Determinants, Part 1

Transcription

1 Determinants, Part We shall start with some redundant definitions. Definition. Given a matrix A [ a] we say that determinant of A is det A a. Definition 2. Given a matrix a a a 2 A we say that determinant of A is det aa 22 a2 a2 a 2 22 A. Important Note about 2 2 matrices. Elementary row operations effect on determinant: Scaling row scales determinant with the same factor, switching rows changes the sign of determinant, and adding one row to another row does not change determinant at all. [Attend the class!] Important Note 2 about 2 2 matrices. Transposing matrix does not change determinant. Important Note 3 about 2 2 matrices. It is obvious that if the columns of A are dependent then the determinant of A is 0. The converse is also true i.e., if the determinant of A is 0 then the columns of A are linearly independent. This is because elementary row operations would change non-zero determinant into non-zero determinant (see previous note). Since the matrix A with independent columns is row-equivalent to I2 whose determinant is non-zero (in fact it is ) then the determinant of A should be non-zero as well. For the rest of this lecture make sure you read appendix first unless you are already familiar with permutations, transpositions, and parity of permutations. Definition 3 Let A be a square n n matrix with n 2. Let p be any permutation of,2,,n. The determinant of matrix is a summation of all possible terms ( sgn p) a a p 2 p2 anpn over all possible permutations of,2,,n. [There are n! of them and therefore n! terms.] Here the sgn p means the parity of permutation p, which is + if it is made of even number of transpositions and - if it is made of odd number of transpositions. Clearly this definition is in accordance with definition and 2 given previously. We use notation A or det A for determinant of matrix A.

2 Example. a a a 2 3 a a a a a a a a a a a a a a a a a a a a a a a a Another way to see this determinant would be by writing first two columns additionally to the right of the matrix: a a2 a3 a a2 a2 a22 a23 a2 a 22 a 3 a32 a33 a3 a 32 and then simply assembling products across diagonals, positive products from top down, negative products from bottom up a a a a a a a a a a a a a a a [This is known as Sarrus Rule, Pierre Frederic Sarrus] Example det Theorem. Determinant of transpose of A is the same as determinant of A. Shortly 2 A T A. A term ( sgn p) a a p 2 p2 anpn of determinant of A becomes ( sgn p) a a p p2 a 2 pn when the entries of the matrix A are transposed. Using n commutative property of multiplication of real numbers we can rearrange the factors in this product so that we get ( sgn p ) a a a and the only thing remains to p 2 p 2 np n check is that sgn p sgn p which is obvious since is composition of the same transpositions as p only in reverse order. Therefore transposing matrix every term of determinant of A becomes a term of determinant of transpose of A. p

3 Example. (again) Now notice that red diagonals correspond to ( ) 2 3 cycle 32 ( cycle) which are all three even permutations. The blue diagonals correspond to transpositions therefore odd permutations. Theorem 2. If matrix B is made by interchanging two rows (columns) of matrix A the determinant changes a sign, i.e., B A. With a i-j column swap a product a a p ipi a jp j anpn becomes a a a a. The permutation relative to this product is simply p ip j jp i np n i j p which changes the sign. In case of row swap one can make the same conclusion using Theorem. Theorem 3. If a matrix A has two identical rows (columns) then A 0. Obvious consequence of previous theorem Theorem 4. (a) If matrix B is made by multiplying a row (column) of matrix A by real number k then B k A. [This is obvious given the definition of determinant.] (b) If matrix B is made by adding a row (column) of matrix A to another row (column) of A then B A. Proof of (b): A product in determinant of B as a a p ipi aipi a jp j a becomes a np n sum of two products a a a a p ip i jp j npn a p aipi aipi anpn where the second term would correspond to a product in determinant of matrix made from matrix A with j-th row being replaced with i-th row and the conclusion now follows from Theorem 3. 3

4 Corollary. If a matrix B is made from matrix A by adding a multiple of one row to another the determinant of B is the same as determinant of A. This is easy to see since if we add say 3 times the second row to the first then this is equivalent to multiply second row with 3 (the determinant triples) then adding such a second row to the first (no change in determinant), then multiplying second row with /3, which makes determinant /3 of the one in previous step. All in all, no change to determinant. Theorem 5. If E is elementary matrix (made by either (a) exchanging two rows of identity matrix, or (b) multiplying one row of identity matrix with non-zero real number, or (c) adding a multiple of one row to another then EB E B det det det In (a) case this is obvious since here det (E) is simply so we can use Theorem 2. The same with (b) using Theorem 4. And again using corollary we get the case (c). Corollary 2. If two matrices A and B are row equivalent then if A 0 implies B 0. We can see from Theorem 5 that elementary row operations cannot produce determinant 0 if the starting matrix did not have determinant 0. Theorem 6. Matrix A is invertible if and only if A 0. If A is invertible then we know it is row equivalent to identity matrix and now we simply use Corollary 2 for the conclusion that A 0 since I 0. If A is not invertible then it is row equivalent to a matrix with at least one row with zeros and the conclusion is obvious again by Corollary 2. Corollary 3. Square n n matrix A has rank n if and only if A 0. 4

