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 $c_1, \ldots, c_n \in \mathbf{F}$)?


3 Answers 3


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!

  • $\begingroup$ 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 $b_1w_1 + \ldots + b_nw_n$? $\endgroup$
    – gorgonolo
    Jul 24, 2020 at 13:34
  • 1
    $\begingroup$ 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. $\endgroup$
    – user810049
    Jul 24, 2020 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 $T$.


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