TITLE: Intuition for the invariance of the determinant under change of basis
QUESTION [5 upvotes]: $$A' = PAP^{-1}$$
$$\det(A')=\det(P)\det(A)\det(P^{-1})=\det(A)$$
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 $\det(A)=\det(A')$?

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

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