Matrix Theorems Addendum

Matrix Theorems Addendum PhL 9.9.11 This subject has cooled off for a while, I have my fat "matrix binder" which is a collection of matrix theorems, but every once in a while I come across something I did not include there. So I will add those new theorems in this addendum document. A. Some cofactor-related theorems. 1 1. Expressions for the cofactor and determinant 1 2. An even fancier formula for cofactor 3 3. The famous matrix inversion theorem and cofactor transpose identity 5 4. Theorem 1: cof(Rab) = Rab for rotation in N dimensions 6 5. Application of the cofactor expressions to generalized cross products. 6 5A. Application of the cofactor expressions to generalized cross products (Std Notation) 8 B. Traceless Matrix Theorem #1: 9 A. Some cofactor-related theorems. 1. Expressions for the cofactor and determinant This item appears in matrix binder section 2 under the section 10.3 Minors and Cofactors (M&M). I show that the following is one way to write a cofactor cof(apq) = ij.....q..... a1i a2j ...... ap-1x (apy) ap+1z .............. // in general Remember that the "cofactor of a matrix element" is a signed minor = a signed determinant = a number (not a matrix). In the above line, q and p are some fixed numbers. On the ε, q appears in the pth position. Stealing from my tensor.doc Appendix A, here is a more precise way to write the above cof(apq) = εiii...[i→q]...i a1i a2i ...... api.......... aNi The factor api is missing from the set of factors (it is crossed out), so there are only N-1 a factors, and the index that would normally be called ip has been replaced with the number q. I am not proving this right here, but a reasonable derivation is given by me in the above mentioned notes. Notice how p appears on the RHS only implicitly: it is the missing a factor, and index ip replaced by constant q. Now here is one way you can verify that the above must be correct. Multiply both sides by apq and do an implied sum on q. On the right, shove this factor into the missing a position, so we have apq cof(apq) = εiii...[i→q]...i a1i a2i .... apq......... aNi Now just on the RHS, rename the summation index q to be ip (since any dummy summation name is allowed). We then have apq cof(apq) = εiii... i...i a1i a2i .... api......... aNi But now summation index ip is in its exact right position, and factor api is in its exact right position, so we recognize the RHS as being just det(a) and we have shown that Σq apq cof(apq) = det(a) and the fascinating fact is that this result is true for any integer p in the range 1...N ! The index p is the row we are moving across to make this form of the determinant. The formula is also true if you If you accept the above expression for cofactor, then this is a proof that det(a) can be written in this manner. As I then argue in those notes, the sum down columns variation is easily obtained by replacing a → aT in the above formula Σq aTpq cof(aTpq) = det(aT) which we rewrite as Σq aqp cof(aqp) = det(a) and now we are summing on the first indices instead of the second on the LHS. Now suppose we take our opening line above apply it to aT cof(aTpq) = εiii...[i→q]...i aT1i aT2i ...... aTpi.......... aTNi Then we find that cof(aqp) = εiii...[i→q]...i ai1 ai2 ...... aip.......... aiN where now we have swapped the indices on all the a factors everywhere and removed all T's. So this is another way to write the cofactor. Then of course the det(a) formula in terms of cof is the same Summary of all results: The cofactor can be written two ways cof(apq) = εiii...[i→q]...i a1i a2i ...... api.......... aNi cof(aqp) = εiii...[i→q]...i ai1 ai2 ...... aip.......... aiN The determinant evaluation rule can be written in terms of the cofactor in two ways Σq apq cof(apq) = det(a) for any p sum on second indices Σq aqp cof(aqp) = det(a) for any p sum on first indices 2. An even fancier formula for cofactor (a) We have established this fact so far, where [i→Q] is in the Pth subscript position on ε, cof(APQ) = εiii...