# 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. Dec 16 '18 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? – Pancake_Senpai Dec 16 '18 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 $1$). – user293794 Dec 16 '18 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. – Pancake_Senpai Dec 16 '18 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. – Pancake_Senpai Dec 16 '18 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. – user293794 Dec 16 '18 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.