5 Straightforward consequence from the previous theorem and the Theorem 4 from lecture on Inverse Matrix. Lemma. Let A, B be two n n matrices. If matrix A is not invertible then matrix AB is not invertible. Linear transformation given by A is not - by Theorem 4 from lecture Inverse Matrix. That means that linear transformation given by AB is not -. This implies that AB is not invertible by the same theorem. Note. If B is not invertible then AB is not invertible. That is because linear transformation given by B is not onto and thus linear transformation given by AB is not onto so again we can use Theorem 4 from Inverse Matrix lecture. Theorem 7. AB A B det det det If A is invertible then it is a product of elementary matrices A EE 2 E n thus by multiple application of Theorem 5 we get as needed. AB EE 2 EnB E E2 En B Adet B det det det det det det det If A is not invertible AB is not invertible and thus the statement of theorem is simply a statement 0 0 by Theorem 6. Corollary 4. If A is invertible matrix the following is true. A A. Obvious consequence of the previous Theorem since I n. Definition 4. Matrices A and B are said to be similar if there is an invertible matrix P such that B P AP. 5

6 Note 4. This relation is a relation of equivalence. Show that! Corollary 5. Similar matrices have the same determinants. [Obvious consequence of Theorem 7.] The reverse is not true as shown in the following example. Example 3. The following matrices 0 0, 0 0 have the same determinant but they are not similar since the first matrix can only be similar to itself. Definition 3. Let A be a square matrix, then we define the minor of the entry a i, j (also called the (i,j)- minor, or a first minor) as the determinant of the submatrix A i, jformed by deleting the i-th row and j-th column from matrix A. This number is denoted det A i, j or just M i, j. The submatrix itself is denoted by A i, j. The (i, j)-cofactor is obtained by multiplying the minor by i j. We shall denote (i,j)-cofactor by C i, j. The following figure is the map of signs of cofactors. Fig. Map of cofactor signs Example 4. M M A A 22, A C det A ,2 2,2 2,2 23 C det A ,3 2,3 2,3 6

7 Theorem 8. [Pierre-Simon, marquis de Laplace] Let A be a square n n matrix with n 2. det A a C a C a C a C (.) 2 2 n n j j j n Example det Proof of Theorem 8 (difficult so you can skip on the first reading): Since every term in det A contains one (and only one) of the entries from the first row then we can factor them out and the result is something that looks like det A ad a2 D2 a nd n The terms in D do not have any entry from the first row nor from the first column but they do look like a a 2 p 2 3 p3 anpn over all the permutations p of,2,,n 2 3 n meaning and since the parity of is the same as parity of p2 p3 pn 2 3 n the we have D M. We can also (switching the first and p2 p3 pn the second column of A) show that D2 M2. [It is important to notice that parities of permutations are affected by switch while the determinant M 2 of (i,j)-minor is not affected by this switch since one of the columns that is switched is not in the minor and the columns that are switched are adjacent.] We can continue this switching adjacent columns and using previously proven to show that D3 M3, D4 M4, which shows that Dj Cj as we needed to show. Theorem 9. Let A be a square n n matrix with n 2. det A a C a C a C a C (.2) i i i2 i2 in in i j i j j meaning that determinant of A can be calculated by using any row. n 7

8 According to theorem #2 switching two adjacent rows like the first and the second only changes the sign of determinant but since in the new position the signs of cofactors change as well while the minors stay the same then the determinant does not change at all. Thus (.2) holds for i 2. But then we can continue by induction for example switching rows 2 and 3 follows the same rules (determinant of A changes but minors do not) etc. Example Calculate determinant of A (from example 2) and convince yourself it is 4 again regardless which row you would use for the expansion (.2). Note 5. Using always the last row you can easily see that upper triangular matrix has determinant that is simply a product of diagonal entries. The same with lower triangular matrix (using always the first row) and obviously with diagonal matrix. Theorem 0. Let A be a square n n matrix with n 2. Then the following holds, det A a C a C a C a C (.3) j j 2 j 2 j n j n j i j i j i meaning that determinant can be calculated by using any column as well. Follows trivially from the previous theorem and from Theorem since the cofactor map of signs does not change with transposition. n We refer to the - map of We also write the following notation: Appendix Permutations of,2,,n,2,,n onto,2,,n as a permutation of,2,,n. 8

9 2 n p p2 pn for permutation p. Example is a permutation sending to 2, 2 to 3, 3 to, and 4 to 4. We say that permutation is a transposition when it is made of exchange of two elements from,2,,n, for example [or the simplest way 3 as in swapping and 3 ] is a transposition because it exchanges and 3. Any permutation can be written as a product (meaning composition) of transpositions. Example = 3 2 meaning we achieved permutation first swapping and 2 and then and 3. Example 3. [Straightforward decomposition of a permutation into transpositions. We have a better way than this and about that we shall learn later.] can be achieved by the following steps (in that order). [Keep in mind that we are not looking for the most efficient way but for algorithmic approach!] We start with Put 3 at the first position by the following transpositions: swap 3 and 2, then swap 3 and. The result is 9

