- #1
graphking
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- Summary
- in R^n, why the jordan measurement of linear transformation(A) of a square(V) is det(A)*m(V)?
nothing more need to be add(smiling face)
thanks for helping! But I cant agree with you in some point:The shortest answer is the Cauchy-Binet formula, known as the multiplication property of the determinant.
The Jordan volume is a generalization of Riemann sums. So the volume of ##V## can be normalized to be ##1=\operatorname{Vol}( x_1\wedge x_2\wedge \ldots \wedge x_n)## for a suitable orthogonal basis ##(x_1,\ldots,x_n)##. Now the volume of ##A(V)## is
$$
\operatorname{Vol}(Ax_1\wedge \ldots \wedge Ax_n)=|\det(A)|\cdot \operatorname{Vol}(x_1\wedge x_2\wedge \ldots \wedge x_n)=|\det(A)|
$$
thanks！I choose several ideas from your answer and I think a conduction on the dim n would be a good way to understand(adding a new edge, consiser the orthogonal directions, and splitting the paralellopiped(thanks for the word!) to many layers like you say), though I am still confuse at this things, it's just out of the directness of geometry(4-dim and more)I think the typical way this is proven is by showing that the determinant is the only function on matrices which satisfies all the regular properties (scales proportional to each row, invariant under adding one row to another, etc), and they vol(A) also satisfies those properties. It's not super satisfying though, so here's a bit of a handwavy proof by direct geometry:
The way you compute the volume of a parallelopiped is the following:
Start with ##l_1##, the length of edge 1.
Take edge 2. Shift it along the edge 1 direction so that it's orthogonal to edge 1. The length of that is ##l_2##. Then the area of the 2-d parallelopiped is ##l_1 l_2##, because the area is just ##l_1## layered over and over again over a span of length ##l_2##.
Then take edge 3, and take the orthogonal component of it to the span of edges 1 and 2. If that has length ##l_3###, then The 3-d volume of this 3-d parallelopiped is ##l_3 A## where ##A## is the 2-d area generated by edges 1 and 2, which can be computed as ##l_1 l_2## as described above.
This goes all the way to the end, where you compute the volume of the whole thing as - take the nth edge, compute the length of the component of it that is orthogonal to the other n-1 edges, and multiply that by the n-1 volume of the other edges.
Note the determinant and the volume are both unchanged by applying a rotation to the coordinates. You can always rotate the parallelopiped so that the first edge is of the form ##(a_{11},0,0....,0)##, the second edge is of the form ##(a_{21},a_{22},0,0,....)##, etc. Then the orthogonalization scheme for computing the volume and the determinant both evaluate to ##\prod a_{ii}##, the product of the diagonal terms.
oh thanks! i forget the rotate, know i understand it better, you need to rotate in order to make the edge except the first one(orthogonal one) all lies in the span of the first n-1 basis element, and from det(AB)=det(A)*det(B) is obvious rotate is ok.Yeah I think you could really clean this up by an induction. If you orthogonalize the first edge, and rotate so that this orthogonal thing lies along the first axis, then the matrix is a block matrix with a number in the top left corner and an n-1xn-1 matrix in the bottom right. By induction the volume of the n-1 parallelopiped of edges 2-n is the determinant of that bottom right matrix, and then both the volume and the full determinant are computed by multiplying by the number in the top left.
I dont think it's necessary to rotate the vectors except the first one. cause we could use the induction assume for the n-1 dimension part.If almost feels like this is ripe for an inclusion-exclusion formula - given any n vectors, let the first vector be ##(a_,1....,a_n)##. Then you can compute the volume by assuming the first vector is actually ##(a_1,0,...,0)##, then subtract the volume you get by assuming it's ##(0,a_2,0,0,...,0)## then add back the volume you get by assuming it's ##(0,0,a_3,0,0,...,0)##, but I don't see how to actually show that adds up.
I dont think it's necessary to rotate the vectors except the first one. cause we could use the induction assume for the n-1 dimension part.
I dont think it's necessary to rotate the vectors except the first one. cause we could use the induction assume for the n-1 dimension part.
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I just understand you wrongly... I think we know the det is just directly added up, so I dont think here the inclusion-exclusion formula would engage in. in 2-dim and 3-dim I would think this kind of adding up is simply a cut and complement of a wierd area(in 2-dim would be a triangle, which is not weird)
I think the adding up equals to the formal volume, which is simply a translation of some weird area(in 2-dim is some triangle). also, I need to note that we have come to the oriental area(i.e the area could be minus), which is define simply the det of a set of ordered vectors(or edges).I agree, I was thinking about how you might prove this without any rotations, which would feel really motivating for why the determinant formula is what it is, other than "amazing, it has all these great properties"