r r Prof. Inder K. Rana Room 112 B Department of Mathematics - - PowerPoint PPT Presentation
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
Room 112 B Department of Mathematics IIT-Bombay, Mumbai-400076 (India) Email: ikr@math.iitb.ac.in Lecture 7
Department of Mathematics, IIT - Bombay
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 In.
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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.
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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 axis in I R2). 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.
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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(λu, v)) = λ(area(P(u, v))) and area (P(u, λv)) = λ(area(P(u, v))), i.e., if a side is magnified the area gets magnified by the same factor.
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(iii) area (P(u1 + u2, v)) = area (P(u1, v)) + area (P(u2, v)) and area (P(u, u1 + u2)) = area (P(u, u1)) + area (P(u, u2)), which is the additive property of the area, i.e.,
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area (P(u1 + u2, v)) = Area of the parallelogram APRC, = area of the parallelogram APQB + area of the parallelogram BQRC = area(P(u1, v)) + area (P(u2, v)).
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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 R1(A), R2(A), . . . , Rn(A) denote its row vectors and let A be represented as A = R1(A) . . . Rn(A). . Then P(i): D(A) = 0 if Ri(A) = Rj(A) for some i = j. P(ii): If Ri(A) = αR
′
i + βR
′′
i for some α, β ∈ I
R, then D(A) = αD(B) + βD(C),
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Definition where B := R1(A) . . . Ri−1(A) R
′
i
Ri+1(A) . . . Rn(A) ; and C := R1(A) . . . Ri−1(A) R
′′
i
Ri+1(A) . . . Rn(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.
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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
For a matrix A, determinant of A is also denoted by |A|.
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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: Rk(B) := Rk(A) if k = i or k = j, Ri(A) + Rj(A) if k = i, j < Rk(C) := Rk(A) if k = i, j, Ri(A) if k = i, j , Rk(P) := Rk(A) if k = i, j, Rj(A) if k = i, j. . Then, by property P(i), D(B) = D(C) = D(P) = 0, and using property P(ii), we have
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0 = D(B) = D(A) + D(C) + D(P) + D(A
′) = D(A) + D(A ′).
Hence, D(A) = −D(A
′).
Theorem If Ri(A) = 0 for some i, then, D(A) = 0. Proof: if Ri(A) = 0 and B is the n × n matrix with Ri(B) = Ri(A) + Ri(A), Rj(B) = Rj(A) for j = i, then D(A) = D(B) = D(A) + D(A) = 2D(A), implying that D(A) = 0.
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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 Ri(A) =
αkRk(A). Thus, if Bk, k = i, is the n × n matrix with Rj(Bk) = Rj(A) for j = i, and Ri(B) = Rk(Ak), then D(Bk) = 0 for every k = i and using property (iii) of determinant function, D(A) =
αkD(Bk) = 0.
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Theorem If B is obtained from A by replacing jth row of A by αRi(A) + Rj(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) = αD(P) + D(A), where P is the n × n matrix with Pi(A) = Pj(A) = Ri(A), and Pk(A) = Rk(A) for k = i. Since D(P) = 0, we have D(B) = D(A).
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Lemma
1
D(Ei↔j) = −1, D(Ei+j) = 1 and D(Eα(i)) = α.
2
Let E be any elementary matrix and A ∈ M(n × n; I R). Then D(EA) = D(E)D(A).
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Theorem For every A ∈ M(n × n; I R) and E an elementary matrix, D(EA) = −D(A) if E = Ei↔j, D(A) if E = Ei+j, αD(A) if E = Eα(i) . Proof: If E = Ei↔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.
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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 = EkEk−1 . . . E1A, where E1, E2, . . . , En and elementary matrices, then D(A) = 0, if A is not invertible, 1 D(Ek)D(Ek−1) · · · D(E1), 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)rD(U).
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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.
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In case both A and B are invertible, AB and BA are both invertible. Further, there exist elementary matrices E1, . . . , Ek and F1, . . . , Fr such that EkEk−1 . . . E1A = Id, i.e., A = E−1
1
. . . E−1
k−1E−1 k
and FrFr−1 . . . F1B = Id, i.e., B = F −1
1
. . . F −1
r−1F −1 r
. Thus AB can be written as
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AB = E−1
1
. . . E−1
k F −1 1
. . . F −1
r
. Now using properties of determinants of elementary matrices, D(AB) = D(E−1
1 )D(E−1 2 ) . . . D(E−1 k ) D(F −1 1 ) . . . D(F −1 r
) = [D(E1)D(E2) . . . D(Ek)]−1 [D(F1) . . . D(Fr)]−1 = D(A)D(B).
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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 D(A−1 = 1 D(A).
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Let A be a square n × n real matrix. We define |A|- the determinant of A (also written det A) in an inductive manner. Definition If A is 1 × 1, A = [a] say, we define |A| = a. To define n × n determinants inductively (induction on n) we suppose that (n − 1) × (n − 1) determinants have already been defined. To carry out the induction we first define minors of A. Definition (Minors of A) The (jk)th minor Mjk = Mjk(A) of A is defined to be the determinant of the (n − 1) × (n − 1) sub-matrix of A obtained by deleting its jth row and the kth column.
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Finally, the definition of |A|, determinant is in terms of its minors: Definition (Determinant of A) The determinant |A| is defined as the sum |A| =
n
(−1)j+kajkMjk (Expansion by the jth row) =
n
(−1)j+kajkMjk (Expansion by the kth column) Remark: For convenience we denote (−1)j+kMjk as Cjk = Cjk(A) and call it the (jk)th-cofactor of A.
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Thus we can re-write |A| =
n
ajkCjk (Expansion by the jth row) =
n
ajkCjk (Expansion by the kth column) Example:Let A = a b c d
C11 = d, C12 = −c, C21 = −b, C22 = a.
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Expanding by the : first row gives |A| = ad + b(−c) = ad − bc, second row yields |A| = c(−b) + da = ad − bc, first column gives |A| = ad + c(−b) = ad − bc, second column gives |A| = b(−c) + da = ad − bc. All are same showing consistency and |A| = ad − bc. Theorem (Uniqueness) Suppose D and D
′ are two determinant functions on M(n × n; I
R). Then D(A) = D
′(A) for all A ∈ M(n × n; I
R).
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Theorem For a square matrix A, A ∈ M(n × n; I R) D(AT) = D(A). Proof: Let A be invertible. Then A = EN...E1 is a product of matrices and |A| =
AT = ET
1 ...ET N is again a product of elementary matrices
(Exercise: each Ej and its transpose ET
j
have equal determinant. Thus
=
1
N
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Corollary For a square matrix A, expansion of the determinant by the rows implies that by the columns. The determinant function is column-wise linear and skew-symmetric. In other words, the column analogues of the properties D-1,D-2 hold. Some more properties Theorem (Invertibility via determinant) Let A be an n × n matrix. A is invertible if and only if |A| = 0. Proof: A is invertible = ⇒ ∃B s.t. AB = I = ⇒ |A||B| = 1 = ⇒ |A| = 0. Conversely, by the lemma below |A| = 0 = ⇒ |A|−1AdjA is the inverse
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Let M = M(A) be the matrix of minors of A and C = C(A) be the matrix of cofactors of A. Definition The transpose of the cofactor matrix is called the adjoint of A denoted AdjA. