# Prove that there exists $T \in \mathcal{L}(V, W)$ such that $\operatorname{null}(T) = U$ if and only if $\dim (U) \ge \dim (V) - \dim (W)$

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

The answer to this question partially makes sense, particularly the forward direction in which we assume $$\operatorname{null}(T) = U$$. However, the other direction does not, here is the answer given:

Suppose that $$\dim(U) \ge \dim(V) - \dim(W)$$. Let $$(u_1, \ldots, u_m)$$ be a basis of $$U$$. Extend to a basis $$(u_1, \ldots, u_m, v_1, \ldots, v_n)$$ of $$V$$. Let $$w_1, \ldots, w_p$$ be a basis of $$W$$. For $$a_1, \ldots, a_m, b_1, \ldots, b_n \in \mathbf{F}$$ define $$T(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n)$$ by:

$$T(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n) = b_1w_1 + \ldots + b_nw_n$$

Clearly $$T \in \mathcal{L}(V, W)$$ and $$\operatorname{null}(T) = U$$.

I don't see why $$\operatorname{null}(T) = U$$. Additionally, I do not see the reasoning for defining the linear mapping as it is - what is the thought process behind choosing it to be that? Why does it map to $$b_1w_1 + \ldots + b_nw_n$$ and not use another constant (i.e. $$c_1w_1 + \ldots + c_nw_n$$ where $$c_1, \ldots, c_n \in \mathbf{F}$$)?

Another way to write it:

Define $$T : V \to W$$ by $$T(u_i) = 0_W$$ for $$i \in \{1,\dots,m\}$$, $$T(v_j) = w_j$$ for $$j \in \{1,\dots,n\}$$ and extend it by linearity.

Note that the hypotheses implies that $$p = \dim(W) \geq \dim(V) - \dim(U) = (m+n)-m = n,$$ so choosing $$w_1,\dots,w_n$$ of $$w_1,\dots,w_p$$ makes sense.

Also, note that this satisfies your definition, for if $$a_1,\dots,a_m,b_1,\dots,b_n \in \mathbf F$$, then \begin{align} T(a_1u_1 + & \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n) \\ &= a_1T(u_1) + \cdots + a_mT(u_m) + b_1T(v_1) + \cdots + b_nT(v_n) \\ &= b_1w_1 + \cdots + b_nw_n. \end{align}

Now, in one hand, it is easy to see that $$U \subseteq \operatorname{null}(T)$$ since every $$u \in U$$ can be written as a linear combination of $$u_1,\dots,u_m$$. On the other hand, let $$v \in V$$ and write it as $$v = c_1u_1 + \cdots + c_mu_m + d_1v_1 + \cdots + d_nv_n$$ for some $$c_1,\dots,c_m,d_1,\dots,d_n \in \mathbf F$$. If $$v \in \operatorname{null}(T)$$, then $$0_W = T(v) = d_1w_1 + \cdots + d_nw_n$$ and since $$w_1,\dots,w_n$$ are linearly independent, $$d_1 = \cdots = d_n = 0$$. So $$v = c_1u_1 + \cdots + c_mu_m \in U.$$

Let's first consider this definition: $$T(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n) = b_1w_1 + \ldots + b_nw_n.$$ This definition only really makes sense because $$(u_1, \ldots, u_m, v_1, \ldots v_n)$$ is a basis for $$V$$. So, any vector $$x \in V$$ can be expressed uniquely in the form $$x = a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n.$$ Now, if $$x \in U$$, then $$x$$ must uniquely take the form $$x = a_1u_1 + \ldots + a_mu_m = a_1u_1 + \ldots + a_mu_m + 0v_1 + \ldots + 0v_n,$$ since $$(u_1, \ldots, u_m)$$ is a basis for $$U$$. So, according to our definition of $$T$$, for $$x \in U$$, we have \begin{align*} T(x) &= T(a_1u_1 + \ldots + a_mu_m) = T(a_1u_1 + \ldots + a_mu_m + 0v_1 + \ldots + 0v_n) \\ &= 0w_1 + \ldots + 0w_n = 0. \end{align*} So, $$U \subseteq \operatorname{Null} T$$.

