▲✐♥❡❛r ❆❧❣❡❜r❛ Prof. Inder K. Rana Room 112 B Department of Mathematics IIT-Bombay, Mumbai-400076 (India) Email: ikr@math.iitb.ac.in Lecture 7 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Recall We looked at the invertibility of a square matrix: Theorem A n × n matrix A is invertible if and only if it is of full rank, i.e., rank ( A ) = n . In fact we gave a way of checking that a given matrix A is invertible of not and a way of computing its inverse: A is invertible iff the reduced row echelon form of the n × n matrix A is the identity matrix I n . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra Let u = ( a , c ) , v = ( b , d ) be vectors in the plane spanning the parallelogram APQB . We call it the parallelogram spanned by the vectors u and v denote it by P ( u , v ) . ]pause We know from school geometry that the area of P ( u , v ) = ad − bc . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra Note that, if we interchange the order of u and v , then the area of the parallelogram is given by − ( ad − bc ) , which is the negative of the earlier expression. (This relates to the choice of left-handed or right-handed system of R 2 ). axis in I In any case, the expression | ad − bc | can be treated as the area of the parallelogram, and ( ad − bc ) can be treated as the signed area of the parallelogram generated by the vectors u and v . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra It is easy to see from figure below that areas ( P ( u , v )) has the following properties: (i) area ( P ( u , v )) = 0 , This corresponds to the fact that when two edges of a parallelogram coincide, the area reduces to zero. (ii) area ( P ( λ u , v )) = λ ( area ( P ( u , v ))) and area ( P ( u , λ v )) = λ ( area ( P ( u , v ))) , i.e., if a side is magnified the area gets magnified by the same factor. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra (iii) area ( P ( u 1 + u 2 , v )) = area ( P ( u 1 , v )) + area ( P ( u 2 , v )) and area ( P ( u , u 1 + u 2 )) = area ( P ( u , u 1 )) + area ( P ( u , u 2 )) , which is the additive property of the area, i.e., Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra area ( P ( u 1 + u 2 , v )) = Area of the parallelogram APRC , = area of the parallelogram APQB + area of the parallelogram BQRC = area ( P ( u 1 , v )) + area ( P ( u 2 , v )) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra Definition (Determinant function) Let D : M ( n × n ; I R ) → I R be a function with the following properties: For A ∈ M ( n × n ; I R ) , let R 1 ( A ) , R 2 ( A ) , . . . , R n ( A ) denote its row vectors and let A be represented as R 1 ( A ) . . A = . . R n ( A ) . Then P(i): D ( A ) = 0 if R i ( A ) = R j ( A ) for some i � = j . ′ ′′ P(ii): If R i ( A ) = α R i + β R i for some α, β ∈ I R , then D ( A ) = α D ( B ) + β D ( C ) , Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra Definition where R 1 ( A ) R 1 ( A ) . . . . . . R i − 1 ( A ) R i − 1 ( A ) ′ ′′ B := R ; and C := R i i R i + 1 ( A ) R i + 1 ( A ) . . . . . . R n ( A ) R n ( A )) . In other words, if we treat for each A , D ( A ) as a function of its n-row vectors then D is linear in each row. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
From Geometry to Algebra Definition P(iii): If I ∈ M ( n × n ; I R ) is the n × n identity matrix, then D ( I ) = 1 . This is called normalization. Using these defining properties alone, we can deduce other properies of a determinant function. For a matrix A , determinant of A is also denoted by | A | . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Properties of determinant function Theorem If A be any n × n matrix, n ≥ 2 , and B is obtained from A by interchanging two of its rows then | B | = −| A | . Proof: Without loss of generality let i < j . Consider the matrices B , C and P defined as below: � R k ( A ) if k � = i or k � = j , R k ( B ) := < R i ( A ) + R j ( A ) if k = i , j � R k ( A ) if k � = i , j , R k ( C ) := , R i ( A ) if k = i , j � R k ( A ) if k � = i , j , R k ( P ) := if k = i , j . . R j ( A ) Then, by property P(i) , D ( B ) = D ( C ) = D ( P ) = 0 , and using property P(ii) , we have Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Properties of determinant function ′ ) = D ( A ) + D ( A ′ ) . 