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

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

enter image description here

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 $\det(A)=\det(A')$?

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    $\begingroup$ 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 $j'=2j$. $\endgroup$
    – A.Γ.
    Commented Dec 16, 2018 at 22:01

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Dec 16, 2018 at 22:10
  • $\begingroup$ 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 $1$). $\endgroup$
    – user293794
    Commented Dec 16, 2018 at 22:42
  • $\begingroup$ 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. $\endgroup$ Commented Dec 16, 2018 at 23:29
  • $\begingroup$ 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. $\endgroup$ Commented Dec 16, 2018 at 23:30
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    $\begingroup$ 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. $\endgroup$
    – user293794
    Commented Dec 16, 2018 at 23:30
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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.

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