1
$\begingroup$

Excerpt from text:

3.109 The range of T'

Suppose V and W are finite-dimensional and T $\in$ L(V,W). Then

range T' = $(null\;T)^0$

Proof

First suppose $\phi$ $\in$ range T. Thus there exists $\psi$ $\in$ W' such that $\phi$ = T'($\psi$). If v $\in$ null T, then

$\phi$(v) = (T'($\psi$))v = ($\psi$ $\circ$ T)(v) = $\psi$(Tv) = $\psi$(0) = 0

Hence $\phi$ $\in$ $(null\;T)^0$. This implies that range T' $\subset$ $(null\;T)^0$.

We will complete the proof by showing that $\mathbf range\;T'$ and $(null\;T)^0$ have the same dimension. To do this, note that

dim range T' = dim range T

dim range T' = dim V - dim null T

dim range T' = dim $(null\;T)^0$

Questions: I am not able to understand the intuition behind this theorem. $\mathbf range\; T'$ is the set of functionals from $\mathbf W'$ which take any $\mathbf w$ from $\mathbf W$ to $\mathbb F$ ($\mathbf w$ is produced by $\mathbf T$ here; T'($\phi$) = $\phi \circ T$ for $\phi$ $\in$ W'. $\mathbf Tv$ gives a $\mathbf w$ which is then given to $\phi$() ).

On the other hand, annihilator of $\mathbf (null\;T)$ is the set of all functionals from $\mathbf V'$ which take any vector from the set $\mathbf (null\;T)$ to 0. How can these two different sets be equal? Would'nt it mean that all functionals from T' are functionals that produce zero?

I am confused about the subset notation in the first part of the proof; " range T' $\subset$ $(null\;T)^0$ " . My current understanding says all functionals in V' which would take vectors from (null T) to zero will be part of range T' . That is, $(null\;T)^0$ $\subset$ range T' . What am I missing here?

(Note: I am assuming $\mathbf (null\;T)$ $\subset$ $\mathbf V$, since annihilator is defined with a subset/superset relation. Have I taken the right superset? )

Also, in the second part of the proof, the author just proves the dimensions of range T' and $(null\;T)^0$ are same. But, how can proving the dimensions being equal prove both spaces are equal and have same elements?

Could anyone kindly explain what I have misunderstood here?

$\endgroup$
1
$\begingroup$

Note that, since $T : V \to W$, we have $T' : W' \to V'$, and hence $\operatorname{range} T'$ is a subset of $V'$, not $W'$. In order for $T'(\psi) = \psi \circ T$ to make sense, the domain of $\psi$ must be from the codomain of $T$, i.e. $W$, not from $V$. The resulting functional therefore has a domain of $V$.

Other than that, I'm not really sure what the source is of your confusion. I usually find people respond well to a worked example, at least to make certain the concepts involved are being understood.

Let's take, for example,

$$T : \mathbb{R}^3 \to \mathbb{R}^2 : (x, y, z) \to (x - y, 3x + y + z).$$

In this case, $V = \mathbb{R}^3$ and $W = \mathbb{R}^2$. Let's start by computing $(\operatorname{null} T)^0$. We have \begin{align*} &(x, y, z) \in \operatorname{null} T \\ \iff \, &T(x, y, z) = (0, 0) \\ \iff \, &(x - y, 3x + y + z) = (0, 0) \\ \iff \, &\exists t \in \mathbb{R} \, : \, (x, y, z) = (t, t, -4t). \end{align*} That is, $$\operatorname{null} T = \lbrace (t, t, -4t) : t \in \mathbb{R} \rbrace = \operatorname{span} (1, 1, -4).$$ Our nullspace is simply this line. As for the annihilator, let us suppose that $\psi(x, y, z) = ax + by + cz$ is a functional in $(\operatorname{span} (1, 1, -4))^0$. That is, $$\psi(1, 1, -4) = 0 \iff a + b - 4z = 0.$$ That is, $$(\operatorname{null} T)^0 = \lbrace (x, y, z) \mapsto ax + by + cz : a + b - 4z = 0 \rbrace.$$ One way to think about functionals is in terms of their nullspaces, all of which are hyperplanes (in this case, just planes, since we are in $\mathbb{R}^3$). The annihilator of $\operatorname{null} T$ is just all of the functionals in $(\mathbb{R}^3)'$ that contain the line $\operatorname{span} (1, 1, -4)$ in their nullspace. Think about all the planes you can form by rotating the plane about this line. These planes represent the functionals in our annihilator.

Now, let's compute $\operatorname{range} T'$. If we take an arbitrary functional $\psi \in (\mathbb{R}^2)'$. Then, there exist $a, b \in \mathbb{R}$ such that $\psi(x, y) = px + qy$. We then have \begin{align*} (T(\psi))(x, y, z) &= (\psi \circ T)(x, y, z) \\ &= \psi(x - y, 3x + y + z) \\ &= p(x - y) + q(3x + y + z) \\ &= (p + 3q)x + (q - p)y + qz. \end{align*} So, the above functional belongs to $(\operatorname{null} T)^0$ if and only if $$(p + 3q) + (q - p) - 4q = 0,$$ which is true, so $\operatorname{range} T' \subset (\operatorname{null} T)^0$.

