Editing Cholesky decomposition (section)

== Geometric interpretation ==
{{See also|Whitening transformation}}
[[File:Cholesky decomposition with two ellipses.svg|thumb|The ellipse is a linear image of the unit circle. The two vectors <math display=inline>v_1, v_2</math> are conjugate axes of the ellipse chosen such that <math display=inline>v_1</math> is parallel to the first axis and <math display=inline>v_2</math> is within the plane spanned by the first two axes.]]
The Cholesky decomposition is equivalent to a particular choice of [[Conjugate diameters|conjugate axes]] of an [[ellipsoid]].<ref>Pope, Stephen B. "[https://tcg.mae.cornell.edu/pubs/Pope_FDA_08.pdf Algorithms for ellipsoids.]" Cornell University Report No. FDA (2008): 08-01.</ref> In detail, let the ellipsoid be defined as <math display=inline>y^TAy = 1</math>, then by definition, a set of vectors <math display=inline>v_1, ..., v_n</math> are conjugate axes of the ellipsoid iff <math display=inline>v_i^T A v_j = \delta_{ij}</math>. Then, the ellipsoid is precisely<math display="block">\left\{ \sum_i x_i v_i : x^T x = 1 \right\} = f(\mathbb S^n)</math>where <math display=inline>f</math> maps the basis vector <math display=inline>e_i \mapsto v_i</math>, and <math display=inline>\mathbb S^n</math> is the unit sphere in n dimensions. That is, the ellipsoid is a linear image of the unit sphere.

Define the matrix <math display=inline>V := [v_1 | v_2 | \cdots | v_n]</math>, then <math display=inline>v_i^T A v_j = \delta_{ij}</math> is equivalent to <math display=inline>V^TAV = I</math>. Different choices of the conjugate axes correspond to different decompositions.

The Cholesky decomposition corresponds to choosing <math display=inline>v_1</math> to be parallel to the first axis, <math display=inline>v_2</math> to be within the plane spanned by the first two axes, and so on. This makes <math display=inline>V</math> an upper-triangular matrix. Then, there is <math display=inline>A = LL^T</math>, where <math display=inline>L = (V^{-1})^T</math> is lower-triangular.

Similarly, [[principal component analysis]] corresponds to choosing <math display=inline>v_1, ..., v_n</math> to be perpendicular. Then, let <math display=inline>\lambda = 1/\|v_i\|^2</math> and <math display=inline>\Sigma = \mathrm{diag}(\lambda_1, ..., \lambda_n)</math>,  and there is <math display=inline>V = U\Sigma^{-1/2}</math> where <math display=inline>U</math> is an orthogonal matrix. This then yields <math display=inline>A = U\Sigma U^T</math>.