m n which are arranged in the following form 11 12 13 21 22 23 mn Quite often the form in which the complex numbers are given is denoted by itself The complex number ij is called the ij th entry of By the th row of we mean the part of the matr ID: 25598 Download Pdf

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m n which are arranged in the following form 11 12 13 21 22 23 mn Quite often the form in which the complex numbers are given is denoted by itself The complex number ij is called the ij th entry of By the th row of we mean the part of the matr

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MTH 102N — LINEAR ALGEBRA 1 Matrices By an matrix we mean the following data: Complex numbers ij ,i = 1 ,...,m ,...,n which are arranged in the following form: 11 12 13 21 22 23 ... ... ... ... ... mn Quite often the form in which the complex numbers are given is denoted by itself. The complex number ij is called the ( i,j )th entry of . By the th row of we mean the part of the matrix formed by the entries ,a ,...,a in . Similarly, by the th column we mean the part of the matrix formed by the entries ,a ,...,a mj . Thus, an matrix has rows and columns. To simplify the notation

we write = [ ij or even = [ ij ] . Given two matrices = [ ij ] and = [ ij ] we deﬁne another matrix, denoted by , by taking its ( i,j )th entry to be ij ij . Thus = [ ij ij If = [ ij ] is an matrix and and is a complex number then the matrix [ ca ij ] is denoted by cA If = [ ij ] is an matrix and = [ ij ] is an matrix, we deﬁne an matrix, denoted by AB , by taking its ( i,j )th entry to be in nj . Thus AB = [ =1 ir rj Examples. 1. An matrix [ ij ] is called integral, rational or real according as all ij ’s are integers or rational numbers or real numbers. 2. An matrix is called a

square matrix. 3. The matrix, all of whose entries are zero is called the zero matrix and is denoted by 4. The matrix [ ij ] such that ij = 0 if and ii = 1 is called the identity matrix and is denoted by or simply by . Note that if is an matrix, then AI 5. Let = [ ij ] be an matrix. The entries ii ’s are called diagonal entries of and all these entries together are said to form the (principal) diagonal of 6. An matrix [ ij ] such that ij = 0 if is called a diagonal matrix. 7. An matrix [ ij ] such that ij ji for all if and is called a symmetric matrix.

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8. An matrix [ ij ] such

that ij ji for all and is called a skew symmetric or antisymmetric matrix. Note that in a skew symmetric matrix all the diagonal entries are 0. 9. An matrix [ ij ] such that ij = 0 for all i < j is called a lower triangular matrix. Also, an matrix [ ij ] such that ij = 0 for all i > j is called an upper triangular matrix. Remark 1.1. (a) The matrix addition is commutative and is associative. (b) The matrix multiplication is associative and distributes over addition of matrices. Deﬁnition 1.2. Invertible Matrices. An matrix is called invertible if there exists a matrix such that AB BA .

In this case the matrix is called an inverse of Lemma 1.3. Let be an matrix. If is invertible then there exists a unique matrix such that AB BA Proof : Suppose there exist matrices and such that AB BA and AC CA . Then CI AB ) = ( CA Deﬁnition 1.4. In view of above Lemma we denote the inverse of an invertible matrix by Problems. 1. Show that matrix multiplication is associative and distributes over addition of matri- ces. 2. Let and be two matrices. Show that ( if is deﬁned and AB if AB is deﬁned. 3. Show that every square matrix can be written as a sum of a symmetric and a

skew symmetric matrices. Further show that if and are symmetric then AB is sym- metric i AB BA 4. Let and be two invertible matrices. Show that ( AB More generally if ,...,A are invertible matrices then ( ...A ...A 2 Row and Column Operations on a Matrix There are three basic row operations which can be performed on a matrix. They are (1) Multiplying a row (i.e. all the entries of a row) of a matrix by a nonzero constant . (2) Interchanging two rows of a matrix and (3) Adding a constant times a row of the matrix to another row. Interestingly these operations correspond to pre-multiplying the

given matrix by a suitable square matrix. More precisely if is an matrix then the matrix by which we need to pre-multiply in order to perform a row operation on is an

