Volume of a sphere under a linear transformation R3->R4.

BluFoot
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Homework Statement


So there's a linear transformation T: ℝ3 → ℝ4, standard matrix A that satisfies

det(A e1) = 5, det (A e2) = 4, det (A e3) = 5 and det (A e4) = 5

If S is the unit sphere, find the 3-dimensional volume of T(S).

Homework Equations


Volume of sphere = 4/3 * pi * r^3
Volume based on determinants = det(ATA)


The Attempt at a Solution


I know the determinant of a matrix can be seen as the scaling factor for the volume change of a transformation. So the answer will be 4/3 * pi * (something). The (something) will probably be some kind of combination of the determinants above (5,4,5,5), but I have no clue how to find its value. I have spent hours on this question, it's driving me crazy. Help!
 
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Thinking about the ℝ1 → ℝ2 suggests a root-sum-squares relationship. Can't see how to generalise it yet.
 
Have now proved the following for both ℝ1 → ℝ2 and ℝ2 → ℝ3, but still don't see how to prove it in general.
Denoting the square matrices formed by appending the basis vectors as Ai, det(ATA) = Ʃ(det(Ai))2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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