[i→Q]...i A1i A2i ...... APi.......... ANi If we move [i→Q] to the first position on ε, we pick up a factor (-1)P-1 from doing P swaps, = (-1)P-1 εQiii...i...i A1i A2i ...... APi.......... ANi (b) Let {a,b,c...x} be a list of N-1 elements which are taken from the set of N integers {1,2,3....N} such that every number in the list {a,b,c...x} is different. Let P be the omitted number. (c) Let {a,b,c,....P....x} be a permutation of the integers {1,2.3.....N} where P is as described above. We place P in the Pth position in this list {a,b,c,....P....x}. (d) Consider the following object FPQabc..x ≡ (-1)P-1εQiii...i...i Aai Abi ...... APi.......... Axi where the A factors (including the struck-out one) bear the labels {a,b,c,....P....x} as shown. Observation #1: Although we said all the items in the list {a,b,c...x} are different, if we were to allow two of them to be the same, the RHS above vanishes from the usual symmetry rule of contracting a symmetric object (like Aai Abi) against an antisymmetric object like the ε. Thus the observation is that the quantity FPQabc..x vanishes if two subscripts are the same. Observation #2: If we swap any pair of subscripts on FPQabc..x , the RHS changes sign, which is easy to show. Observation #3. FPQ123..P..N = cof(APQ) from (a) above Therefore, we can express any FPQabc..x as a sign times FPQ123..P..N . The correct sign is given by using the ε tensor in this manner FPQabc..x = εabc..P..x FPQ123..P..N where ε has N indices, and P appears in the Pth position. As an example, if N=4 and P=3, F3Qabc = εab3c F3Q124 and if we set abc = 124, we get F3Q124 = ε1234 F3Q124 = F3Q124 which is the desired result. Moving the P to the first position, we then have FPQabc..x = (-1)P-1εPabc..x FPQ123..P..N = (-1)P-1εPabc..x cof(APQ) Therefore we have shown that (-1)P-1 εQiii...i...i Aai Abi ...... APi.......... Axi = (-1)P-1εPabc..x cof(APQ) or εQiii...i...i Aai Abi ...... APi.......... Axi = εPabc..x cof(APQ) Basically if the labels 1,2,3... P..N in our cofactor structure have been permuted into a,b,c....x, this formula tells us how to unpermute them. (e) At this point, we can rename the N-1 summation indices in this way {i1i2,... ip ..iN} → {a',b',c'...x'} so the above becomes εQa'b'c'...x' Aaa' Abb' ... Axx' = εPabc..x cof(APQ)  As noted earlier, if any two indices in {a,b,c...x} are the same, the above equation is true because both sides vanish. Assuming that all the indices in {a,b,c...x} are different, then, we can replace the right hand side with a sum as follows εPabc..x cof(APQ) = ΣS εSabc..x cof(ASQ) This is true since only the term S=P makes a contribution, all other terms in the sum have two indices being the same. So we arrive then at this final result which we state this way If in the list of N-1 letters {a,b,c...x} each letter takes on some value from the set {1,2,3..N}, then the following is true: εQa'b'c'...x' Aaa' Abb' ... Axx' = εSabc..x cof(ASQ)  where now we have implied sums on all repeated indices including S. (f) We can restate the main results above replacing A → AT. We then find: cof(AQP) = εiii...[i→Q]...i Ai1 Ai2 ...... Aip.......... AiN = (-1)P-1 εQiii...i...i Ai1 Ai2 ...... Aip.......... AiN εQiii...i...i Aia Aib ...... Aip.......... Aix = εPabc..x cof(AQP) εQa'b'c'...x' Aa'a Ab'b ... Ax'x = εSabc..x cof(AQS)  3. The famous matrix inversion theorem and cofactor transpose identity [a-1]pq = cof(aqp)/det(a) where again cof(aqp) is a number. We could write this also as [a-1]pq = cof(aTpq)/det(a) One can define a cofactor matrix this way [cof(aT)]pq ≡ cof(aTpq) and then the above matrix inversion theorem would read a-1 = cof(aT)/ det(a) Then here is a theorem about this cofactor matrix cof(aT) = [cof(a)]T // both sides are matrices Proof: LHSpq = [cof(aT)]pq = cof(aTpq) = cof(aqp) RHSpq = [cof(a)]Tpq = [cof(a)]qp = [cof(bT)]qp b ≡ aT = cof(bTqp) = cof(aqp) QED 4. Theorem 1: cof(Rab) = Rab for rotation in N dimensions For an NxN rotation matrix R = RT and det(R) = 1, we claim the following is true cof(Rab) = Rab The thing on the left is a linear combination of N-1 Rij factors, the right is just a single element. Proof: (1) Since RTR = 1, we know det(R) = ±1, let assume +1 and we are doing a rotation. (2) We also know that R-1 = RT and therefore R-1,T = R (3) Therefore: [ note that det(R-1) = det(RT) = det(R) = 1 ] Rab = RTba = (R-1)ba = cof(RTba) / det(R-1) = cof(Rab) QED If we set det(R) = σ = ± 1, we can restate item (3) as (3a) Therefore: [ note that det(R-1) = det(RT) = det(R) = σ ] Rab = RTba = (R-1)ba = cof(RTba) / det(R-1) = cof(Rab)/σ = σ cof(Rab) and then Rab = σ cof(Rab) 5. Application of the cofactor expressions to generalized cross products. Consider the cross product of N-1 vectors, each of which is an N-vector, Qa = εabc...x BbCcDd.....Xx Suppose all the RHS vectors transform according to a linear transformation B'a = Raa'Ba' Then we can define a "transformed object" as (implicit sums on all indices except a) Q'a = εabc...x B'bC'cD'd.....X'x = ( εabc...x Rbb'Rcc'Rdd' ... Rxx') Bb'Cc'Dd'.....Xx' In Section 2 (f) we showed that εQa'b'c'...x' Aa'a Ab'b ... Ax'x = εSabc..x cof(AQS)  We now make a set of changes to this result. First, take A → R and swap lower case primes with non-primes to get εQabc...x Raa' Rbb' ... Rxx' = εSa'b'c'..x' cof(RQS)  Now increment all the lower case letters by one letter (we will still use x to represent the last letter) εQbcd...x Rbb' Rcc' ... Rxx' = εSb'c'..x' cof(RQS)  Now replace Q by a, and S by a' εabcd...x Rbb' Rcc' ... Rxx' = εa'b'c'..x' cof(Raa')  We have now shown that Q'a = ( εabc...x Rbb'Rcc'Rdd' ... Rxx') Bb'Cc'Dd'.....Xx' = ( εa'b'c'..x' cof(Raa') ) Bb'Cc'Dd'.....Xx' = cof(Raa')  εa'b'c'..x'Bb'Cc'Dd'.....Xx' = cof(Raa')  εa'bc..xBbCcDd.....Xx = cof(Raa') Qa' This then shows how a generalized cross product transforms under a linear transformation. Q transforms under a linear transformation Maa' = cof(Raa'), but in general M is not the same as R. If the linear transformation R satisfies RRT = 1 (which says detR = σ = ± 1), then from Section 4 above we know that cof(Raa') = σ Raa' and then we have Q'a = σ Raa'Qa' and finally if R is a true rotation with det(R) = 1, we get Q'a = Raa'Qa' This says that if R is a rotation matrix, then the generalized cross product of N-1 vectors itself transforms as a vector (all in N dimensions). Moreover, in this case we have εabcd...x Rbb' Rcc' ... Rxx' = εa'b'c'..x' cof(Raa')  = εa'b'c'..x' [ σ Raa'] = σ εa'b'c'..x' Raa' For N = 3 this says for example εabc Rbb' Rcc' = σ εa'b'c'Raa' and when R is a rotation with detR = σ = 1, this says εabc Rbb' Rcc' = εa'b'c'Raa' This obscure result sometimes comes in handy! Since this is true for any rotation R, it is also true for the rotation RT so we can write εabc RTbb' RTcc' = εa'b'c'RTaa' or εabc Rb'b Rc'c = εa'b'c'Ra'a Note added 11.13.11. Staying on "developmental notation", consider again Q'a = cof(Raa') Qa' Write cof(Raa') = (R-1)a'a det(R) = Sa'a det(R) = STaa' det(R) then we can write the result as Q'a = det(R) STaa'Qa' = J-1 STaa'Qa' But Qa is a contravariant vector by intension, so should have R and not ST here, and J is the wrong power. ******************************* Try repeating the above converting to Standard Notation, 5A. Application of the cofactor expressions to generalized cross products (Std Notation) Consider the cross product of N-1 vectors, each of which is an N-vector, Qa = εabc...x BbCcDd.....Xx Suppose all the RHS vectors transform according to a linear transformation B'a = Raa'Ba' Then we can define a "transformed object" as (implicit sums on all indices except a) Q'a = ε'abc...x B'bC'cD'd.....X'x = εabc...x B'bC'cD'd.....X'x = ( εabc...x Rbb'Rcc'Rdd' ... Rxx') Bb'Cc'Dd'.....Xx' In Section 2 (f) we showed that εQa'b'c'...x' Aa'a Ab'b ... Ax'x = εSabc..x cof(AQS)  We now make a set of changes to this result. First, take A → R and swap lower case primes with non-primes to get εQabc...x Raa' Rbb' ... Rxx' = εSa'b'c'..x' cof(RQS)  Now increment all the lower case letters by one letter (we will still use x to represent the last letter) εQbcd...x Rbb' Rcc' ... Rxx' = εSb'c'..x' cof(RQS)  Now replace Q by a, and S by a' εabcd...x Rbb' Rcc' ... Rxx' = εa'b'c'..x' cof(Raa')  We have now shown that Q'a = ( εabc...x Rbb'Rcc'Rdd' ... Rxx') Bb'Cc'Dd'.....Xx' = ( εa'b'c'..x' cof(Raa') ) Bb'Cc'Dd'.....Xx' = cof(Raa')  εa'b'c'..x' Bb'Cc'Dd'.....Xx' = cof(Raa')  εa'bc..xBbCcDd.....Xx = cof(Raa') Qa' = J-1 STaa'Qa' // see earlier This says that Qa transforms as a vector density with weight +1. So this is consistent with my work elsewhere in the tensor doc support docs. You get the same result with either approach: ε'abc...x = εabc...x // used in the above ε'abc..x = J-1 Sa'a Sb'b ...... Sx'x εa'b'c'...x' B. Traceless Matrix Theorem #1: Claim: (implied sums on all repeated indices) εjmn SijSnm + εijkSmjSkm = 0 for i = 1,2,3 where Skk= 0 Proof: This is in mws of this name in the matrix folder. First, define eps in the usual way. Then You can see that these are right. Now here are the three evaluations of the LHS: In each case the result is 0, QED. This is a "proof by direct evaluation of all cases".