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matrix which depends only on the type of row operation to be performed on and not on the entries of . Let us look at the following examples: 0 1 #" a b c d e f αa αb αc d e f (2.1) 0 1 1 0 #" a b c d e f d e f a b c (2.2) 1 0 #" a b c d e f a b c αa d αb e αc (2.3) In the ﬁrst case the matrix on the right hand side is obtained by multiplying the ﬁrst row of the matrix a

b c d e f by a constant . The resultant matrix can be seen to have been obtained by pre-multiplying by the matrix 0 1 If one looks at this matrix carefully then one notices that this matrix is obtained by mul- tiplying the ﬁrst row of by . Similarly, in the second case the matrix by which is pre-multiplied is obtained by interchanging the ﬁrst and the second rows of and the eﬀect is that we have interchanged these rows of In the third case we pre-multiply by a matrix obtained by adding times the ﬁrst row of to the second row of . Again we see the same eﬀect

on Thus in general, to eﬀect a particular row operation on an matrix we need to pre-multiply by the matrix obtained from by performing the same row operation on Similar to the elementary row operations we also have elementary column operations, that is, (1) Multiplying a column (i.e. all the entries of the column) of a matrix by a nonzero constant . (2) Interchanging two columns of a matrix and (3) Adding a constant times a column of the matrix to another column. Performing a particular column operation on a given matrix amounts to post-multiplying the given matrix by the matrix

obtained by performing the same operation on identity matrix. Deﬁnition 2.1. Elementary Matrices. There are three type of elementary ma- trices. They are 1. The matrix ), where is a nonzero constant, is obtained by multiplying the th row of by . If an matrix is pre-multiplied by ) then the resultant matrix is same as obtained by multiplying th row of by . Similarly if an matrix is post-multiplied by ) then the resultant matrix is same as obtained by multiplying th column of by 2. The matrix ij , is the matrix obtained by interchanging the th and the th rows of . If an matrix is

pre-multiplied by ij then the resultant matrix is the

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same as that obtained by interchanging the th and the th rows of . Similarly, if an matrix is post-multiplied by ij then the resultant matrix is same as that obtained by interchanging the th and the th columns of 3. The matrix ij ), where is a constant, is obtained by adding times the th row of to its th row. If an matrix is pre-multiplied by ij ) then the resultant matrix is the same as that obtained by adding times the th row of to its th row. Similarly if an matrix is post-multiplied by ij ) then the resultant matrix is

same as obtained by adding times the th column of to its th column. Lemma 2.2. An elementary matrix is invertible with an elementary inverse. Proof (1 /c ) = (1 /c ); ij ij and ij ij ) = ij ij ). Lemma 2.3. Let be an matrix. Then by a sequence of row and column operations can be reduced to the following form 0 0 n. Proof : Suppose = [ ij If then there is nothing to prove. Suppose , choose ij = 0. Interchange the ﬁrst and the th columns and then the ﬁrst and the th rows. Multiply the ﬁrst row of this matrix by 1 /c and then by column operations make all the remaining

entries of the ﬁrst row zero and by row operations make all the remaining entries of the ﬁrst column zero. The resultant matrix is of the form 1 0 0 11 12 ,n ... ... ... ... ... ,n If any of the ij ’s is nonzero then bring it to the (2 2) position by interchanging rows and columns and make it one and the remaining entries of 2nd row and 2nd column zero by suitable row and column operations. Now the matrix has the following form 1 0 0 0 1 0 0 0 If necessary repeat the process with the part of the matrix not in the right form. Lemma 2.4. Let be an matrix. Then is invertible if and

only if it is a product of elementary matrices.

