1. Bases and coordinates

This section is somewhat long and technical: I explain essentially everything about bases and coordinates in linear algebra.

It’s often useful for intuition and concreteness to work things out in terms of a basis. However, you should also be cautious about bases as they obscure the theory and the actual structure: if you can do something in terms of a basis, doing it without choosing a basis will generally clarify matters and deepen your understanding.

The most important theoretical task when doing a construction in terms of a basis is to ask what happens when we change the basis. Why does this matter so much? Because if we do something in terms of a basis, and know what happens when we change the basis, then we know the whole story independent of choice of basis.

1.1. Notation. We potentially have a great many bases at once: to understand all the possibilities, we’ll want to understand two spaces, V and W, and maybe two bases on each, and for any basis we also have a dual basis on V^{∗}, so we could get 8 bases! It’s easy to get confused.

Further, the “correct” mathematical notation can obscure matters further: one often denote the basis ⟨e_{1}, …, e_{n}⟩ by B_{e} or some such, which is pretty verbose.

There’s lots of conventions, especially for duals.

One might use e_{1}, …, e_{n} as a basis for V, and f_{1}, …, f_{m} as a basis for W.

Or e_{1}, …, e_{n} as a basis for V, and f_{1}, …, f_{n} as another basis for V. Or e_{1}, …, e_{n} as a basis for V, and e′_{1}, …, e′_{n} as another basis for V. For duals, one might use e_{i} as a basis for V, and e′_{i} (or e_{i}^{∗}, or f_{i}, or

f^{i}) as the dual basis for V^{∗} FIXME: be consistent below! Lastly, I use ϕ and ψ for the coordinates maps in analogy with

manifolds, where we use these for coordinate patches.

1.2. Categorical background. The key to understanding the relationship between (abstract) linear algebra and (coordinate) matrix algebra is:

“R^{n} and matrices are a skeletal subcategory of vector spaces and linear maps”

Given a category C, a skeletal subcategory S is specified by a collection of objects {A_{i}} (called the skeleton), with exactly one from each isomorphism class of C. FIXME: Do we also need to specify isomorphisms ϕ_{X}: A_{i} )^{∼} X for all objects, or is this not part of the data? IE, for all object X ∈ C, there is a unique A_{i} such that X ≅ A_{i}. We let S be the full subcategory on {A_{i}}: that is, the objects are the {A_{i}}, and the maps are all maps^{1} between them.

Given a skeletal subcategory, you can “reduce most questions” to questions inside the skeleton. This statement is intentionally vague: it’s rather like “first-order logic”: any statement about a specific diagram can be so reduced, but questions about the category as a whole can’t be reduced: see below under “Limitations of skeletons”.

How does this work?

Given an object X ∈ C, we have ϕ_{X}: A_{i} ) X, so any question about the object X can be answered by answering the (transfered) question about its isomorphic object A_{i}. EG, the space of (orthonormal) 1frames in R^{n} is (naturally) homeomorphic to S^{n−1}, so the space of 1frames in any real vector space V of dimension n is homeomorphic to S^{n−1}.

Now given two objects X, Y ∈ C, with ϕ_{X}: A_{i} ) X and ϕ_{Y}: A_{j} ) Y, we have

(ϕ_{X}^{−1} × ϕ_{Y})_{∗} : Hom(A_{i}, A_{j}) )^{∼} Hom(X, Y)

so any question about the Hom-object can be reduced to within the skeleton.

For instance, every linear map between finite-dimensional vector spaces has a rank, by defining the rank of a matrix and representing the linear map as a matrix. We can also prove this directly though, and this is a good example of how intrinsic definitions offer more insight.

Similarly any other diagram can be reduced into the skeleton.

A caution: note that we also have EndX ≅ EndA_{i} by

(ϕ_{X}^{−1} × ϕ_{X})_{∗} : Hom(A_{i}, A_{i}) )^{∼} Hom(X, X)

but that we need to use the same identification each time; this was either part of the data of a skeletal subcategory, or is necessary for this statement to make sense.

1.3. Matrices as a skeletal subcategory for linear maps. We now work out what this abstraction means for the specific case of linear algebra.

So for vector spaces and linear maps, every vector space V is isomorphic to R^{n} for some n: in fact, for n = dim V.

1In the notion of a subcategory, we also want to allow ourselves to throw out maps. So for instance the category of vector spaces and injective linear maps is a subcategory of vector spaces and all linear maps

Similarly, Hom(V, W) is isomorphic to Mat(m, n) (m × n matrices, IE, m rows and n columns), where n = dim V, m = dim W (note the reversal: rows is output, columns is input), via R^{n} ≅ V and R^{m} ≅ W, and maps R^{n} ) R^{m} have natural matrix representations.

