polar decomposition

Polar Decomposition PhL 1.31.12 The claim is that, if F is any matrix with detF ≠ 0, you can write it as F = RU "the right polar decomposition" where R is a rotation matrix and U is symmetric and positive definite, and this decomposition is unique. In Section 1 I find an R and U that works for F = RU and then in Section 2 I show that R and U with their properties as stated are unique. This uniqueness proof requires the use of a theorem which states that the square root of a pos def sym matrix which is itself pos def sym is unique. I prove this theorem in another doc called Crystal...doc, but I comment on it here in Section 3. Then in Section 4 I do a similar treatment of the left polar decomposition which I write as V = VR' and again I get expressions for V and R' and again they are unique. Finally in Section 5 I show that the R of the right decomposition is the same as the R' of the left decomposition. I have not dwelled much on this subject at all beyond just getting these things proved. The proofs were non-trivial and took me a day or two! 1. First, find a solution for U and R in F = RU (Right Polar Decomposition) 1 2. Show the solution is unique 3 3. Comments on the Required Theorem 3 4. Same polar decomposition with V on the other side (Left Polar Decomposition F = VR' . 4 5. Show that R = R' 5 1. First, find a solution for U and R in F = RU (Right Polar Decomposition) Assume such orthogonal R and symmetric positive definite U exist, so that F = RU. [ If we can compute R and U and show they have these properties, then this assumption is justified.]. Then FTF = (RU)T(RU) = URRTRU = UTU = UU = U2 S ≡ FTF = U2 Then the problem of determining U is merely one of finding the square root of the matrix S ≡ FTF, which is manifestly symmetric. We know something else about such an S: the eigenvalues are positive. Here is why: λ|φλ|2= <φλ|λφλ> = <φλ|Sφλ> = <φλ| FTF φλ> = <Fφλ|Fφλ> = | Fφλ|2 QED Therefore S is symmetric and positive definite, and we seek a square root of S, called U, which also has these properties. We know this theorem: A symmetric matrix S (S = ST) can be diagonalized by a real orthogonal matrix (QT = Q-1). This is an application of the more general theorem that a Hermitian matrix (H = HT*) can be diagonalized by a unitary matrix (QT* = Q-1). Since S is symmetric, we therefore know that an orthogonal Q exists such that Q-1SQ = Λ // a diagonal matrix => S = QΛQ-1 where the diagonal elements of Λ are the eigenvalues of S which we showed are positive. So here is our candidate for U: U = QΛ1/2Q-1 where the meaning of Λ1/2 is that we take the positive square root of all the diagonal elements of Λ, so then Λ1/2 has all positive diagonal elements. First, let's show that in fact U2 = S: U2 = QΛ1/2Q-1 QΛ1/2Q-1 = QΛ1/2Λ1/2Q-1 = QΛ Q-1 = S Next, since Q-1= QT we find that UT = (QΛ1/2QT)T = QΛ1/2QT = U so we find that indeed U is symmetric. Looking at our form for U, we have UQ = QΛ1/2 We know that the columns of orthogonal Q must be the normalized eigenvectors of the problem Uφn = λn1/2φn and these columns are in the right order to match Λ and Λ1/2. Thus, the above line really says U {.... φn ....} = {.... φn ....} Λ1/2 = {.... λn1/2φn ....} => U φn = λn1/2φn λn1/2 > 0 Therefore our candidate matrix U is both symmetric and positive definite, which was part of our starting requirement. U is also invertible because detU = product of the λn1/2 > 0. Although we don't use this fact, we can apply U to the above equation to show that U and S have exactly the same eigenvectors, and that while S has eigenvalues λn>0, U has eigenvalues λn1/2 > 0. Meanwhile, our candidate for R is this. R = F U-1 How do we know this R is orthogonal? RT = (F U-1)T = U-1,T FT = UT,-1 FT = U-1 FT Then RTR = U-1 FTF U-1 = U-1 U2 U-1 = 1 Thus R is indeed orthogonal. Thus, we have found explicit expressions for U and R which have the required properties. We must first find the normalized eigenvectors and the eigenvalues of FTF. We know these eigenvalues are all positive, and we know those eigenvectors comprise the columns of orthogonal matrix Q. So our first step then is to obtain Λ (in some order, perhaps monotonic descending) and a corresponding Q, where Q-1 (FTF) Q = Λ Our solution for U and R is then this: U = QΛ1/2Q-1 // U is symmetric and positive definite R = F U-1 // R is a rotation matrix. 