I think I'll leave it there. The other subset inclusion holds too, but I'll leave you to work it out if you're interested. The theorem is very simple, but the concepts behind them aren't. Hopefully an example has shed some light on these concepts, which might help you grasp the theorem.

$\endgroup$
  • $\begingroup$ I have updated the question. Could you kindly have a look at it now? Basically, my current understanding says (null T)^0 should be subset of range T'. How is it the other way round? And the second part of the proof is done by proving the dimensions are same. How does dimensions being same imply the spaces being equal? $\endgroup$ – Satheesh Paul Jun 30 '18 at 15:23
  • $\begingroup$ @SatheeshPaul The proof shown shows that $\operatorname{range} T' \subset (\operatorname{null} T)^0$, because it shows that an arbitrary element of $\operatorname{range} T'$ is in $(\operatorname{null} T)^0$. This is what $\subset$ means. $\endgroup$ – Theo Bendit Jun 30 '18 at 15:59
  • $\begingroup$ As for your other question, if you have $W$ is a subspace of finite-dimensional $V$ and their dimensions agree, then $W = V$. This is a consequence of two facts: every linearly independent list can be extended to a basis, and every basis for a space has the same length. So, a basis for $W$, that has $\operatorname{dim} V$ elements in it, can be "extended" to a basis, which must have $\operatorname{dim} V$ elements in it. So, no vectors could have been added, which means the basis for $W$ was a basis for $V$ all along. Its span must be $V$ and $W$, so $V = W$. $\endgroup$ – Theo Bendit Jun 30 '18 at 16:01
  • $\begingroup$ @SatheeshPaul In future, I would ask another question instead of editing like this. You can always link back to your old question and ask follow-up questions. You can also include a link to the new question as a comment on answers, so that the same people who answered you before might answer again. $\endgroup$ – Theo Bendit Jun 30 '18 at 16:03
  • $\begingroup$ I seem to slowly understand the example you had worked out, but still one question remains. The operation of T' seems to filter out the planes which do not pass through (null T), right? How did the other planes vanish? For example; let's say T produces (1,1) [ (x−y,3x+y+z)=(1,1) ], how will this transform in the T' transformation [ (p+3q)x+(q−p)y+qz ]? I get that in T', we are going from 3D to 2D and back to 3D in the T' map hence not everything from the original space will remain at the end. But, guess I am confused with the details. $\endgroup$ – Satheesh Paul Jul 1 '18 at 16:56
1
$\begingroup$

$\text{range}(T')$ is the space of all functionals of the form $\phi \circ T$.
$(\text{null}(T))^{0}$ is the space of all functionals which map all of $\text{null}(T)$ to $0 \in \mathbb{F}$.
Note that the whole point of $\text{null}(T)$ is that it is the maximal subspace of $V$ which is sent to $0 \in W$ by $T$.

The inclusion $\text{range}(T') \subset (\text{null}(T))^{0}$ says "every functional of the form $\phi \circ T$ sends $\text{null}(T)$ to $0$.". But clearly, $T$ sends everything in $\text{null}(T)$ to $0$, and $\phi$ is linear so it sends $0$ to $0$ no matter what.

Does that help?

Response to the edit: When you write "My current understanding says all functionals in V' which would take vectors from (null T) to zero will be part of range T' ." it is not clear to me why you think this. The proof says, summarised:
Let $\phi \in \text{range}(T')$. Then [...] $\phi \in (\text{null}(T'))^{0}$.
This is equivalent to saying $\text{range}(T') \subset (\text{null}(T))^{0}$.
If I misunderstand you, and you mean that you believe the reverse containment is true for some other reason, then you may be correct. Indeed, the statement is true, and you may or may not believe it for a correct reason (I can't tell).

The remainder of the proof: Take for granted that the proven inequality really is $\text{range}(T') \subset (\text{null}(T))^{0}$. The author then shows that $\text{dim } \text{range}(T') = \text{dim} (\text{null}(T))^{0}$. This implies that $\text{range}(T') = (\text{null}(T))^{0}$.
Why? Because of this theorem:

Let $W$ be a finite dimensional vector space and $V \le W$. If $\text{dim } V = \text{dim } W$ then $V=W$.

Proof: Suppose $V$ is a strict subspace of $W$. Choose a vector $w \in W$ with $w \notin V$. Take a basis of $V$, and add $w$ to it. Since $w \notin V$, this set is linearly independent in $W$ (can you prove this?). Therefore $\text{dim }W \ge \text{dim }V + 1$.

$\endgroup$
  • $\begingroup$ I have updated the question. Could you kindly have a look at it now? Basically, my current understanding says (null T)^0 should be subset of range T'. How is it the other way round? And the second part of the proof is done by proving the dimensions are same. How does dimensions being same imply the spaces being equal? $\endgroup$ – Satheesh Paul Jun 30 '18 at 15:24
  • $\begingroup$ @SatheeshPaul I've tried to address your edits. $\endgroup$ – preferred_anon Jun 30 '18 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.