{p=1}^ma_{p1}w_p=0 \ \ ; \ \ \ldots \ \ ; \ \ \sum_{p=1}^ma_{p\ell}w_p=0\\ &\implies a_{11}=0,\cdots,a_{m1}=0 \ \ ; \ \ \ldots \ \ ; \ \ a_{1\ell}=0,\cdots,a_{m\ell}=0. \ \ \ \ \ \ \ \ \ (***) \end{align} Substituting $(***)$ into $(*)$, we get $$T=\sum_{p=1}^m\sum_{q=\ell+1}^na_{pq}T_{pq}.$$ This means that $\{T_{pq}\vert p=1,\ldots,m \ \text{and} \ q=\ell+1,\ldots,n\}$ is a basis for $\mathcal E$, because $\{T_{pq}\vert p=1,\ldots,m \ \text{and} \ q=\ell+1,\ldots,n\}$ spans $\mathcal E$ and it is linearly independent. Therefore, $$\dim\mathcal E=m(n-\ell)\implies \dim\mathcal E=\dim (W)\left(\dim (V)-\dim(U)\right).$$