Meaning of A^T A and A A^T

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The matrices [tex]A^{T}A[/tex] and [tex]AA^{T}[/tex] come up in a variety of contexts. How should one think about them - is there a way to understand them intuitively, e.g. do they have a geometric interpretation?
 

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  • #2
chiro
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The matrices [tex]A^{T}A[/tex] and [tex]AA^{T}[/tex] come up in a variety of contexts. How should one think about them - is there a way to understand them intuitively, e.g. do they have a geometric interpretation?
In general there is no real interpretation that comes to mind. With some specific groups the transpose matrix is actually the inverse matrix. Matrices that have this property include the rotation matrices that have a determinant of one.

Geometrically rotation matrices conserve length. So if you had a vector with the tail at the origin (in other words a point), then when you apply a rotation it preserves length of that vector: if you apply it to a set of points it preserves area/volume etc.

There are of course other uses for transpose matrices like least squares, but off the top of my head I can't give you geometric descriptions or interpretations for those.
 
  • #3
Hurkyl
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is there a way to understand them intuitively,
Lots of practice. :smile:



One thing I intuit about them is that it is often the "right" thing to use when you might have used x2 in an analogous situation with real numbers, or used a norm with an analogous situation with a vector.

Also, you can think of them as being a way to turn a matrix into a symmetric, square matrix that does the "least damage" in some sense.

Of course, these are all algebraic ways to intuit them, rather than the geometric one you asked about. :frown:
 
  • #4
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I think about it that way:

For n=3 for exemple,
Think at A as a linear transformation on R3: then you can think at A^T A as a convex bilinear transformation on R3xR3 giving you the scalar product of two vectors A(x1) and A(x2)

In particular, when x1 = x2 = x, I think at it as a "lengh" fonction on A(x) (hence positive defiinte)
 
  • #5
Landau
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Its eigenvalues are the singular values of A. In the basis of the eigenvectors of A*A and AA*, the matrix A is 'almost' diagonal. This is the Singular value decomposition.

The Singular value decomposition has some geometric implications, but I don't know whether this qualifies as a geometric explanation of A*A itself.
 
  • #6
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One thing from differential geometry comes to mind: If [tex] \gamma : \mathbb{R}^k \to \mathbb{R}^N [/tex] is a parametrization of a [tex] k [/tex]-manifold [tex] M \in \mathbb{R}^N [/tex], and [tex] A = [D\gamma] [/tex] is its Jacobian, then the matrix [tex] A^T A [/tex] is the metric induced on [tex] M [/tex] by the embedding in [tex] \mathbb{R}^N [/tex], which is a _very_ geometric object. This is just a reflection of the fact that [tex] A^T A [/tex] is the matrix of inner products of the columns of [tex] A [/tex] (which is a nice geometric interpretation in and of itself), and in our particular case, the columns of [tex] A [/tex] are the basis vectors of the tangent space to [tex] M [/tex] in the coordinates we've chosen. This fact sometimes comes up in slightly disguised form in the context of multivariable calculus, in the formula for the volume element of a manifold with parametrization [tex] \gamma [/tex]: [tex] dV_M = \sqrt{ [D\gamma]^T [D\gamma] } dV_k [/tex], where "[tex] dV_k [/tex]" is the volume element in [tex] \mathbb{R}^k [/tex]. Since [tex] A^T A [/tex] is the metric, this is just a version of the usual formula [tex] dV_M = \sqrt{g} dV_k [/tex], where [tex] g [/tex] is the determinant of the metric.
 

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