10 Put 5 at the second position by the following transpositions: swap 5 and 4, then swap 5 and 2, then swap 5 and. The result is Put 2 at the third position by the following transposition: swap 2 and. The result is This gives us the decomposition of our permutation into 6 transpositions: [we are composing functions!] Pictorially (follow the direction of arrows for the composition of transpositions): The previous should give you a general idea how to approach decomposition of any permutation into product (composition) of transpositions. First you accomplish to fill the (eventual) first position, then (eventual) second, then (eventual) third etc., as much as needed. It is obvious that this can be done always with transpositions. The parity of permutation is the parity of the number of transpositions the permutation is made of. The permutation in example 3 is made of 6 transpositions thus the parity is even. A permutation can be decomposed in different compositions of transpositions. Example 4. 0

11 Therefore we need to show that the parity as we defined is always the same, meaning for example if you can write permutation in 6 transpositions then every other decomposition will also have even number of transpositions. p p 2 p n We call permutation p 2 n Obviously any transposition is its own inverse. inverse of permutation p. Theorem. Assume that permutation p can be written as a product of transpositions p 2 r 2 s then r and s have the same parity. Since p p 2 r s 2 it suffices to prove that identity can only be written as a product of even number of transpositions. So let a b a b a b (.4) 2 2 k k product (i.e., composition of transpositions, and we need to show that k is even. Clearly k cannot be since () is identity. Thus is k is 2 we are done. Suppose that k 3 and suppose that we know that any product of less than k transpositions that is identity would have even number of transpositions. It is obvious that a has to be present in another transposition in (.4) beyond the first one. Let s call this a just a. Since roles of two elements in a transposition can be reversed we can assume that that is another of the a s. Because of the following cd ab ab cd bc ab ac bc

12 we can always bring that second occurrence of a next to ab without changing the number of transpositions. Meaning we can assume that we have ab ab2 akbk. Neither b nor b 2 are equal to a. Now we have two cases: Case. b b2 In this case the first two transpositions cancel each other and we have by induction that remaining number of transpositions is even. This our claim is proven. Case 2. b b2 Now we have and we have ab ab ab b b ab b b a b a b k k. Note that we got rid of second occurrence of a. Thus we have another a in a b a b and we can repeat the same procedure again. Thus it is obvious that at 3 3 k k some point we shall reach Case (cancelation of two transpositions) because we cannot end up exhausting the length of decomposition without finding another a. Theorem 2. There is equal number of even and odd permutations of,2,,n if n 2. For every even permutation p we have 2 p to be odd and whenever permutations p q then 2 p 2 q. [Assume opposite and use inverse 2which is 2 ] We call permutation a cyclic permutation or a cycle when it is made of portion of elements of,2,,n in a cyclic swap: i j k i while other indices do not change. Example as 3 5 2

13 We also write this shorter as 35, swap cycle of 3 elements. We call this number of elements the length of the cycle. We shall label the length of cycle using notation of absolute value, for example This type of permutation is also its own inverse. Every permutation can be written as a product of disjoint cycles. [This should be obvious.] For example is itself cyclic since but we can have more than one cycle in a permutation (i.e., the permutation made of more than one cycles) for example [ Disjoint means there are no common elements in the short notation.] It should be obvious that cyclic decomposition for any permutation into disjoint non-repeating cycles is unique since cycles do not have any common elements. It should also be obvious that any cycle of length k will be possible to decompose into k transpositions in the manner of the following example. Example Note the following example for the inverse of a cycle. Example Theorem 3. The parity of permutation p is the same as the parity of n c c2 c m where c, c2,, c m are cycles in the decomposition of permutation p into non-repeating disjoint cycles. Proof (the idea): Just take a look at the example 6. 3

14 Theorem 4. p p2 pn Inverse permutation p 2 n Since every permutation is a product of cycles then has the same parity as p. p c c c c c c 2 n n n and the rest is obvious from the previous theorem. [We could also use the same argument with c s being the transpositions.] The following theorem is known Theorem 5. The number of permutations of,2,,n is n!. The number of permutations of is which is! Assume that the number of permutations of,2,, n, where n, is n!. That means that the number of bijections (bijection is - and onto mapping) of,2,, n onto a,, an is also n!. This means that the number of permutations p of,2,,n such that pn i is also n!. Since we have n choices for such an i then the we have n! n n! permutations of,2,,n. The proof of the theorem follows by induction. Now by virtue of the previous theorem and the theorem #2 we have the following, Corollary The number of even and odd permutations of,2,,n when n 2 is the same, namely n!. 2 4