Conversely, suppose $$x \in \operatorname{Null} T$$. We still know $$x$$ is in the form $$x = a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n,$$ but this time we know that $$0 = T(x) = b_1w_1 + \ldots + b_nw_n.$$ Therefore, $$x = a_1u_1 + \ldots + a_mu_m + 0 \in U,$$ completing the proof that $$U = \operatorname{Null} T$$.

Why define it with $$b_1, \ldots, b_n$$ instead of $$c_1, \ldots, c_n$$? Well, remember that $$b_1, \ldots, b_n$$ are not constants, they are placeholder variables. $$T$$ is defined by expansion with respect to the basis $$(u_1, \ldots, u_m, v_1, \ldots, v_n)$$, and the way that the author has chosen to denote such an expansion is by $$x = a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n.$$ Thus, $$b_1, \ldots, b_n$$ are defined implicitly as (linear) functions of $$x$$, taking the vector $$x$$, and returning the coordinate of the corresponding basis vector $$v_i$$.

To simply replace them with $$c_1, \ldots, c_n$$, without defining them somehow, would mean the transformation is ill-defined. What are $$c_1, \ldots, c_n$$ in this context? How does the value of $$x$$ change them?

You could replace $$b_1, \ldots, b_n$$ with certain functions of $$b_1, \ldots, b_n$$ to obtain an equally valid construction $$T'$$ such that $$\operatorname{Null} T' = U$$. For example, the following $$T'$$ will also work: $$T'(a_1u_1 + \ldots + a_mu_m + b_1v_1 + \ldots + b_nv_n) = b_nw_1 + \ldots + b_1w_n.$$ Such functions are usually not unique!

• I don't quite understand this part: $T(x) = T(a_1u_1 + \ldots + a_mu_m = a_1u_1 + \ldots + a_mu_m + 0v_1 + \ldots + 0v_n) = 0v_1 + \ldots + 0v_n = 0$. Could you explain the steps to get this? Additionally, I don't understand $0 = T(x) = b_1v_1 + \ldots + b_nv_n$. Why do both equal $b_1v_1 + \ldots + b_nv_n$ and not $b_1w_1 + \ldots + b_nw_n$? Jul 24 '20 at 13:34
• This answer was really a mess, sorry. I've edited some points. I don't know if it's the multiple errors in the line you're quoting, but the first part is simply the definition of $T$. Remember, $x$ can be expressed uniquely in terms of the basis of $V$, and$$a_1u_1+\ldots+a_mu_m+0v_1+\ldots+0v_n$$is one such linear combination. The rest is according to the definition of $T$ (well, now it is, given I've corrected it). Hopefully the other point is clear now that I'm using the $w$s instead of the $v$s. Jul 24 '20 at 16:06

First of all your proof does not emphasis where the assumption is used.
To define a linear map $$T$$, it is sufficient to assign the images of members of a basis. Now we want $$\text{null}(T)=U$$, so first we take a basis $$\mathcal{B}_0$$ of $$U$$. Then extend $$\mathcal{B}_0$$ to a basis $$\mathcal{B}$$ of $$V$$. Now to have a linear map $$T$$ with $$\text{null}(T)=U$$ we have to assign each member of $$\mathcal{B}_0$$ to $$0$$ and rest of the members of the $$\mathcal{B}$$ have to assign so that $$\{T(v):v\in\mathcal{B}\smallsetminus\mathcal{B}_0\}$$ is linearly independent in $$W$$. Now the assumption $$\dim(U)\geq\dim(V)-\dim(W)$$ $$\implies \dim(W)\geq\dim(V)-\dim(U)$$ $$\implies \dim(W)\geq|\mathcal{B}\smallsetminus\mathcal{B}_0|$$. Thus this assures there are $$|\mathcal{B}\smallsetminus\mathcal{B}_0|$$ many linearly independent vectors in $$W$$. Which guarantees the existence of required $$T$$.