Tricky Problem: Prove range T = null ##\phi## when null T' has dim 1

[zenterix]

Homework Statement

Suppose $V$ and $W$ are finite-dimensional, $T\in L(V,W)$, and there exists $\phi\in W'$ such that null $T'$ = span($\phi$). Prove that range $T$ = null $\phi$.

Relevant Equations

null $T'$ = span($\phi$)

This is problem 28 from chapter 3F "Duality" of Axler's Linear Algebra Done Right, third edition.

I spent quite a long time on this problem, like a few hours. Since there is no available solution, I am wondering if my solution is correct.

One assumption in this problem is that $\text{null}(T')=\text{span}(\phi)$ where $\phi\in W'$

We know from a previous theorem that

$$\text{null}(T')=(\text{range}(T))^0\tag{2}$$

The reason for this is as follows

$\text{null}(T')$ is formed by $\phi\in W'$ such that

$$\phi(\text{range}(T))=0\tag{3}$$.

Thus, $$\text{null}(T')=\{\phi\in W':\phi v=0\ \forall v\in\ \text{range}\ T\}=(\text{range}(T))^0\tag{4}$$

Moving on, since a single non-zero vector is linearly independent (l.i.) then $\phi$ is a basis for $\text{null}(T')$. All other $\alpha\in\ \text{null}(T')$ are a multiple of $\phi$.

Note that for each such $\alpha\in\ \text{null}(T')$, (4) tells us that every element in $\text{range}\ T$ is in $\text{null}\ \phi$. That is, $$\text{range}(T) \subseteq\ \text{null} (\phi)\tag{5}$$

Claim: all $\alpha\in\ \text{null}(T')$ have the same nullspace.

Proof

By assumption, $\text{null}(T')=\ \text{span}(\phi)$.

Then, $\forall\ \alpha\in\ \text{null}\ T'\implies \alpha=\lambda\phi,\ \lambda\in\mathbb{F}$.

Let $w\in\ \text{null}(\alpha)$. Then

$$\alpha (w) = (\lambda\phi)(w) = 0\implies \phi(w)=0\tag{6}$$

$$w\in\ \text{null}(\phi)\tag{7}$$

Now, suppose that $w\in \text{null}(\phi)$. Then

$$\phi(w)=0 \implies \alpha(w) = (\lambda \phi)(w) = 0$$

$$w\in\ \text{null}(\alpha)\tag{8}$$

Therefore, we can infer that $\text{null}(\alpha)=\text{null}(\phi)$, and since this is true for a generic $\alpha\in\ \text{null}(T')$ then it is true for them all.

Suppose, for proof by contradiction, that $\text{range}(T) \neq\ \text{null}(\phi)$. That is, there is some $w\in\ \text{null}(\phi)$ that is not in range$(T)$.

Then, $\phi(w)=0$, which means that not only does $\phi$ annihilate range$(T)$, it also annihilates the subspace span$(w)$.

Now, as we proved above, all $\alpha\in\ \text{null}(T')$ have the same nullspace and thus

$$\forall\ \alpha\in\ \text{null}(T) \implies\ \alpha w=0\tag{9}$$

Consider a $\beta\in W'$ such that $\forall\ x\in\ W$ we have

$$\beta x = \begin{cases} \phi x,\ \text{if}\ x\in\ \text{range}(T) \\ 1,\ \text{if}\ x\in\ W \backslash\text{range}(T) \end{cases}\tag{10}$$

Then, $\beta\in\ (\text{range}(T))^0=\text{null}(T')$.

But $w\notin\ \text{null}(\beta)$, which contradicts the fact that all linear functionals in $\text{null}(T')$ have the same nullspace which includes $w$ since by assumption $\phi (w)=0$.

Thus, but proof by contradiction, we infer that $$\text{range}\ T =\ \text{null}\phi$$.

 

[nuuskur]

I don't have Axler on hand, but I assume
$$T' : W' \to V',\quad (T'w^*)v := w^* (Tv),\quad w^*\in W', v\in V.$$

(6) You use that $\alpha (\phi w) =0$ for ALL $\alpha$, therefore $\phi w = 0$. Otherwise, proof of Claim is correct.

You defined $\beta$ in (10), but is it well defined? It's not obvious to me. For instance, if $x+y\in T(V)$, then $\beta (x+y) = \phi (x+y) = \phi x + \phi y$ ..but then what?

 

[pasmith]

Your proposed $\beta$ is not linear (what is $\beta(2x)$ if $\newcommand{\range}{\operatorname{range}}x \notin \range T$?) and is therefore not in $W'$.

