OK, back to the drawing board. If we have a singular value decomposition of X,

with u and v orthogonal, then

where we have defined

.

That is, if we have an SVD of X, then we can construct an eigendecomposition of .

(conceptually, the derivation of the SVD is nothing more than reversing that implication: given an eigendecomposition of , get an SVD for X. the challenge is to construct u.)

Note that i did not say the SVD of X, nor the eigendecomposition of . neither is unique. sometimes i may forget, so if i do refer to the SVD of a matrix or the eigendecomposition of a matrix, please forgive me. of course, “the SVD” as such means any decomposition of the form

We can also construct an eigendecomposition of . you should expect that it uses u instead of v.

.

where we have defined

.

since w is not square in general, . That is, (they’re not even the same size), but both are diagonal, and both have the same nonzero entries. In fact, and we want to know this,

the nonzero entries of D or are the squares of the nonzero entries in w.

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