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I've taken a linear algebra course in the past, but I feel my understanding of coordinate change is very superficial. For example this exercise (4.7.1 from Lay's "Linear Algebra and its Applications" 4. edition):

Let $\mathcal{B}={\{ \bf{b_1,b_2} \}}$ and $\mathcal{C}={\{ \bf{c_1,c_2} \}}$ be bases for a vector space V, and suppose $\bf{b_1}=6\bf{c_1}-2\bf{c_2}$ and $\bf{b_2}=9\bf{c_1}-4\bf{c_2}$.

a. Find the change-of-coordinates matric from $\mathcal{B}$ to $\mathcal{C}$.

b. Find $[\bf{x}]_\mathcal{C} =-3\bf{b_1}+2\bf{b_2}$.

I can solve a. because I remember that $P_{\mathcal{C}\leftarrow\mathcal{B}}=\bf{[[b_1]_\mathcal{C}\,\,[b_1]_\mathcal{C}]}$ and that $\bf{[b_1]_\mathcal{C}}$ is the 'weights on the linear combination of $\mathcal{C}$-vectors'.

I can also solve b. because I remember that $[\bf{x}]_\mathcal{C}=P_{\mathcal{C}\leftarrow\mathcal{B}}[\bf{x}]_\mathcal{B} $

So I know how to apply the equations, but I feel that if I had a good mental model of what's going on here, it would be obvious that $\left(\begin{array}{rr} 6 & 9 \\ -2 & -4 \end{array}\right)$ is the matrix I'm looking for.

In my own experience, usually, if I learn something well I can construct the formulas I don't remember from reasoning, but here I fail, so I've literally spent all day trying to think up some analogy that would let me do the abovementioned exercise without remembering the formulas, here's what I've come up with so far:

$[\bf{x}]_\mathcal{B} $ is like a recipe for how much of the vectors of $\mathcal{B}$ you need to get to some point. So, how much of $\mathcal{C}$ do you need to get to $\bf{b_1}$? You need $\left[\begin{array}{rr} 6 \\ -2 \end{array}\right]=[\bf{b_1}]_\mathcal{C}$. How much of $\mathcal{C}$ do you need to get to $\bf{b_2}$? You need $\left[\begin{array}{rr} 9 \\ -4 \end{array}\right]=[\bf{b_2}]_\mathcal{C}$. Thus any "$\mathcal{B}$-recipe" $\left[\begin{array}{rr} r_1 \\ r_2 \end{array}\right]$ will in our case become a "$\mathcal{C}$-recipe" when we multiply it with $\left(\begin{array}{rr} 6 & 9 \\ -2 & -4 \end{array}\right)$, because the result will tell us "how much of $\mathcal{C}$ to get to $\bf{b_1}$ and $\bf{b_2}$, and then how much of these two to get to some new point". But it's been 'how much to use of $\mathcal{C}$-vectors' all the way this time, so our result has to be some $[\bf{x}]_\mathcal{C}$.

My question: What is a nice way to think about change between bases when neither is the standard one? Just looking for any way to think about this at any level of sophistication.

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Given a linear transformation $T:V \to W$ and ordered bases $B$ and $C$ for $V$ and $W$ respectively, you know how to write down $[T]_B^C$. It's the most concise way to describe $T$, you just write down what $T$ does to each basis vector in $B$. Now consider the case where $W = V$ and $T = I$ (the identity mapping) to get the change of basis matrix from $B$ to $C$. – littleO Oct 12 '13 at 4:39
up vote 2 down vote accepted

Suppose you have a vector $\mathbf{x}$ whose components are $(x_B, y_B)$ in the $B$-basis. This means that $$ \mathbf{x} = x_B\mathbf{b}_1 + y_B\mathbf{b}_2 $$ But we know that $\mathbf{b_1}=6\mathbf{c_1}-2\mathbf{c_2}$ and $\mathbf{b_2}=9\mathbf{c_1}-4\mathbf{c_2}$, so $$ \mathbf{x} = x_B(6\mathbf{c_1}-2\mathbf{c_2}) + y_B(9\mathbf{c_1}-4\mathbf{c_2}) $$ Rearranging gives $$ \mathbf{x} = (6x_B + 9y_B )\mathbf{c_1} + (-2x_B - 4y_B)\mathbf{c_2} $$ But, by definition, the coefficients of $\mathbf{c_1}$ and $\mathbf{c_2}$ on the right-hand side are just $x_C$ and $y_C$, so $$ x_C = 6x_B + 9y_B \\ y_C = -2x_B - 4y_B $$ The "change of basis" matrix is just a slick way of writing these last two equations as one: $$ \left[\matrix{ x_C \\ y_C}\right] = \left[\matrix{ 6 & 9 \\ -2 & -4}\right] \left[\matrix{ x_B \\ y_B}\right] $$ And the formula you memorized is just a slick way of getting the results I obtained from first principles. There's no magic in the matrix, it's just a convenient notation that saves you some writing.

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It's easier to understand this as a special case of more general transformations.

Let $f(x) = x'$ be an arbitrary, perhaps nonlinear, transformation of the position vector $x$. Given some basis $b_i$, this could be written as

$$f(x^1 b_1 + x^2 b_2) = (x^1)' b_1 + (x^2)' b_2$$

where $(x^1)', (x^2)'$ are functions of $x^1, x^2$.

Let $\ell(t)$ be some parameterized curve. The curve has an image under the transformation, $\ell' \equiv f \circ \ell$ (this prime should not be confused with differentiation with respect to $t$). The curve has a tangent vector $d\ell/dt$. The chain rule tells us that the tangent vector transforms as

$$\frac{d\ell'}{dt} = \left. \frac{d\ell}{dt} \cdot \nabla f \right|_{x=\ell(t)}$$

For any vector $a$, the quantity $(a \cdot \nabla) f$ is called the Jacobian map. This is sometimes denoted $J$ or $J_f$, so the above would be written

$$\frac{d\ell'}{dt} = J_f \left[ \frac{d\ell}{dt} \right]_{x=\ell(t)}$$

This is the transformation law for what we usually call "vectors". The map $J_f$ is usually expressed as a matrix in some basis. When the Jacobian map is constant, it is represented by a single matrix everywhere, and that is the kind of problem you're being asked to do.

Now, there are some matters of convention here that must be worked out. For instance, I can tell rather quickly that what your text is calling the change of basis matrix must be the inverse of the Jacobian. Given an equation $b_1 = 6 c_1 - 2 c_2$, it seems more natural to go from the $\mathcal C$ frame to the $\mathcal B$ frame, but this is the opposite of what they want you to do. But that's not uncommon; it's well known that coordinates transform opposite of basis vectors.

Still, you can use the above reasoning to imagine a set of coordinate axes in space. Each of these axes can be considered a curve $\ell(t)$, and so the above logic applies to see that, for instance,

$$b_1 = J_f (c_1)$$

and similarly for $b_2, c_2$. This naturally expresses $b_1, b_2$ in terms of $c_1, c_2$ and so it best matches the information given in your problem.

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Thanks for taking the time to write this out. What is meant by $(a\cdot \nabla)f$? This notation is unfamiliar to me. – john.abraham Oct 12 '13 at 13:10
It's a directional derivative. $(a \cdot \nabla)$ is a differential operator, equal to $a^1 \partial_{x^1} + a^2 \partial_{x^2}$. – Muphrid Oct 12 '13 at 13:42

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