Extract 8.7
Context: The tutor says that students Cathy and George generally did not have many problems with LA7. Particularly in LA7.35 Cathy, he adds, offered a 'beautiful solution'. Cathy says she is not very contented with the amount of 'rigour' she used. The tutor invites her to the b/b and asks her to present her proof fully.
The Episode:
Cathy writes on the b/b her definition of ImT, where T is a linear transformation of a vector space V on V( V 'transforms itself' says Cathy).The tutor sounds discontented with her definition (Note: unfortunately I only have the modified and final version of Cathy's writing and not what she initially wrote on the b/b) because she has been using the same letter for the elements of ImT and V. 'They all come from V' says Cathy, but the tutor explains that this might lead to confusing situations such as Tx=x where x is meant differently. He also stresses that what she has been saying is not what she has been writing: 'for v in V' usually means 'for all v in V'. He then defines ImT while Cathy is writing the definition on the b/b:
ImT={wÎV: w=T(v) for some vÎV}
She then says that her solution ' might be wrong now' and defines ImT2 on the b/b as
ImT2={T(w) for wÎV: T(w)=T2(v) for some wÎV}
The tutor then asks for her notion of T2. ' Isn't T square applying T twice?' she says. He suggests thinking of T2 as just one transformation of V on V. He then turns to her definition of T2: 'If you read it in English what does it mean?' he asks. Silence. To him her definition 'doesn't make much sense'. She changes her writing to
ImT2={sÎV: s=T2v for some vÎV}
with which he agrees. She then claims that 'ImT2 has less vectors than ImT' but when asked to prove her claim she says she is not sure; 'kerT is easier to do formally' she claims. The tutor asks her to prove kerT£kerT2 first, if she likes. Cathy then defines kerT as
kerT={vÎV: T(v)=0}
and says:
C1: If T sends some v to zero, then T2 will send the same vector to zero, so it can never be less but it could be more... Is that formal enough?
The tutor says it is ' beautifully literary' and explains that the reason that all they need to do is prove that kerTÍkerT2 is because both kerT and kerT2 are subspaces of V. He then proves that kerTÍkerT2 and the students similarly prove that ImT2ÍImT (by taking a vector in ImT2 and proving that it belongs to ImT).
Now they turn to the part of LA7.35 which is about the equivalence of propositions a, b and c. Cathy in her writing had suggested proving that aÞbÞcÞa. The tutor says her suggestion is sufficient and the student nods in agreement. George says he proved that aÛc but could not decide how to go on and stopped at aÞb. The tutor recommends perseverence. He also points out that b and c look more or less the same whereas a looks different. That means, he adds, that they are probably of the same degree of difficulty and he suggests proving bÞc, cÞb, bÞa, aÞb. He then asks whether they know anything about the dimension of image and kernel. Cathy replies: the Rank and Nullity Theorem (RNT). The tutor writes
dimkerT+dimImT=dimV
and asks: 'What this can tell us about the dimension of T2?'. Silence follows. The tutor suggests that they 'have to combine somehow' the information they have about images and kernels. Cathy then rearranges the RNT in terms of dimImT and sees that b implies that dimImT=dimImT2. George says that if a subspace is contained in another then the dimension of one is £ the dimension of the other. This is obvious, replies the tutor, and reminds them of ImT2ÍImT and kerTÍkerT2. Cathy then points out that then ImT2=ImT. By the same argument, says George, cÞb.
About bÞa Cathy says:
C2: If the intersection wasn't zero, then kerT2 would be bigger, wouldn't it?
Silence follows. George tries to start another suggestion but the tutor stays with Cathy's idea, that is to start from ImTÇkerT¹0, which he asks her to pursue. Cathy then says:
C3: If ImT intersection kerT was not zero... then kerT2 would be bigger than kerT.
The tutor says he approves and asks for a proof. Cathy hesitantly says 'It seems true but... '. The tutor asks for a proof again and Cathy suggests taking vÎImTÇkerT, v¹0. Then Tv=0. She also says that then T2v=0 but the tutor points out that this is not helpful and that it is true anyway because kerTÍkerT2. George then points out that 'v is special. It is in ImT'. The tutor asks what this means but the students remain silent. Then George suggests 'apply translation T to... '. He is interrupted by the tutor who reminds him that vÎImT is an 'existential' statement: there exists wÎV such that v=Tw. Cathy then suggests:
C4: Transform it again and then it is nought so they are not equal.
The tutor rephrases that to: 'then wÎkerT2 but wÎkerT'. This contradicts b and therefore Cathy's instinct was right. She also says that they have to prove V=kerT+ImT but the tutor points out this is trivial because if two subspaces intersect trivially then the dimension of the sum is the sum of the dimensions. So bÞa.
For aÞb Cathy suggests:
C5: Go backwards, not backwards as such but if it equals the direct sum and then equate this to the dimension of the brackets, so by the inequality, the intersection is zero.
The tutor says that he thinks that 'the first half of your strategy works. Second half is not clear'. It's not 'what he had in mind', he adds, but he suggests pursuing Cathy's idea.
C6: If the intersection is zero then surely kerT cannot be bigger than kerT2...
T1: Every time you say 'surely'... you give an argument-by-sweet-'n-reasonable Cathy. I mean an argument by voice is an argument by intimidation but you make me say: where is the proof? Surely I agree with you but...
She then suggests: 'Can you say let w be a vector in kerT2 but not in kerT and show that this is a contradiction?'. He agrees but he also adds that he 'can see no negative fact implied by her suggestion so there is no point pursuing contradiction'. Contradiction, he explains, 'only works well if you can use well a negative fact'. He then presents the 'positive' proof, that is one which is not by contradiction, he had in mind (fig.7).
Return to Appendices for Chapter 8.