Intuition for the invariance of the determinant under change of basis

$$A' = PAP^{-1}$$ $$\det(A')=\det(P)\det(A)\det(P^{-1})=\det(A)$$

Now, that makes sense algebraically, but consider the below diagram:

This a geometric representation of the two 'normal' basis vectors $$\bf i$$ and $$\bf j$$ (I will denote this set by $$B$$) in $$\Bbb R^2$$, and my choice of two new basis vectors $$\bf i'$$ and $$\bf j'$$ (I will denote this set by $$B'$$ ). The determinant preserves the area of of the unit square, which is determined by our choice of basis vectors. The unit square area in the basis $$B$$ is different to the unit square area in basis $$B'$$.

The determinant gives the area of the image of the unit square. The image of the black B unit square will likely be different to the image of the red $$B'$$ unit square, so why is $$\det(A)=\det(A')$$?

• Two matrices $P$ and $P^{-1}$ mean that one does two conversions: to the new basis and back. Here you do only one. To make it easy: consider $i'=2i$ and $j'=2j$.
– A.Γ.
Commented Dec 16, 2018 at 22:01

When people say that the determinant is the area of the image of the unit square, the unit square it taken to mean the square given by sides $$e_1$$ and $$e_2$$, the standard basis vectors. Another way of thinking about this is to note more generally that the determinant is the scaling factor of the image of a square so that $$\det A$$ is the volume of $$\text{Vol}(A(\text{square}))/\text{Vol(square)}$$. For instance, if $$A$$ is the identity, then the square given by $$i'$$ and $$j'$$ still satisfies $$\text{Vol}(A(\text{square}))/\text{Vol(square)}$$ since the identity doesn't do anything to the square. From this perspective, it should be clear the the determinant doesn't depend on basis, because the area of a square doesn't depend on how you choose to write its sides. If this is confusing, think about the fact that if you write the cube given by $$i,j$$ in the basis $$i'=2i,j'=2j$$ then its sides are given by $$\frac{1}{2}i',\frac{1}{2}j'$$ so that it's area is $$\frac{\text{area}(i',j')}{4}=1$$. Essentially, the coordinates for the sides scaled in the opposite direction that the basis vectors did.

• So the determinant always gives the area of the standard unit square (or cube, etc.)? I'm not sure how I feel about that. What makes i and j so special other than our regular usage of them? Commented Dec 16, 2018 at 22:10
• What's special about it is that we agree that it's area is one. Think about the second characterization of the determinant that I gave: it's the amount by which a linear map scales the area of a square. It's thus natural to look at the image of a cube with area one since the amount by which it is scaled is the same as the area of the image (as in the denominator of $\frac{\text{Vol(A(cube)}}{\text{Vol(cube)}}$ is just $1$). Commented Dec 16, 2018 at 22:42
• It's just clicked. The determinant gives the scale factor by which the area of the red square changes during the transformation. For the standard unit vectors i and j this is the area of the image of the black square, because the area of the black square is just 1. So the determinant is clearly then independent of basis. Commented Dec 16, 2018 at 23:29
• This is a very good video I found that can help explain this to anyone finding this thread in the future: youtube.com/watch?v=Ip3X9LOh2dk. Commented Dec 16, 2018 at 23:30
• Yes exactly. All this shows is that the determinant is fundamentally geometric, and geometry doesn't care about the coordinates we choose to describe it. Commented Dec 16, 2018 at 23:30

It basically means that to see what happens with your area, say, in square meters ($$\det(A')$$ [m$${}^2$$]) you may

1. Convert the units to feet by $$P^{-1}$$,
2. Do the calculations in ft$${}^2$$ ($$\det(A)$$),
3. Convert the units back to meters by $$P$$.

The result will be the same. Of course, the unit square in meters and in feet look different and have different area (1 m$${}^2\ne$$ 1 ft$${}^2$$), but area calculations in both systems are consistent.