0 = D ( B ) = D ( A ) + D ( C ) + D ( P ) + D ( A ′ ) . Hence, D ( A ) = − D ( A Theorem If R i ( A ) = 0 for some i , then, D ( A ) = 0 . Proof: if R i ( A ) = 0 and B is the n × n matrix with R i ( B ) = R i ( A ) + R i ( A ) , R j ( B ) = R j ( A ) for j � = i , then D ( A ) = D ( B ) = D ( A ) + D ( A ) = 2 D ( A ) , implying that D ( A ) = 0 . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Properties of determinant function Corollary If two of the rows a matrix are are linearly dependent, then the determinant vanishes. Proof: by the given hypothesis, there exists some i such that � R i ( A ) = α k R k ( A ) . k � = i Thus, if B k , k � = i , is the n × n matrix with R j ( B k ) = R j ( A ) for j � = i , and R i ( B ) = R k ( A k ) , then D ( B k ) = 0 for every k � = i and using property (iii) of determinant function, � D ( A ) = α k D ( B k ) = 0 . k � = i Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Additional properties Theorem If B is obtained from A by replacing j th row of A by α R i ( A ) + R j ( A ) for some i � = j , and α ∈ I R , then D ( B ) = D ( A ) . Proof: Note that by property of linearity of a determinant function, D ( B ) = α D ( P ) + D ( A ) , where P is the n × n matrix with P i ( A ) = P j ( A ) = R i ( A ) , and P k ( A ) = R k ( A ) for k � = i . Since D ( P ) = 0 , we have D ( B ) = D ( A ) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Determinant of Elementary matrices Lemma 1 D ( E i ↔ j ) = − 1 , D ( E i + j ) = 1 and D ( E α ( i ) ) = α. Let E be any elementary matrix and A ∈ M ( n × n ; I R ) . Then 2 D ( EA ) = D ( E ) D ( A ) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Determinant of Elementary matrices Theorem For every A ∈ M ( n × n ; I R ) and E an elementary matrix, − D ( A ) if E = E i ↔ j , D ( EA ) = D ( A ) if E = E i + j , . α D ( A ) E = E α ( i ) if Proof: If E = E i ↔ j , then − D ( A ) = D ( EA ) is already proved in lemma. Other follow straightaway from the properties P(i) and P(ii) of the determinant function. Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Determinant of Elementary matrices Theorem Let A ∈ M ( n × n ; I R ) . Then the following hold: (i) If ˜ A is the reduced row-echelon form of A with ˜ A = E k E k − 1 . . . E 1 A , where E 1 , E 2 , . . . , E n and elementary matrices, then 0 , if A is not invertible , D ( A ) = 1 D ( E k ) D ( E k − 1 ) · · · D ( E 1 ) , if A is invertible . (ii) If to obtain U , the row echelon form of A , the number of row interchanges required is r , then D ( A ) = ( − 1 ) r D ( U ) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Determinant of a product contd. Corollary For every elementary matrix E , D ( E − 1 ) = 1 / D ( E ) . Theorem (determinant of Product) Let A and B be n × n matrices then | AB | = | A || B | Proof: If either A is not invertible or B is not invertible, then AB is also not invertible (Exercise). Hence, D ( AB ) = D ( A ) D ( B ) = 0 . Interchanging A and B , D ( BA ) = D ( B ) D ( A ) = D ( AB ) = 0 . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Determinant of a product contd. In case both A and B are invertible, AB and BA are both invertible. Further, there exist elementary matrices E 1 , . . . , E k and F 1 , . . . , F r such that i.e., A = E − 1 . . . E − 1 k − 1 E − 1 E k E k − 1 . . . E 1 A = Id , 1 k and i.e., B = F − 1 . . . F − 1 r − 1 F − 1 F r F r − 1 . . . F 1 B = Id , . r 1 Thus AB can be written as Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Proof... AB = E − 1 . . . E − 1 k F − 1 . . . F − 1 . r 1 1 Now using properties of determinants of elementary matrices, D ( E − 1 1 ) D ( E − 1 2 ) . . . D ( E − 1 k ) D ( F − 1 1 ) . . . D ( F − 1 D ( AB ) = ) r [ D ( E 1 ) D ( E 2 ) . . . D ( E k )] − 1 [ D ( F 1 ) . . . D ( F r )] − 1 = = D ( A ) D ( B ) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
Invertibility We summarize all the conditions for invertibility of a square matrix: Theorem Let A be an n × n matrix. The following statements about A are equivalent: A is invertible. A is of full rank i.e. ρ ( A ) = n . The determinant | A | � = 0 and in that case 1 D ( A − 1 = D ( A ) . Prof. Inder K. Rana Department of Mathematics, IIT - Bombay
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