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!


4 Answers 4


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.)

  • 1
    $\begingroup$ 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, 2019 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, 2019 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, 2019 at 17:48
  • 1
    $\begingroup$ 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, 2019 at 17:51
  • $\begingroup$ Yes, I always introduce those fairly early too. $\endgroup$
    – kcrisman
    Mar 23, 2019 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, 2019 at 13:21
  • 2
    $\begingroup$ 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, 2019 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 $W$. $\endgroup$ Apr 16, 2019 at 9:26
  • 1
    $\begingroup$ Hmm, I'm two and a half years late to the party, but since this has just popped up on the frontpage: I'm a bit skeptical about the point of view that a basis is a set and that one thus has to distinguish it from an ordered basis: I looked up a few linear algebra books which I had at hand: Axler (1997), Strang (2007) and Garcia and Horn (2017) all define a basis itself as an ordered object (calling it a "list" or a "sequence"). The only book I've readily found which defines a basis as a set is Lang (1987), but ironically this results in a false statement on the top of the very next page. [...] $\endgroup$ Dec 7, 2021 at 10:41
  • 1
    $\begingroup$ [...] Similarly, I have some doubts about the sentence "if one's definition of a basis is a linearly independent set that spans", because a "linearly independent set" seems like a very unhandy concept to me: if one defines linear independence for sets rather than for lists of vectors, it is e.g. not possible to make claims like "an $n$-by-$n$ matrix is invertible if and only if its columns are linearly independent" (since the same column can occur multiple times within a matrix). $\endgroup$ Dec 7, 2021 at 10:41

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".

  • 4
    $\begingroup$ +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, 2019 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 $\neg$. $\endgroup$ Apr 16, 2019 at 9:16
  • $\begingroup$ Wait, you don't write all your fractions like 1/2? :-) $\endgroup$
    – kcrisman
    Apr 16, 2019 at 17:27

You may use the division symbol. This division by a base has an important meaning: the duality. This notation has a lot of benefits, including reducing the amount of computations and explicitly showing the logic of linear algebra.

If $\alpha = (\alpha_1 ,\ \alpha_2 ,\ \cdots ,\ \alpha_n)$ and $\beta = ( \beta_1 ,\ \beta_2,\ \cdots ,\ \beta_n)$ are bases, which will be regarded formally as rows, then the matrix representing the change-of-base from $\alpha$ to $\beta$ is denoted by $$\frac \alpha \beta \ \ \ \ \text{or}\ \ \ \ \beta \backslash \alpha \in K^{n\times n},$$ (but not $\alpha / \beta$) where $K$ is the base field. As I mentioned below, This notation is still valid even if $\alpha$ is not a basis, but just an ordered tuple of vectors. In this case $\beta\backslash \alpha$ may not be a change-of-base matrix, though.

This notation has some benefits: for example we may write $v = \alpha x$ to say that $v$ is the linear combination $x_1 \alpha_1 + x_2 \alpha_2 + \cdots + x_n \alpha_n$ of the base $\alpha$ by the coefficient (column) vector $x = [x_1 \ x_2 \ \cdots \ x_n]^T$. Then it is natural to use symbols like $$ x = \frac v \alpha \ \ \ \ \text{or} \ \ \ \ \alpha \backslash v. $$ Now suppose $y = [y_1 \ \cdots \ y_n]^T$ is the coordinate column vector of $v$ w.r.t. the base $\beta$. Then now we can write $$ y = \frac \alpha \beta x, \ \ \ \ \text{that is,} \ \ \ \ \frac v \beta = \frac \alpha \beta \frac v \alpha \ \ \ \ \text{or} \ \ \ \ \beta \backslash v = (\beta \backslash \alpha)(\alpha \backslash v). $$ Be careful: perform cancellations only along the line $\backslash$, not $/$.

The above notaions $\beta \backslash \alpha$ and $\beta \backslash v$ are compatible in the sense that $$ \frac \alpha \beta = \frac {(\alpha_1,\ \cdots,\ \alpha_n)} \beta = \left[ \frac {\alpha_1} \beta \ \frac {\alpha_2} \beta \ \cdots \ \frac {\alpha_n} \beta \right]. $$

We may define several similar symbols to make things involving change-of-base easy. Let $T: V \to V$ be a linear map. Denote by $T\alpha$ the ordered tuple $(T\alpha_1 ,\ \cdots, \ T\alpha_n) \in V^n$ of vectors. Then the matrix representation of $T$ is $\alpha\backslash T\alpha$.