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Proof : First suppose is invertible. By the previous Lemma there exist elementary matrices ...E AF ...F 0 0 Since the left hand side of above is invertible (being a product of invertible matrices) hence . Thus ...E ...F a product of elementary matrices. Conversely if is product of elementary matrices then is obviously invertible as a product of invertible matrices is invertible. Lemma 2.5. An matrix is invertible if and only if by a sequence of elementary row operations can be reduced to Proof : First suppose is invertible. Then

...E . Hence ...E , that is, by row operations can be reduced to . Conversely, suppose by a sequence of row operations can be reduced to . Then there are elementary matrices ,...,E such that ...E . Hence ...E and therefore is invertible. Remark 2.6. The proof of above Lemma also contains a method of ﬁnding the inverse, if it exists, of an matrix . This method is known as the Gauss row-reduction method or simply the row-reduction method. The method not only determines the inverse of if it exists but also determines if it does not exist. The method works as below: From the above Lemma we

see that if is invertible then there exist elementary matrices ,...,E such that ...E . Hence ...E and therefore is invertible and further ...E ...E . This means if is reduced to be a sequence of row operations and if these row operations, in the same order, are performed on , then the matrix thus obtained is the inverse of

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3 Determinant Permutations. Let denote the set ,...,n . Any one-one onto mapping of to itself is called an -permutation or simply a permutation. Given a permutation , its inverse is denoted by , which is also a permutation. Also given two permutations and ,

the composition is also a permutation. We denote by , the set of all -permutations. We denote by , the identity permutation. Now we prove the following two results: (1) and (2) If is a ﬁxed permutation then To prove (1) we only need to show that every permutation is the inverse of some per- mutation. But as the inverse of the inverse of a function is the function itself, we are done. To prove (2) we need to show that given a permutation there exists a permutation such that . Here we can take Let be a permutation and let . Then is called a ﬁxed point of or ﬁxed by if ) = .

Otherwise is said to be moved by A permutation is called a transposition if it moves exactly two points of Let be a transposition and let and be two distinct point moved by . Then for all \{ x,y we have ) = . We claim that ) = . Suppose . Then \{ x,y and hence )) = ). This implies ) = as is one-one. Similarly ) = . Hence is completely determined by the points x,y it moves. For this reason = ( x,y ). It is fairly easy to show that Any -permutation can also be considered to be an ( + 1)-permutation by assigning the value +1 to +1, that is, by deﬁning +1) = +1. In fact all the

-permutations can be viewed as those ( + 1)-permutations ﬁxing + 1 There following two results are of great importance to us. Theorem 3.1. 1. Every permutation can be written as a product (composition) of trans- positions. 2. If is a permutation and ,..., , ,..., are transpositions such that ... and ... , then either and are both even or are both odd. Proof : The proof of the second result is quite involved, therefore we shall skip it here. We now prove the ﬁrst result. Let be a permutation and let denote the number of points of moved by . We prove the result in the following

inductive manner: We show that if every permutation which moves strictly less than points of , is a product of transpositions then so is . Note that if = 0 then and hence = (1 2)(1 2) and the result holds. Now suppose 1. One can easily see that then 2. Let ,x ,...,x be all the points moved by . Then ) is not ﬁxed by . Else we have )) = ) and in turn ) = , a contradiction. Assume ) = . Put = ( ,x ), a transposition. Consider . Then ) = )) = ) = . Also ) = for all \{ ,x ,...,x . Hence ﬁxes at least + 1 points of . This means points moved by are at maximum 1. Now by induction

hypothesis ... for some transpositions ’s. This implies

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... A permutation is called an even permutation if it can be written as a product of an even number of transpositions. Otherwise it is called an odd permutation. Note that is an even permutation and a transposition is an odd permutation. Sign of a permutation is deﬁned to be +1 if it is an even permutation. Otherwise it is deﬁned to be 1. It is denoted by sign( ). We see that 1. sign( ) = 1 and sign( ) = 1 for every transposition 2. Let and be permutations. If both are even or both are odd then is even and

if one is odd and other is even then is odd. This implies that sign( ) = sign( ) sign( ). Determinants: Let = [ ij ] be an matrix. Its determinant is denoted by and it is the quantity sign( (1) (2) ...a n where the summation is taken over all -permutations. This deﬁnition of determinant appears to be quite complicated and in fact it is so. The origin of this deﬁnition is in the th exterior power of an -dimensional vector space on which a linear transformation is given. But this deﬁnition has certain advantages– it allows us to prove certain properties of the determinants

quite easily. Let = [ ij ] and = [ ij ] be matrices. Then (a) If is obtained by interchanging two rows of then −| (b) If has two identical rows then = 0 (c) If is obtained by multiplying a row of by a constant then (d) Suppose = [ ij ]. Further assume that A, B and diﬀer only in the th row for some and kj kj kj for all . Then show that (e) If is obtained by adding times the th row of to its th row then (f) If is an elementary matrix then EA || (g) = 0 i is not invertible. (h) AB || (i) If then Proof (a) Suppose the th row of is the th row of and the th row of is the th row of ,

that is, pj qj and qj pj and ij ij for p,q and for all . Let denote the transposition ( p,q ). Then is an -permutation . Hence

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sign( (1) (2) ...b p ...b q ...b n sign( ) sign( (1) (2) ...b p ...b q ...b n = sign( sign( (1) (2) ...a q ...a p ...a n −| as sign( ) = (b) Suppose the th and the th rows of are identical, that is, pj qj for all and let denote matrix obtained by interchanging the th and the th rows of . Then and by the ﬁrst part −| . Therefore = 0. (c) Suppose th row of is multiplied by to obtain . Then pj ca pj and ij ij for and for all sign( (1)

(2) ...b p ...b n sign( (1) (2) ...ca p ...a n sign( (1) (2) ...a p ...a n (d) sign( (1) (2) ...c k ...c n sign( (1) (2) ... k k ...a n sign( (1) (2) ...a k ...a n sign( (1) (2) ...b k ...b n (e) In this case for all ij ij for and qj qj ca pj . Let = [ ij ] where for all ij ij if and qj ca pj . Then A,B and diﬀer only in th row and qj pj qj . Hence by (d) . Further = 0 by (c) and (b). (f) By deﬁnition it is clear that = 1. Hence from (a) ij 1, from (c) ) = and from (e) ij ) = 1. Again by (a), (c) and (e) we see that EA || (g) First suppose that is not invertible. Then ...E where

s are ele- mentary matrices and is a matrix which has its last row identically zero. Then || ... || = 0 as = 0. Conversely suppose is invertible. Then ...E a product of elementary matrices. Hence by (f) || ... |6 = 0 as the determinant of an elementary matrix in nonzero.

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(h) If is not invertible then AB is also not invertible. Therefore by (g), 0 = AB || = 0. Now suppose is invertible. Then ...E a product of elementary matrices. Therefore AB ...E || ... || || by (f). (i) Put = [ ij ]. Then ij ji for all i,j . Consider the set is an -permutation This set contains all the

-permutations and every -permutation can be written as an inverse of an -permutation uniquely. Furthermore sign( ) = sign( ). Hence sign( (1) (2) ...b n sign( (1) (1)) (2) (2)) ...b )) sign( (1)1 (2)2 ...b sign( (1) (2) ...a n The results (a)—(e) are valid if the word row is replaced by the word column. As mentioned earlier the deﬁnition of determinant is useless when it comes to computing the determinant. But the properties of determinants provide us with an inductive method for computing the determinant of matrices. First we prove the following lemma: Lemma 3.2. (a) Let = [ ij such

that nj = 0 for = 1 ,...,n and nn = 1 Let = [ ij 1) 1) where ij ij for all i,j = 1 ,...,n . Then (b) Let = [ ij and let ij denote the matrix obtained by removing the th row and the th column of . Then =1 1) ij ij Proof : (a) sign( (1) (2) ...a n If is a permutation such that then n = 0 and hence its contribution to above sum is zero. Therefore , )= sign( (1) (2) ...a n Moreover as nn = 1 we have , )= sign( (1) (2) ...a 1) 1) The set ) = is precisely the set of all ( 1)-permutations and hence the summation on the right hand side equals (b) Let = [ ij ] be an matrix such that there exist and

with the following properties: ij = 1 and ik = 0 for all = 1 ,...,n,k . Let denote the matrix