A choice of isomorphism R^{n} )^{∼} V is exactly the same as a choice of basis. The correspondence is given by:

• given a basis e_{1}, …, e_{n}, we get a map R^{n} ) V by (a_{1}, …, a_{n}) ∈ R^{n} ) a_{1}e_{1} + ⋯ + a_{n}e_{n} ∈ V. This is an isomorphism by definition of basis.

• given an isomorphism ϕ: R^{n} )^{∼} V, we get a basis e_{1}, …, e_{n} of V by e_{1} = ϕ(1, 0, …, 0), e_{2} = ϕ(0, 1, 0, dots, 0), etc. – just push the standard basis forward. This is a basis by definition of isomorphism.

Okay, so given a map T : V ) W, how do we express it in terms of a basis, from this point of view? A basis e_{1}, …, e_{n} of V corresponds to ϕ_{e} : R^{n} )^{∼} V, and a basis f_{1}, …, f_{m} of W corresponds to ψ_{f} : R^{m} )^{∼} W.

So we get a diagram

\xymatrix{R^{n} \ar[d]_{ϕ_{e}} & R^{m} \ar[d]^{ψ_{f}} \\ V \ar[r]^{T} & W}

and to get a matrix we need a map M: R^{n} ) R^{m}, IE, we need a map

\xymatrix{R^{n} \ar[d]_{ϕ_{e}} \ar@{-->}[r]^{M} & R^{m} \ar[d]^{ψ_{f}} \\ V \ar[r]^{T} & W}

“such that the diagram commutes”, IE it doesn’t matter which path you follow: the result is the same – this requirement just means that M represents T: it’s not some arbitrary map.

This is easy to get: it’s

M = ψ_{f}^{−1}Tϕ_{e}

just reverse ψ_{f}! This formula tells us a lot: see how changing the basis of the domain acts on the right, while changing the basis of the codomain acts on the left – but with an inverse. Each of these is reflected in the nuances of the computations.

Also, note that it says “our input is in terms of the basis e; our output is in terms of the basis f”: input comes in on the right and comes out on the left. Yes, this is confusing: it’s because we want to write T(x) (we want to write T on the left), which means the x comes in the right side of T and comes out the left: think of U(T(x)). In some places they do composition and matrix multiplication the other way.

Now write T in terms of other bases, e′ and f′, getting M′; clearly M′ = ψ_{f′}^{−1}Tϕ_{e′} How can we get M′ from M? Let’s look at fixing the base on V: algebraically, we need to multiply M on the right by B_{e′)e} = ϕ_{e}^{−1}ϕ_{e′}.

How can we interpret this? B_{e′)e} is the identity on V, with input written in basis e′, and output written e.

Note that this is just a map R^{n} ) R^{n}: we cannot interpret it as a map T : V ) V, written in some basis ⟨g_{i}⟩, because it would have to be of the form: ϕ_{g}^{−1}Tϕ_{g}.

However, what makes this confusing is that this very matrix admits two more interpretations, if we interpret it as a map V ) V in the basis ⟨e_{i}⟩ or ⟨e′_{i}⟩:

In the basis ⟨e_{i}⟩, we get:

B_{e′)e} = ϕ_{e}^{−1}ϕ_{e′} = ϕ_{e}^{−1}ϕ_{e′}ϕ_{e}^{−1}ϕ_{e}

In the basis ⟨e′_{i}⟩, we get:

B_{e′)e} = ϕ_{e}^{−1}ϕ_{e′} = ϕ_{e′}^{−1}ϕ_{e′}ϕ_{e}^{−1}ϕ_{e′}

IE, the map T_{e)e′} : V ) V, where T = ϕ_{e′}ϕ_{e}^{−1}, such that T(e_{i}) = e′_{i} (it sends e_{i} to the ith basis vector of R^{n}, which it sends to e′_{i})

FIXME: ack!!!!! Okay, so this matrix tells you how to rewrite coordinates for e′ as coordinates for e, but it represents (in both the basis e and the basis e′???) the map that sends e ) e′???

We work this out in much, much more detail below.