With some more work, one may easily verify the formula $$ \frac {T\beta} \beta = \frac \alpha \beta \frac {T\alpha} \alpha \frac \beta \alpha, $$ which means that the matrix of $T$ w.r.t. $\beta$ is the conjugate of the matrix of $T$ w.r.t $\alpha$ by $\beta\backslash\alpha$.

Even more benefit: If the space $V$ is just $\mathbb R^n$, then the tuples $\alpha$ and $\beta$ are merely the matrices, and the three notations below denotes completely the same thing: $$ \frac \alpha \beta = \beta \backslash \alpha = \beta ^{-1} \alpha, $$ and this is the reason why I recommended you not to use the symbol $\alpha / \beta$.

Again more benefit: We may denote the coordinate map $V \to \mathbb R^n$ w.r.t the base $\alpha$ by just $$ \frac 1 \alpha : v \mapsto \frac v \alpha, $$ and its inverse by just $$ \alpha: x = [x_1 \ \cdots \ x_n]^T \mapsto \alpha x = \sum x_i \alpha_i. $$ In this viewpoint we find the relations: $$ \alpha \frac 1 \alpha \ \ \ \text{ equals the identity map } V \to V, $$ and $$ \frac 1 \alpha \alpha \ \ \ \text{ equals the identity matrix } I. $$

Note that the notations in this answer enjoys the associative law.

Here is another benefit. For a base $\alpha$, we may promise to denote the dual base $(\alpha_i^*)$ of $V^*$ by $\alpha \backslash 1$: $$ \frac 1 \alpha := [ \alpha_1^* ,\ \alpha_2^* ,\ \cdots ,\ \alpha_n^* ]^T, $$ where the dual base is regarded as a column. This makes no conflict with `Again more benefit'. And our first notation $$ \frac \beta \alpha $$ now may be regarded as a multiplication of column and a row, producing a square matrix: $$ \frac \beta \alpha = \frac 1 \alpha \beta = \begin{bmatrix} \alpha_1^* \\ \alpha_2^* \\ \vdots \\ \alpha_n^* \end{bmatrix} [ \beta_1 \ \ \beta_2 \ \ \cdots \ \ \beta_n ], $$ which is exactly equal to the change-of-base matrix. This is the genuine meaning of our division by a basis.

So let's take a look at an example. How this notation may be used? Let me prove a famous theorem in linear algebra:

Theorem. Let $V$ be a fin dim vector space over a field $K$. Let $V^*$ be the dual space of $V$, and $V^{**}$ the double dual. Then there is a linear isomorphism of $V$ onto $V^{**}$ that is independent of the choice of base of $V$.

Proof. Let $n$ be the dimension of $V$ over $K$. Let $\alpha = (\alpha_1 ,\ \cdots,\ \alpha_n) \in V^n$ and $\beta \in V^n$ be bases of $V$, which are regarded as a row form. Then they induces the dual bases $\alpha^* = \alpha \backslash 1 = [\alpha_1^* ,\ \cdots,\ \alpha_n^*]^T$ and $\beta^* = \beta \backslash 1$ of $V^*$, which are regarded as columns. We use a lemma:

Lemma. The change-of-base matrix $$ \frac {\alpha^*} {\beta^*} $$ from $\alpha^*$ to $\beta^*$ equals $\alpha \backslash \beta$: $$ \frac 1 \alpha = \frac \beta \alpha \frac 1 \beta. $$ Remark: It is well-known that the resulting matrix must be the transposed inverse, but the reason why we got just the inverse is that we are regarding the dual bases as columns, not rows. So the lemma does not defy your background knowledge; if you regard the dual bases as rows, then you'll get the familiar transposed inverse.

So let's continue to working with the theorem. If we apply the lemma twice, then we know that the changeofbase matrix from $\alpha^{**}$ to $\beta^{**}$ equals $\beta \backslash \alpha$.

Now define a linear isomorphism $T:V \to V^{**}$ by the linear extension of the assignment $\alpha_i \mapsto \alpha_i^{**}$. We want to show that $T$ is independent of choice of the base; that is, $T\beta_i = \beta_i^{**}$ for all $i$. This is equivalent to showing that the matrix representation $\beta^{**} \backslash T\beta$ of $T$ equals the identity matrix. This is easily done by our notations: $$ \frac {T\beta} {\beta^{**}} = \frac{\alpha^{**}}{\beta^{**}} \frac {T\alpha}{\alpha^{**}} \frac {\beta} {\alpha} = \frac \alpha \beta \cdot I \cdot \frac \beta \alpha = I. $$ This completes the proof. //

So our division notation allows us not to do enormous scalar computations. This is the other great benefit, I think.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.