As far as I can tell, it's only a slight exaggeration to say that every text has a different notation for a change of basis matrix from (say) $\mathcal{B}$ to $\mathcal{C}$. That's not even to talk about the "standard" matrix for representation, e.g., changing from $\mathcal{B}$ to the standard basis $\mathcal{E}$ in $\mathbb{R}^n$.

My question is whether anyone has any ideas for why any particular notation is superior in the classroom (or other educational outlet) to other such notations. I'm not asking for a list of actual notations currently used - that would be more appropriate for math.se anyway.

I want to know the specific notation that you feel, in your experience, helps students understand this notably thorny concept. (Yes, if you wrote a book, feel free to explain why you chose that notation!)

Aside: Seriously, at this point, I need a dictionary just to translate from one text to another. A web search just now showed three or four notations I'd never even seen before, and I teach this course regularly!

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    $\begingroup$ To the person who edited: It was nice of you to edit for fairly trivial grammar issues, and I approved them. In the future, however, one might consider that changing the author's "voice" without any real need might also change how people perceive the question. I find it very hard to believe there was anything actually unclear, and I intentionally used more informal language/run-ons to indicate what style of answer would be appropriate. Thank you for reading. $\endgroup$ – kcrisman Mar 23 '19 at 1:08

I like to use

$ {}_{\mathcal{C}}A_{\mathcal{B}}$

for the change from $\mathcal{B}$ to $\mathcal{C}$ because then the subscripts match up when you try to compose the matrices (in the usual convention) - you put a vector in $\mathcal{B}$ coordinates into the right hand side, and get out one in $\mathcal{C}$ coordinates. If you combine multiple matrices (which is the point), you need the subscripts to match up in the same way as the matrix sizes:

$$ {}_{\mathcal{B}}D_{\mathcal{C}}\,{}_{\mathcal{C}}M_{\mathcal{C}}\,{}_{\mathcal{C}}A_{\mathcal{B}}\, v_{\mathcal{B}} $$

Here I can see what I'm doing: changing into $\mathcal{B}$ coordinates, doing the linear map using $\mathcal{C}$ coordinates, and then convert back to $\mathcal{B}$ coordinates.

(The use of $A$ is not important in this notation.)

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    $\begingroup$ Clarification question - do you distinguish between a "change of basis" matrix and a "matrix written in a basis" and/or a "matrix written from one basis to another" like the full composition in your example? $\endgroup$ – kcrisman Mar 22 '19 at 11:48
  • $\begingroup$ On a separate note, how did you choose the letters A or D (I'll assume M is a linear transformation written fully in the basis C)? $\endgroup$ – kcrisman Mar 22 '19 at 11:50
  • $\begingroup$ I chose A because I usually start there, B for basis, C because it's next, D as the next letter not used, and M for a matrix not in the same series as A and D. I'm usually teaching maths students, so they need to be able to handle varying notation. I teach it very much as my notation, not as standard, because, as you say, there isn't a standard. $\endgroup$ – Jessica B Mar 22 '19 at 17:48
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    $\begingroup$ I use the phrase 'change of basis matrix', but will usually at some point connect it with being the matrix of the identity map wrt two bases. I don't have a fixed idea about which should be introduced first. Actually, I think it's helpful to first try with bases that are not subsets of R^n, such as polynomials, as it's clearer then why you would want another basis, and the examples can actually be easier. $\endgroup$ – Jessica B Mar 22 '19 at 17:51
  • $\begingroup$ Yes, I always introduce those fairly early too. $\endgroup$ – kcrisman Mar 23 '19 at 1:01

I think it is best to use no notation for change of basis! Let me explain what I mean.

One defines the matrix of a linear map $T:V \to W$ with respect to (ordered) bases $\{v_{1}, \dots, v_{n}\}$ and $\{w_{1}, \dots, w_{m}\}$ of $V$ and $W$ to be the $m \times n$ matrix whose $i$th column comprises the coordinates of $T(v_{i})$ with respect to the (ordered) basis $\{w_{1}, \dots, w_{m}\}$.

When $V = W$, the "change of basis matrix" is then simply the matrix of the identity transformation $\operatorname{Id}_{V}:V \to V$ with respect to the basis $\{v_{1}, \dots, v_{n}\}$ in the domain and $\{w_{1}, \dots, w_{n}\}$ in the codomain. More prosaically, the $i$th column of this matrix are the coordinates of $v_{i}$ in the (ordered) basis $\{w_{1}, \dots, w_{n}\}$.

Note that I have been fussily writing "ordered" in parentheses. This is because these notions do not depend simply on the basis (which is a set). They depend on the ordering of this basis implicit in the choice of indices. Reordering a given basis leads to a nontrivial change of coordinates! (By a permutation matrix). So when one speaks of the "change of basis" matrix one should really speak of the "change of ordered basis matrix". Of course no one does this, but this sometimes causes confusion for students, particularly if one writes something like ${}_{B}M_{C}$ and speaks of bases $B$ and $C$. Are $\{(1, 0, 0), (0, 1, 0), (0, 0, 1\}$ and $\{(1, 0, 0), (0, 0, 1), (0, 1, 0\}$ different bases? Not if one's definition of a basis is a linearly independent set that spans. But they are different ordered bases, and the usual conventions for change of basis matrices assign to them the permutation matrix \begin{equation*} \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}. \end{equation*} Notation like ${}_{B}M_{C}$ tends to obscure this subtle point.

