I don't understand what this theorem for a characterization of the determinant actually means, and what use it has. Could you please explain it? Let D be a function mapping from the set of square matrices M with n rows/columns over the field F, to a field F. Also let D be a function that is multilinear and alternating in columns. Then D(A)=D(I)detA

Sonia Elliott 2022-10-20 Answered
I don't understand what this theorem for a characterization of the determinant actually means, and what use it has. Could you please explain it?
Let D be a function mapping from the set of square matrices M with n rows/columns over the field F, to a field F. Also let D be a function that is multilinear and alternating in columns. Then
D(A)=D(I)detA
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Answers (2)

indyterpep
Answered 2022-10-21 Author has 12 answers
TLDR: the theorem lists some properties that a "measurement of volume" ought to have, and says that the determinant is the only thing that satisfies those properties.
The purpose of the determinant is to answer the question, "How does a matrix transform (signed) volumes?". If we didn't already know about determinants, we might try to build a function that takes in matrices and outputs an answer to that question. Call it "D". We might then try to reason about how volume works to try and figure out some properties of D. The theorem says that, with only a few of those properties in hand (the ones mentioned in kccu's answer), the standard algorithm for the determinant gives the only answer consistent with those properties.
Let's see how these properties relate to volume.
D is a multilinear function from n vectors in Fn to a scalar.
Think of volume in terms of a parallelogram / parallelepiped with one corner at the origin. The vectors will then be the edges based at the origin, and the output will be the (signed) volume of the parallelepiped, which is a scalar. If you scale any one of the sides, the volume scales by that same amount, which should hint that it ought to be multilinear. Consult a good linear algebra textbook for a geometric explanation of additivity and why it makes sense for the output to switch signs when one of the vectors is scaled by a negative number. (Also consult that textbook for an explanation for why we can see what a matrix does to volumes in general just by looking at the volume of this parallelogram/parallelepiped.)
It is alternating.
I'm not sure what definition your source gives for "alternating;" some definitions more intuitively relate to volume than others. One way of saying that a multilinear function is alternating is that its output is zero whenever its inputs are linearly dependent. (If your definition looks different, try to see why this one is equivalent.) That makes sense for volume: if the vectors are linearly dependent, then the parallelepiped will be "flat" and won't have any volume at all.
As described in kccu's answer, the only remaining property is:
It sends the identity matrix to 1.
This last property is somewhat arbitrary. In the "real world," our vectors and scalars have units attached to them. With just the above two properties, it would be consistent for, say, D to take input vectors in meters and output its volume in gallons. Requiring D to send the identity matrix to 1 just means we're enforcing consistency with our units of length and volume: if we have a square/cube/whatever whose sides are all length "one" in whatever units of length we're using, our units for area/volume/whatever ought to treat that shape as having area/volume/whatever of "one", also.
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limfne2c
Answered 2022-10-22 Author has 2 answers
This means that the determinant function is completely determined by three properties:
1. It is multilinear in the columns
2. It is alternating in the columns
3. It maps the identity matrix to 1
After all, if D satisfies all of the above, then by the theorem you stated, we have D(A)=D(I)det(A)=1det(A)=det(A).
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