I think perhaps you meant to construct $\beta$ through its action on a basis $B$ of $W$, obtained by extending a basis of $\range T$ and such that $w \in B$. But really it's only necessary to specify $\beta(\range T) = \{0\}$ and $\beta(w) = 1$.

 

[nuuskur]

You already know that $\mathrm{span}\phi=\mathrm{Ker}T' = T(V)^0 = \{w^*\in W' \mid w^*\vert_{T(V)}=0\}$. By assumption all such $w^*$ are multiples of $\phi$. So you are to conclude that if $\lambda \phi \vert_{T(V)} = 0$ for all $\lambda\in\mathbb K$, then $\mathrm{Ker}\phi = T(V)$.

 

[zenterix]

Your proposed $\beta$ is not linear (what is $\beta(2x)$ if $\newcommand{\range}{\operatorname{range}}x \notin \range T$?) and is therefore not in $W'$.

I think perhaps you meant to construct $\beta$ through its action on a basis $B$ of $W$, obtained by extending a basis of $\range T$ and such that $w \in B$. But really it's only necessary to specify $\beta(\range T) = \{0\}$ and $\beta(w) = 1$.

Indeed. How about if I define it as follows

Let $w_1,...,w_n$ be a basis of range $T$.

Extend this to a basis of $W$: $w_1,...,w_n,w_{n+1},...,w_m$.

Consider $\beta\in W'$ defined by

$$\beta(w_i)=\phi(w_i)=0,\ \text{for}\ i=1,...,n$$

$$\beta(w_i)=1,\ \text{for}\ i=n+1,...,m$$

I think the rest of the proof still holds. That is

1) $\beta$ still annihilates range $T$, thus $\beta\in (\text{range})^0=\text{null}(T')$.

2) This is a contradiction since $\beta$ does not map $w$ to zero but instead maps it to some non-zero number.

 

[zenterix]

I don't have Axler on hand, but I assume
$$T' : W' \to V',\quad (T'w^*)v := w^* (Tv),\quad w^*\in W', v\in V.$$

(6) You use that $\alpha (\phi w) =0$ for ALL $\alpha$, therefore $\phi w = 0$. Otherwise, proof of Claim is correct.

You defined $\beta$ in (10), but is it well defined? It's not obvious to me. For instance, if $x+y\in T(V)$, then $\beta (x+y) = \phi (x+y) = \phi x + \phi y$ ..but then what?

Let me try that portion of the proof again:

By assumption, $\text{null}(T')=\text{span}(\phi)$.

Every $\alpha\in\ \text{null}(T')$ is thus a scalar multiple of $\phi$.

Let $\alpha\in\ \text{null}(T')$.

Then, $\alpha=\lambda\phi$ for some $\lambda\in\mathbb{F}$.

Let $w\in\ \text{null}(\alpha)$.

Then, since $\alpha(w)=0$ we have that $(\lambda\phi)(w)=0$ and by linearity of $\phi$ we have $\lambda\phi(w)=0$ so $\phi(w)=0$.

Thus, $w\in\ \text{null}(\phi)$.

As for the definition of $\beta$, it is indeed incorrect. See the post just above this one with the corrected version.

 

[nuuskur]

But $w\notin\ \text{null}(\beta)$, which contradicts the fact that all linear functionals in $\text{null}(T')$ have the same nullspace which includes $w$ since by assumption $\phi (w)=0$.

This works, since $\beta\vert _{T(V)} = 0$ by definition of $\beta$. Hence, $\beta = \lambda \phi$ and $1=\beta w = \lambda \phi w = 0$, a contradiction.

 

[zenterix]

You already know that $\mathrm{span}\phi=\mathrm{Ker}T' = T(V)^0 = \{w^*\in W' \mid w^*\vert_{T(V)}=0\}$. By assumption all such $w^*$ are multiples of $\phi$. So you are to conclude that if $\lambda \phi \vert_{T(V)} = 0$ for all $\lambda\in\mathbb K$, then $\mathrm{Ker}\phi = T(V)$.

Is this an entire proof or is this a hint. It is not clear if in the last sentence you are concluding a proof or giving a hint.

 

[nuuskur]

Is this an entire proof or is this a hint. It is not clear if in the last sentence you are concluding a proof or giving a hint.

It's a rewording of the problem statement in a way that makes it immediately clear why said equality holds.

As for #5, the take away is to define linear maps on bases, then you do not have to worry about whether they are well defined.

Now retrace your steps and identify where you explicitly made use of finite dimension. You do not require finite dimension to define $\beta$.