1.4. Dual of a basis. One interpretation of duals are as “coordinates”.

More correctly: “a element of V^{∗} is a coordinate on V, and a basis for V^{∗} is a system of coordinates on V”

Given a basis e_{1}, …, e_{n} of V, we get a dual basis f_{1}, …, f_{n} of V^{∗}, given by: “f_{i}(v) is the coefficient of e_{i} when you express v in terms of the basis e_{1}, …, e_{n}”

In symbols, if v = a_{1}e_{1}+⋯+a_{n}e_{n} (where a_{i} ∈ K), then f_{i}(v) = a_{i}. Note that each f_{i} depends on the entire basis e_{1}, …, e_{n}:

“you can’t take the dual of a vector – only the dual of a basis”

More formally, if you change a basis, leaving one vector fixed, the dual change of the dual basis needn’t leave the dual vector fixed.

This will be clearer when we discuss the effect of change of basis on the dual.

This point is particularly important in differential geometry: dx_{i} is canonical, but ∂/∂x_{i} isn’t (it depends on the whole basis – IE, it’s kinda bad notation)

1.5. Duals in coordinates. NB: dual map is transpose

1.6. Change of basis on a vector space. Given two bases e_{1}, …, e_{n} and f_{1}, …, f_{n} for a vector space V, we can ask how to switch between the two.

That, if v = a_{1}e_{1} + ⋯ + a_{n}e_{n} = b_{1}f_{1} + ⋯ + b_{n}f_{n}, then how can we express ⟨b_{i}⟩ in terms of ⟨a_{i}⟩? This is a question about the dual space, IE, it’s about coordinates, so we will demur. You’ll thank me for this

– this can get pretty confusing, with all the back and forth.

FIXME: Let’s first write the matrix M_{e)f;e} ; that is, the matrix that takes ⟨e_{i}⟩ to ⟨f_{i}⟩, expressed in the basis ⟨e_{i}⟩. That is, T_{e)f} , the linear transform of V defined by T(e_{i}) = f_{i} is well-defined; we’re just writing it in a basis.

1.7. Classification of maps V ) W. We will illustrate change of basis by recalling a classification theorem, and showing the analog for duals.

Theorem 1. Recall that maps V ) W are “classified” by their rank. In coordinates, this says that if T : V ) W has rank k, then there are bases for V and W such that the matrix for T has the form like:

(\begin{matrix}1 & 0 \\ 0 & 1 & 0 \\ & 0 & 1 & 0 \\ & & 0 & 0 & 0 \end{matrix})

IE, it has k ‘1’s on the diagonal, followed by ‘0’s on the diagonal, and ‘0’s everywhere else.

Proof.The proof is by row and column reduction. That is, if dim V = n and dim W = m, then a choice of basis for T expresses it as an m × n matrix: m rows (because that’s the output), and n columns (because that’s the input).

mneumonic for left and right: left action is by row operations; right action is by column operations left action act on column vectors; given a set of vectors, it acts on each column in the same way in particular, it doesn’t change the columns around, hence it must be acting by row operations dually for a right action

intrinsically defined □

A better formulation of the above result is:

Definition 1. Say that two diagrams T : V ) W and T′ : V′ ) W′ are equivalent if there are isomorphisms ϕ: V )^{∼} V′ and ψ: W ) W′ such that this diagram commutes:

\xymatrix{V \ar[r]^{T} \ar[d]_{ϕ} & W \ar[d]_{ψ} \\ V′ \ar[r]^{T′} & W′}

Even more formally, consider the category of functors

(· ) ·) ) (Vect, LinMap)

where objects are functors between these categories and morphisms are natural transforms of functors. This is called the quiver category, because it’s a bunch of arrows.

Concretely, an object is just “put a vector space at each dot, and a map on each arrow”, and a map is “maps between the corresponding objects such that everything commutes”.

Then two diagrams are equivalent if and only if they are isomorphic in the quiver category.

first do for two separate spaces, then for one space (Jordan is harder) (now that I have language of quivers, life is good)

Vectors, Covectors and Transposes (cute way of visualizing the duality: horizontal vectors and vertical vectors. Note that this requires a basis cute: this is actually identifying V = Hom(K, V) and V^{∗} = Hom(V, K) – that’s what the matrices mean) (transposes as making sense once you have a basis)

———————————————————– quadratic forms, symmetric, etc. maps preserving a basis are lame; basis gives inner product different bases give same inner product ...fibres are an O(n) foliation of GL(n) ...and get O(n) as subgroup preserving an inner product a choice of basis gives S_{n} < O(n) (heh: a clever student might ask: does every S_{n} < O(n) correspond to a basis? This is a very good question: it leads to representation theory, IE, how can groups (of symmetries) be realized as linear symmetries, IE symmetries of a vector space. Oh, and the answer is no – there’s lots of other interesting ways to realize S_{n} as a subgroup of O(n), and the connection between S_{n} and O(n) is very deep indeed; the “best answer” so far is called “Springer theory”.) affine spaces: space of frames, etc.