Extract 8.2
Context: Same college, tutor and students as Extract 8.1. The tutor explains that a subset S of a vector space V is a subspace of V (S<V) if S is a vector space and that a common way to prove that S<V is to use the Subspace Test (namely prove that 0ÎS and that addition and scalar multiplication are closed in S). The students cannot recall the SubspaceTest at first and the tutor suggests two examples: prove that the n¥n symmetric matrices and the matrices of Trace zero are subspaces of Mn(Â).The third example is to
prove that U={f:®Â, f(0)=f(1)} is a subspace of ÂÂ, the set of all real functions from  to Â.
The students look as if they are not familiar with ÂÂ.
The Episode:
Tutorials 1, 2 and 4
Note: In the following I have summarised Tutorials 1, 2 and 4 and highlighted some instances in each one of them.
In Tutorial 2, even though it turns out that the students do not know what ÂÂ is and have not realised that UÍÂÂ, Abidul mechanically suggests checking out the two conditions of the Subspace Test. The tutor realises their unease with ÂÂ and asks what is the zero element of ÂÂ. She reminds them that the zero element is an element of U and she asks them what is the property it has to satisfy. Silence follows. Eleanor says about the zero element of ÂÂ that 'It will stay the same'. The tutor disagrees and switches to talking about a general vector space. To their silence the tutor writes down a+0=0+a=a and asks them to think in terms of the zero element z in U and ÂÂ. Abidul says z(x)=x and Eleanor says 'nought'. The tutor agrees with Eleanor that it is the function of value 0 everywhere and stresses that it is a function they have been dealing with 'for ages' in Analysis.
In Tutorial 4, Beth remembers that the zero element of ÂÂ is the zero function when the tutor lists the axioms that a vector space has to satisfy (one of them is that there must be an element, called zero element, such that a+0=0+a=a, "aÎÂÂ).
In Tutorial 1 the students remain silent for longer and they cannot recall the property that the zero element of  must satisfy. The tutor switches to asking the same question in a general vector space and Patricia responds that in a general vector space 'if you add zero to any vector you end up with the same vector'. The tutor asks them to apply that in U but, since the students remain silent, she eventually defines the zero function (z: ® z(x)=0) and stresses that 'It is the same function from Analysis in a slightly more abstract context'.
In proving closure, the second condition of the Subspace Test, the students suggest evaluating af+bg, for f, gÎU and a, bÎÂ at 0 and 1 and proving that the two values are equal. In some cases the tutor needs to prompt the students who seem to confuse f with f(x). The sessions close with the tutor repeating the conditions of the Subspace Test.
The Episode: Tutorial 3
Camille is asking about ÂÂ:
C1: Is it the same as ÂÂ?
The tutor defines ÂÂ, 2Â and Â2 (starting from the definition of AB as the set of functions from the set B to the set A). Camille asks whether these are transposes. These are the mappings from one set to the other, replies the tutor, and are vector spaces over addition of functions and over scalar multiplication, both pointwise. She then asks the students what is the zero in ÂÂ but Camille is still trying to understand what ÂÂ consists of.
C2: It's a mapping from  to Â... and each element of  has a correspondent... with the mapping... the graph... it's a mapping from  to  so it's...
T1: No, no, no. Each mapping is a subset of ÂxÂ. It's not...
C3: And U is a subset...
T2: Yes. No. No, you are not looking at the individual f. Yes, it's true that f is a subset of ÂÂ. That's true but I'm not looking at the individual f. I'm lookiing at the set of all fs.[The tutor returns to the question about the zero element of ÂÂ] What's your mapping from  to  with that property? It's not in there. U is not a subset of Â2.
C4: Do you have an example of a function f that isn't in Â2?
T3: Yeah, I mean any of them... if you like you could have... cos2px... this is a function that... I mean to say that something belongs toÂ
2... what you are saying is that f belongs to Â2. I mean that f belongs to Â2. But it doesn't because that... to say that f belongs to Â2 is to say that f is an ordered pair. It's a set of ordered pairs.C5: (x, f(x))... but isn't f an ordered pair?
T4: No, f doesn't belong to Â2. f is a subset of Â2. And it belongs to the set ÂÂ. It belongs to a set of functions.
C6: It's very hard to imagine that... a set is usually a set of elements or matrices...
T5: Ah,... yes, that makes it harder. But I mean you could do it with equations.
The tutor stresses that all the things they are talking about here are casual things in Analysis and that their trouble is with the vector space context.
C7: So  is a subset of ÂÂ...
T6: The elements are...
C8: How about Â2®Â... is it a subset of this?
The tutor repeats the definition of ÂÂ and draws the paralled with AB, the set of functions between sets A and B, where A has k elements and B has m. She stresses that ÂÂ contains the functions they 've been dealing with usually in Analysis and that it is the vector space context used in the example that makes things look more complicated. She then repeats the question about the zero vector in ÂÂ: it has to be a function in ÂÂ that satisfies some kind of property. Then Camille asks:
C9: Is the zero vector a function?
T7: Zero vector is a function because all of them are functions here... all the elements...
C10: They are not vectors?
The tutor repeats the definition of the zero element of a vector space and explains that in this case this element is a function that has the properties of the zero element. The students listen and Camille says:
C11: So we are not looking for the zero vector anymore but for the zero function.
The tutor accepts that z(x)=0 is the zero element of ÂÂ and advises the students to think as if they are in Analysis. So zÎU, concludes the tutor. Camille quickly explains that zÎU because z(0)=0=z(1). She also checks out the condition for closure in U. The proof that U<ÂÂ is thus completed.
Return to Appendices for Chapter 8.