bunk001 ([info]bunk001) wrote in [info]mathematics,
@ 2008-12-27 16:45:00
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Matrices, Determinant, and MANOVA
I realize that the determinant is used to calculate the inverse matrix and that the inverse matrix is useful for solving simultaneous equations.

However, the determinant is also used in MANOVA.

According to a text that I am reading, the determinant is analogous to variance. Is it possible for someone to explain why the determinant is considered analogous to variance?

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[info]cheeser1
2008-12-27 04:59 pm UTC (link)
I don't know this. I will guess that the matrix has entries a(i,j)=Cov(Xi,Xj). I don't know at all though.

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[info]notquitezeus
2008-12-27 05:11 pm UTC (link)
imagine you are working with a single r.v. with gaussian distribution. if i tell you that your r.v. has mean μ and variance σ2, you know that approximately 68% of samples should lie in the interval [μ-σ, μ+&sigma].

now, when you go from scalar r.v. to a vector r.v., a similar situation applies — your vector valued r.v. now lives in box. you can figure out what the box is by looking at the eigen-decomposition of Σ — the eigenvalues are the squares of the lengths of the sides of the boxes and the eigenvectors are the orientations of each side of the box. unfortunately, you don't yet have a single number that describes the total "size" of the box yet. this is where the determinant comes to play. one of the (many) definitions of the determinant is that it is the product of all the eigenvalues, which means that it corresponds to something proportional to the square of the volume of your r.v.'s box.

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