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obtained by removing the th row and the th column of . Perform row operations on in the following sequence: interchange the th and the ( + 1)st rows, then interchange the ( + 1)st and the ( + 2)nd rows and ...., then interchange the ( 1)th and the th rows. Now perform column operations in the following sequence: interchange the th and the ( + 1)st columns, then interchange the ( + 1)st and the ( + 2)nd columns and ...., then interchange the ( 1)th and the th columns. By these

operations transforms to 0 1 By part (a) the determinant of this matrix is . On the other hand this matrix has been obtained by 2 interchanging of rows and columns and hence its determinant is 1) , i.e, = ( 1) Fix an = 1 ,...,n . Put 11 ij nn It is clear that =1 . Note that ij can also be obtained by removing the th row and the th column of . It follows that = ( 1) ij ij . Hence =1 1) ij ij Continuing as above we see that =1 1) kj ij = 0 for as it equals the determinant of a matrix which has th and th rows identical. Put ij = ( 1) ij . We then have the following equations: =1 kj ij = 0 for = 1

,...,i ,i + 1 ,...,n and =1 ij ij . These can be rewritten as 11 12 13 ... ... ... ... ... in ... ... ... ... ... nn ii in Put = [ ij ]. The matrix is called the matrix of co-factors of Put Adj( ) = . Then we have Adj( ) = Similarly, on working along the columns of we get Adj( 10

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These two equations lead to a method of ﬁnding the inverse of a matrix if it exists. This method is known as the determinant method of ﬁnding inverse. We state the result in the following theorem: Theorem 3.3. Let be an matrix. Then is invertible if and only if |6 = 0 Furthermore if is

invertible then = (1 ) Adj( Proof : The ﬁrst part has already been proved. The second part follows from the fact that (1 ) Adj( (1 ) Adj( A. Deﬁnition 3.4. Submatrix and Minor. A matrix is said to be a submatrix of a matrix if can be obtained by removing some rows and columns of . If is a square submatrix of then is called a minor of Deﬁnition 3.5. Determinant Rank of a Matrix. Let be an matrix and let = max | submatrix of with non-zero determinant The number is called the determinant rank or simply the rank of and is denoted by rank( ). Lemma 3.6. Let be a matrix. Then an

elemetary row or column operation does not alter the rank of Proof : Put = rank( ). Let be an elementary matrix. We ﬁrst show that rank( EA rank( ). Let s > r and let be an submatrix of EA (a) If ), then or , for some submatrix of according as the th row of EA is involved in or not. In either case, it is zero, by deﬁnition of rank( ). (b) If ij , then equals determinant of an submatrix of with possibly a diﬀerent sign. But then it has to be zero. (c) Suppose ij ). If the th row of EA is not involved in then is in fact an submatrix of and hence = 0. If both the th and the

th rows of EA are involved in then for some submatrix of and hence zero. If only the th row of EA is involved in or | , where and are submatrices of and therefore = 0, that is, = 0. Therefore we see that if s > r then every submatrix of EA has zero determi- nant. This means rank( EA = rank( ). Replacing by EA and by we get rank( rank( EA ), that is, rank( EA ) = rank( ). Similarly rank( AE ) = rank( ). Corollary 3.7. Let be an matrix and let and be respectively and invertible matrices. Then rank( PAQ ) = rank( Proof : As an invertible matrix is a product of elementary matrices, the result

follows by the above lemma. 11

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Inversions. Let be an -permutation. An ordered pair < i,j > is called an inversion of if i < j but > ). Let denote the number of inversions of . Let ,...,a be distinct real numbers. Deﬁne ) = i Let 1 k < l . If and < ) then ( ) appears in numerator as well denominator of ) and hence cancels out. Suppose > ). Then numerator of ) contains the factor ( ) whereas the denominator contains ( ), hence on simpliﬁcation this contributes 1 to the value of ). Since every inversion of does the same, we have ) = ( 1) In particular, ) is

independent of the choice of ,...,a . Let be another permutation and let . Then ,...,b are also distinct. We now have ) = i i )) )) i i i Moreover if = ( p,q then its inversions are < p,p + 1 >,< p,p + 2 >,...,< p,q >, < p + 1 ,q >,...,< q ,q > . These are 1, which is an odd number. Hence is this case, ) = 1. Now suppose ... , where ’s are transpositions. Then ) = ...S ) = ( 1) Therefore is even, i.e., is even (resp. odd) then so is 12

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