# Determinant and Jacobian

Today I started my linear algebra class by motivating the determinant using a little bit of history and applications. In particular I talked about determinant appearing in the change of variables in integration of multivariate functions in calculus:Then after the class I noticed that my picture of what happens when we linearize the image wasn’t very clear as I had drawn it. So I decided to take a look at an actual example. The following code takes a function $f$ that maps $\mathbb{R}^2$ to $\mathbb{R}^2$ and draws three regions:

• the unit cube in blue,
• the image of the unit cube under $f$ in red, and
• the linearization of the image by Jacobian of $f$ at the origin
```def show_linearization(f,n=100):
# f is a function that maps R^2 to R^2 for example
# f(x,y) = ((x^2+2*(x+2)*(y+1))/4-1,y^2-(x+1)*y)
# The output is three areas:
# the unit cube
# the image of unit cube under f
# the linearization of the image using the Jacobian of f at (0,0)

var('x y')
A = jacobian(f,(x,y))(0,0)
p = f(0,0)
domxy = []
imgxy = []
jacxy = []
for i in range(n+1):
for j in range(n+1):
domxy.append((i/n,j/n))
imgxy.append(f(i/n,j/n))
jacxy.append(p+A*vector([i/n,j/n]))

P = points(domxy,color="blue",aspect_ratio=1, alpha = .3)
Q = points(imgxy,color="red",aspect_ratio=1, alpha = .3)
R = points(jacxy,color="green",aspect_ratio=1, alpha = .3)

(P+Q+R).show()```

Here is a sample run:

```f(x,y) = ((x^2+2*(x+2)*(y+1))/4-1,y^2-(x+1)*y)
show_linearization(f)```

and the output is:

I chose the function in a way that $f(0,0)=(0,0)$ for simplicity. Here it is on sage cell server for you to play around with it: link.

What do you think? Any suggestions or comments?