## Polar Decomposition

Okay, let’s take the singular value decomposition and do something really neat with it. Specifically, we’ll start with an endomorphism and we’ll write down its singular value decomposition

where and are in the unitary (orthogonal) group of . So, as it turns out, is also unitary. And is positive-semidefinite (since is). And, of course, since , we can write

That is, any endomorphism can be written as the product of a unitary transformation and a positive-semidefinite one.

Remember that unitary transformations are like unit complex numbers, while positive-semidefinite transformations are like nonnegative real numbers. And so this “polar decomposition” is like the polar form of a complex number, where we write a complex number as the product of a unit complex number and a nonnegative real number.

We can recover the analogy like we did before, by taking determinants. We find

since the determinant of a unitary transformation is a unit complex number, and the determinant of a positive-semidefinite transformation is a nonnegative real number. If is nonsingular, so is actually positive-definite, then will be strictly positive, so the determinant of will be nonzero.

We could also define , so . This is the left polar decomposition (writing the positive-definite part on the left), where the previous form is the right polar decomposition

What if I don’t want to ” take the singular value decomposition”? Your entire argument collapses.

NS

Comment by notedscholar | August 19, 2009 |

That’s the nice thing about mathematics. My argument doesn’t collapse just because you hold your breath and stamp your feet.

Comment by John Armstrong | August 19, 2009 |

Is the decomposition T=UP unique in any sense?

Comment by Å | August 20, 2009 |

The positive-definite part is unique (I’ll be able to say more about that after today’s post), and from that point is also unique exactly when (and thus ) is nonsingular.

Comment by John Armstrong | August 20, 2009 |

[…] Uniqueness of Polar Decomposition I asserted in a comment that the polar decomposition of a transformation has certain uniqueness properties. First, we can always uniquely recover the […]

Pingback by The Uniqueness of Polar Decomposition « The Unapologetic Mathematician | August 21, 2009 |

Right. But it does collapse if we don’t grant your premises.

NS

Comment by notedscholar | August 21, 2009 |

[…] used this to show in particular if is an endomorphism on an inner product space we can write where is unitary and […]

Pingback by Decompositions Past and Future « The Unapologetic Mathematician | August 24, 2009 |