More problematically, notation like ${}_{B}M_{C}$ is really quite sophisticated. Without going into details, this is essentially a transition function for the frame bundle (this explains why it satisfies cocycle identities like ${}_{B}M_{C} = ({}_{C}M_{B})^{-1}$ and ${}_{B}M_{C}{}_{C}M_{D} = {}_{B}M_{D}$).

Also, most of us can't remember whether ${}_{B}M_{C}$ means what it means or $({}_{B}M_{C})^{-1}$ (that is, on which side do $B$ and $C$ go? Equivalently, when do I write $A = P^{-1}BP$ and when $A = PBP^{-1}$?).

My solution is to use commutative diagrams. This sounds horrible, but my experience is that it works with first year engineering students. I draw diagrams whose vertices are vector spaces and edges are arrows between them labeled to indicate linear transformations. Choices of bases are indicated by writing them at each vector space and writing the label of the matrix determined by the linear transformation and the choices of bases below the corresponding edge. (I would show here what I mean but I can't figure out how to get xypic to work here.)

The relation expressing the change of the matrix of a linear transformation takes a form such as $PA = BP$ rather than $A = P^{-1}BP$ (or $B = PAP^{-1}$), where $P$ is the matrix of the identity transformation. There is no need to memorize a convention for placing inverses (equivalently the labels $B$ and $C$) because the correct choices are forced by the directions of the arrows.

  • $\begingroup$ I think that this is part of the story - I certainly use the diagrams, though I don't think my students really "get" them any more than they "get" whatever notation I use. Really, change of basis is just a hard topic. $\endgroup$ – kcrisman Apr 15 '19 at 13:21
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    $\begingroup$ Also, kudos for bringing up the ordered basis issue :-) yes, probably we are all guilty of sloppiness on a day-to-day basis (!) on this. $\endgroup$ – kcrisman Apr 15 '19 at 13:21
  • $\begingroup$ One could avoid the dependence of the "order" of the basis by defining a matrix not as a collection of scalars indexed by natural numbers, but as a collection of scalars indexed by the bases. Borrowing the notation suggested in my answer: the matrix of the linear map $T\colon V \to W$ would then be $\left(\frac{Tv}{w}\right)_{v\in B, w\in C}$ where $B$ is a basis of $V$ (an unordered set) and $C$ is a basis of $W$. $\endgroup$ – Michael Bächtold Apr 16 '19 at 9:26

We shouldn't need a new notation for this, there's a long established one: partial derivatives.

Let me explain: When people write something like $$ \left(\frac{\partial T}{\partial p}\right)_V \quad \text{ or } \quad \left.\frac{\partial T}{\partial p}\right|_V $$ it means "the coefficient in front of $\text{d} p$ in the expansion of $\text{d}T$ using the basis $\text{d} p,\text{d}V $". This might sound awkward, but that's what it means. (In fact, people used to talk of "differential coefficients" before the derivative terminology became standard. And they used to write $\left(\frac{\text{d} T}{\text{d} p}\right)$ for partial derivatives, before someone thought we'd gain something by changing $\text{d}$ to $\partial$.)

So we're extracting a coefficient from the expansion of a vector wrt. to a basis. That's something meaningful in any vector space, not just the space of differentials and it's useful in any linear algebra class. So let's generalize the notation to arbitrary vector spaces.

If $V$ is a finite dimensional vector space with a basis $B=\{b_1,\ldots,b_n\}$ and $v\in V$, denote with $$ \left(\frac{v}{b_i}\right)_B \quad \text{ or } \quad \left.\frac{v}{b_i}\right|_B $$ the coefficient in front of $b_i$ in the expansion of $v$ wrt. to $B$. In other words $$ v = \sum_{i=1}^n \left.\frac{v}{b_i}\right|_B \cdot b_i. $$ Using Einstein's summation convention we could just write $$ v = \left.\frac{v}{b_i}\right|_B b_i. $$ and if we are lazy/sloppy (as we are with partial derivatives) we could also drop the $|_B$ and claim it's clear from the context which basis $B$ we're using. So we arrive at $$ v = \frac{v}{b_i} b_i. $$ Using this notation, the matrix of a linear map $A\colon V \to W$ with respect to bases $B=\{b_1,\ldots,b_n\}$ of $V$ and $C=\{c_1,\ldots,c_m\}$ of $W$ is written as $$ \left( \left.\frac{A\, b_i}{c_j}\right|_C \right)_{i,j} $$ or in the sloppy variant $$ \left( \frac{A\, b_i}{c_j} \right)_{i,j} $$ The nice thing about this, is that change of basis becomes as natural as the chain rule in multi variabe calculus.

I have to admit that I haven't tested this in class. Mainly because I teach parallel classes with other instructors, who object to using notation that was not used when they were students. But I think it's the natural consequence if you take partial derivative notation seriously and want to make the link between linear algebra and multi variable calculus more explicit.

I also think it is good to have an explicit notation for a concept that's often used, like "coefficients of a vector with respect to a basis".

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    $\begingroup$ +1 for guts to even propose this :-) I would personally confuse it with the Legendre symbol but it's true that "expansion in terms of coefficients - even of Fourier series" is something that a 'traditional' linear algebra class (in the USA, at any rate) does not prepare one adequately for. $\endgroup$ – kcrisman Apr 15 '19 at 16:49
  • $\begingroup$ @kcrisman I agree that there's a risk of confusing it for other things, like fractions (which is the origin of the notation for derivatives), so I would personally modify it slightly. For instance by changing the fraction bar to something like $\neg$. $\endgroup$ – Michael Bächtold Apr 16 '19 at 9:16
  • $\begingroup$ Wait, you don't write all your fractions like 1/2? :-) $\endgroup$ – kcrisman Apr 16 '19 at 17:27

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