The Unapologetic Mathematician

Mathematics for the interested outsider

Polar Decomposition

Okay, let’s take the singular value decomposition and do something really neat with it. Specifically, we’ll start with an endomorphism T:A\rightarrow A and we’ll write down its singular value decomposition

\displaystyle T=V\Sigma W^*

where V and W are in the unitary (orthogonal) group of A. So, as it turns out, U=VW^* is also unitary. And P=W\Sigma W^* is positive-semidefinite (since \Sigma is). And, of course, since W^*W=1_A, we can write

\displaystyle T=V\Sigma W^*=VW^*W\Sigma W^*=UP

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 T is nonsingular, so P is actually positive-definite, then \det(P) will be strictly positive, so the determinant of T will be nonzero.

We could also define P'=W\Sigma W^*, so T=P'U. This is the left polar decomposition (writing the positive-definite part on the left), where the previous form is the right polar decomposition


August 19, 2009 - Posted by | Algebra, Linear Algebra


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


    Comment by notedscholar | August 19, 2009 | Reply

  2. 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 | Reply

  3. Is the decomposition T=UP unique in any sense?

    Comment by Å | August 20, 2009 | Reply

  4. The positive-definite part P is unique (I’ll be able to say more about that after today’s post), and from that point U is also unique exactly when T (and thus P) is nonsingular.

    Comment by John Armstrong | August 20, 2009 | Reply

  5. […] 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 | Reply

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


    Comment by notedscholar | August 21, 2009 | Reply

  7. […] 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 | Reply

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