Jacobian matrix

1. Jan 14, 2010

Rasalhague

Is the following correct, as far as it goes?

Suppose I have a vector space V and I'm making a transformation from one coordinate system, "the old system", with coordinates xi, to another, "the new system", with coordinates yi. Where i is an index that runs from 1 to n.

Let ei denote the coordinate basis for the old system, and e'i the coordinate basis for the new system.

I can define matrices BL and BR (where subscript L and R stand for "left" and "right") such that

$$B_L \begin{bmatrix} \vdots \\ \textbf{e}_i \\ \vdots \end{bmatrix} = \begin{bmatrix} \vdots \\ \textbf{e}'_i \\ \vdots \end{bmatrix}$$

$$\begin{bmatrix} \cdots & \textbf{e}_i & \cdots \end{bmatrix} B_R = \begin{bmatrix} \cdots & \textbf{e}'_i & \cdots \end{bmatrix}$$

and likewise matrices CL and CR, replacing the basis vectors in the above definitions with components of vectors in (the underlying set of) V.

And

$$C_L = \left ( C_R \right )^T = \begin{bmatrix} \frac{\partial y^1}{\partial x^1} & \cdots & \frac{\partial y^1}{\partial x^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial y^n}{\partial x^1} & \cdots & \frac{\partial y^n}{\partial x^n} \end{bmatrix}$$

and

$$\left ( C_L \right )^{-1} = B_L = \left ( B_R \right )^T = \begin{bmatrix} \frac{\partial x^1}{\partial y^1} & \cdots & \frac{\partial x^1}{\partial y^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial x^n}{\partial y^1} & \cdots & \frac{\partial x^n}{\partial y^n} \end{bmatrix}.$$

And some people (e.g. Wolfram Mathworld, Berkley & Blanchard: Calculus) define the Jacobian matrix of this transformation as

$$J \equiv C_L \equiv \frac{\partial \left ( y^1,...,y^n \right )}{\partial \left ( x^1,...,x^n \right )}$$

while others (e.g. Snider & Davis: Vector Analysis) define it as

$$J \equiv \left ( C_L \right )^{-1} \equiv \frac{\partial \left ( x^1,...,x^n \right )}{\partial \left ( y^1,...,y^n \right )}.$$

Last edited: Jan 14, 2010
2. Jan 15, 2010

HallsofIvy

They are really the same thing. Your CL transforms from the "x_i" coordinates system to the "yj" coordinate system while CL-1, of course, goes the other way, transforming from the "yj" coordinate system to the "xi" coordinate system.

3. Jan 15, 2010

Rasalhague

I don't understand how they're "really the same thing". For a given coordinate transformation, won't CL generally be a different matrix from its inverse BL? Also, changing a subscript on one of these matrices from L to R or vice versa transposes it, and in general a matrix is not the same thing as its transpose.

Experimenting with the transformation from 2d Cartesian to plane polar coordinates confirms that using the wrong matrices, or the right ones in the wrong order, gives the wrong answer. In fact in this thread, I did make a mistake (see #4), and if I'd done the multiplication correctly it wouldn't have given the required answer.

I'm thinking if they were literally the same, there'd be no need for Griffel's "Warning. There are two ways to define the change matrix. In our definition, the columns are the B'-components of the B vectors. Some authors define it with B and B' interchanged, giving a matrix which is the inverse of ours. The two versions of the theory look slightly different, but they are equivalent. It does not matter which version is used, provided itis used consistently. Using formulas from one version in a calculation from the other version will give the wrong answers" (Linear Algebra and its Applications, Vol. 2, p. 11).

But maybe you only meant that they're the same sort of thing, or that if we swapped the labels "old" and "new", the same matrices would be playing opposite roles.

Last edited: Jan 15, 2010
4. Jan 17, 2010

Landau

Yes, this is correct.