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My Linear Algebra text has some introductory examples of interpreting the determinant as a scaling factor. These all involve interpreting the columns of a matrix B as coordinates to form a square or triangle, then showing that when left multiplying B by another matrix A that has a determinant, det(A), that the area of the shape formed by interpreting the columns of the resulting matrix as coordinates is scaled by det(A). That part all makes sense.

Later they introduce this proposition:

determinant proposition

The first confusing thing is they're saying a 3x3 matrix is a box -- that's not true if you interpret the columns as coordinates like before; you need more columns. At best 3 points could describe a plane or a triangle in a plane. If I interpret their A matrix to be a matrix that we multiply by to scale another matrix B with columns as coordinates, I can see why (kA)B=C_1 would have factors that are k times longer than those in just AB=C_2, because the rows of A each in essence describe a linear combination for how to produce the new columns of C_1 and C_2, and if that sum is multiplied by k every column vector in C_2 should be k times longer than in C_1. Then if we separately already accept that det(A) is the scaling factor for the area of a new shape produced when left multiplying by A, I can see the proposition. But in that line of reasoning kA never represents a box; you could interpret it to be one in the sense that you can interpret 3 vectors with a length as describing a cube, but that's inconsistent with how they interpreted things before and it's not clear why that interpretation is supposed to help.

Am I interpreting this correctly? Or is there an easier way of looking at it?

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    $\begingroup$ Can you see how two linearly independent vectors in $R^2$ determine a parallelogram? $\endgroup$
    – J. Doe
    Jun 27, 2017 at 0:12
  • $\begingroup$ Yes. I mention near the end I can see how it could describe a cube, just not why it's a useful way to think about it. The determinant is a scaling factor for the area of the shape described by C which was originally the shape described by B, but kA is scaling the matrix you're using to do the transform, so it doesn't seem like a justification for the area of C getting bigger. $\endgroup$ Jun 27, 2017 at 0:15
  • $\begingroup$ In my opinion, the prose in this textbook is vague, indirect, and ambiguous to the point of being misleading. You should find a better textbook to study. $\endgroup$ Jun 27, 2017 at 1:13
  • $\begingroup$ @ChrisCulter any specific suggestions? $\endgroup$ Jun 27, 2017 at 1:14

2 Answers 2

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The columns of the $3\times 3$ matrix are the position vectors of three vertices of the cube. In the case shown above:

$$ v_1 = (1,0,0) $$ $$ v_2 = (0,1,0) $$ $$ v_1 = (0,0,1) $$

With determinant $det([v_1,v_2,v_3]) = 1$. If you multiply each column by $k$, the determinant is $det([kv_1,kv_2,kv_3]) = k^³$.

That is completely general for any parallelepiped in three dimensions. The volume of the parallelepiped is equal to the determinant of the matrix formed by three column vectors along the directions of the three non-redundant edges with length equal to the length of the edge.

It all stems from the fact that the determinant can be writen as this "mixed" product of the columns $det([v_1,v_2,v_3]) = v_1\cdot v_2\times v_3$.

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  • $\begingroup$ I don't think this really answers the question. Like I said at the end, I can see a way to interpret it to get a cube (like you describe), but not why that's helpful to reach the proposition. You want to show that the area of C scales by k^n. That you can interpret kA as scaling a cube doesn't help because that's scaling the transformation matrix A, not C. What was established earlier in the text is the determinant is a scaling factor for the area of going from B to C. $\endgroup$ Jun 27, 2017 at 0:20
  • $\begingroup$ What is $C$ to you? $\endgroup$
    – user410935
    Jun 27, 2017 at 0:23
  • $\begingroup$ C_1 is the resulting matrix from left multiplying by a "transformation matrix" A that has a determinant det(A) by a "coordinate matrix" B whose columns we interpret as coordinates of a shape. If I accept the determinant of A as being area of C_1 over the area of B, the fact that I can separately go and interpret the columns of A in a totally different way to get a cube and can multiply the lengths of As column vectors by k doesn't seem to immediately imply (at least to me) that the area of C_2 (formed by (kA)B) over the area of B must be k^n * det(A). $\endgroup$ Jun 27, 2017 at 0:28
  • $\begingroup$ $B$ defines a shape, $AB$ is the shape in another space with volume $det(A)det(B)$, $kAB = A(kB)$ so you transfer the rescaling factor to the coordinates and you have a $k^n$ times bigger shape in the original space which, it's intuitively easy to see, will give a $k^n$ times bigger shape in the space of $AB$ under a linear transformation, does that solve your issue? $\endgroup$
    – user410935
    Jun 27, 2017 at 0:35
  • $\begingroup$ you're probably right but I think you're using material I'm not exposed to yet. To be clear are the A and B you're referring to the same as mine or do you just mean two matrices in general? I have only read of determinant as a scaling factor, not an actual area. When you say the area of $AB$ is $det(A)det(B)$, do you mean that that's the area of the shape when the shape is the vertices described by the columns as coordinates, or when it's the parallelogram described by the columns as vectors? $\endgroup$ Jun 27, 2017 at 0:41
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I’ll take a stab at this:

First, the easy one. The “boxes” that we’re dealing with here are parallelotopes: paralallograms, paralallepipeds, and so on. Because of all of the parallelism constraints that they have among edges, faces, &c, an $n$-dimensional parallelotope is completely determined by $n$ edges that meet at a common vertex. In particular, three linearly independent edge vectors and a common vertex completely specify a parallelepiped, which means that the three columns of a $3\times3$ matrix are also enough to define a box with vertex at the origin. You don’t need any more columns.

Now, for the proposition at hand, they’re transforming a specific parallelotope—the unit $n$-cube—by the matrix $A$. Going back to your first paragraph, all of this cube’s side lengths are equal to $1$, so the matrix $B$ for the unit cube is just the identity matrix. Left-multiplying this by the transformation matrix $A$ produces $A$, so you can then proceed to interpret its columns as the edges of a parallelotope just as you did with the product $AB$ for some other matrix $B$.

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