2. Show the solution is unique Suppose the decomposition is not unique so we have in fact two distinct solutions: F = R1U1 = R2U2 Then write FTF two ways to get FTF = FTF (R1U1)T(R1U1) = (R2U2)T(R2U2) U1T U1 = U2TU2 => U12 = U22 where Ui are each symmetric and positive definite. or S = S where S = U12 and S = U22 Now we need this theorem: Theorem: Consider U2 = S. If S is a given positive definite matrix, we claim that there is only one positive definite symmetric U which solves this equation. Call this unique solution U0. Assume we have proven this theorem. The problem S = U12 then has only the solution U1 = U0, and the problem S = U22 has only the solution U1 = U0. Therefore we end up with U1 = U2 so our U candidate given above is in fact unique. Now from the start we have R1 = F U1-1 R2 = F U2-1 If the U's are the same, then the R's must also be the same. 3. Comments on the Required Theorem Theorem: Consider U2 = S. If S is a given positive definite matrix, we claim that there is only one positive definite symmetric U which solves this equation. Call this unique solution U0. This theorem is "the hard part" of the uniqueness proof. After various attempts, I was unable to prove this theorem on my own, so I went hunting out on the web and found this slightly more general theorem proved by a master's student named Crystal (see Crystal kth root of a matrix.doc) : First of all, A is positive semidefinite means λi≥ 0 A is positive definite means λi> 0 so the case I need is a subset of her general theorem. Secondly, I only need k = 2, but it is shown for any integer k, so again my need is a subset. Crystal goes beyond the basic theorem I need and shows that: If A = Bk and B is the unique solution of interest, then [A,B] = 0 and B = poly(A). In my positive definite case, A and B both have full rank N, and (c) is obvious. It is this polynomial business that allows the uniqueness proof to be done, something I would never have thought of. 4. Same polar decomposition with V on the other side (Left Polar Decomposition F = VR' . You can also decompose this way (where R is the same as above), it is claimed: F = VR' Let's try imitating some of the steps above, not knowing that it is the same R. I won't repeat all the words, just the basic equations: FFT = VR'(VR')T = VR'R'-1VT = VVT = V2 S' = FFT = V2 S' is positive definite by the same argument as above. Diagonalize S' as above Q'-1S'Q' = Λ' Candidate is V = Q'Λ'1/2Q'-1 Since S ≠S', everything is (can be) different now, the eigenvectors and values are different, Λ' is different, Q' is different. The candidate for R' is R' = V-1F and we show this is orthogonal as before, R'T = FTV R'R'T = V-1F FTV = V-1 V2V = 1 where again V is symmetric and positive definite, and R' is a rotation matrix. 5. Show that R = R' This is completely non-obvious to me! We do know this much: F = VR' F = RU => RU = VR' => R' = V-1RU => V = RUR'-1 Horn verifies that it is true that R = R', but it is given as an exercise to verify it. I tried and flailed and then found the answer in Google books Dover Continuum Mechanics by Peter Chadwick, 1999, 192 pages, $7 Amazon. Here it is: F = RU = (R'R'T) VR' = R' (R'T VR') ≡ R' T Consider the matrix T = R'T VR'. Since V is symmetric, we have TT = T so T is symmetric. Written as T = R'-1 VR', it is trivial to show that T has the same characteristic equation as V and therefore T has the same eigenvalues as V. Therefore, since V is positive definite, so is T. So we have just shown that F = RU = R'T This represents two right polar decompositions of F! Since we know that the two pieces with the usual properties are unique, we must conclude that R = R' T = U => R'T VR' = U => RTVR = U Thus, in fact U and V both have the same eigenvalues since they are related as shown on the right. Thus in fact Λ = Λ' as used above.