Context: The Extract presented here follows Extract 8.4 where the notions of a basis and the dimension of a vector space were exemplified in finding the dimension of the vector spaces given in LA6.26. Moving on to LA6.29 the tutor stresses that behind the proof for this question lies an important theorem about finding bases for a vector space V, two subspaces X and Y and the sum X+Y. To his question whether they can recall the theorem he is talking about, students Jack and Andrew protest about the absence of examples in both Continuity and Differentiability and Linear Algebra courses ('we've never found a basis for anything'). The tutor insists on his question about recalling the theorem and Jack dictates

Part a: The tutor reads LA6.29ii and asks what they know about X+Y. Jack replies that it 'has a greater dimension' but is then silent. The tutor asks them to consider the information they have: X and Y are of dimension n-1 in a space of dimension n. They are still silent. The tutor prompts them: 'Jack was right about "greater" dimension. It is strictly so'. Silence. What does it mean 'bigger than n-1'? Silence. They both whisper it could be 'anything' and the tutor reminds them that X+Y cannot be of dimension greater than n. Andrew then deduces that dim(X+Y)=n and the tutor says that then X+Y=V. By replacing the known dimensions in the above formula, dim(XÇY)=n-2. The tutor says that 'neither of them looks terribly pleased'. They 'look as if they do not accept it', he says. Jack replies: 'No, I accept it. It's just based on a formula we haven't proved'. In a while they will, promises the tutor.

Part b: First the tutor draws their attention to 'a mistake there that everybody makes' including Andrew: given a vector space Z of dimension n and two subspaces X and Y of dimension n-1, Andrew considered a basis for Z, and then produced 'the' basis for X by removing one vector and produced 'the' basis for Y by removing another. The tutor proves (employing the counterexample of ²⁷and ²⁶) that a basis for a vector space does not necessarily contain a basis for a subspace; on the contrary it is true that any linearly independent subset of elements of Z can be extended to a basis for Z. Also he stresses that phrasing such as 'the basis of a subspace' is misleading because a basis is not unique.

Part c: The tutor then suggests trying to construct a basis for X+Y that contains a basis for X and a basis for Y. Jack suggests 'We choose a basis for X and Y first' but the tutor asks him to 'forget about Y' for now and start from the idea that we can extend any linearly independent set to a basis for Z. Andrew wonders how starting from a basis for X will allow them to find a basis for Z that contains both a basis for X and a basis for Y. Jack whispers a suggestion: start from a basis for X and then consider the intersection. The tutor agrees: they consider a basis for XÇY and extend to a basis for X and to a basis for Y ( fig.5). Then Jack claims that {u₁,..., u_p, x₁,..., x_q, y₁,..., y_r} will be a basis for X+Y. The tutor agrees and they prove that this set is linearly independent.

Once linear independence is proved Jack notes that this actually was a 'quite straightforward' proof for the formula. The tutor agrees and adds that this not only verifies the formula but provides with a method to extend linearly independent sets to bases for the vector space. Then Jack wonders: 'Can we extend it for more than two subspaces?'. Andrew is sceptical: he is concerned whether 'the situation cannot be treated the same way. He suggests considering X to be the sum of two subspaces A and B and check out the consequences. Jack replies that his question was of a more general nature: what happens if we take three subspaces A, B and C instead of two. Andrew asks him to 'show me you can do it then'. The tutor thinks it is an interesting question and asks them about the dimension of A+B+C. Andrew thinks it will be 'the three dimensions minus the dimensions of the intersections'. The tutor responds that Andrew's suggestion sounds reasonable but reminds them that A, B and C are subspaces, not subsets and therefore they 'behave more complexly'. So the 'analogue' might not work so perfectly. He nevertheless invites them to experiment with, for instance, ²⁷ and requests they move on to other problems.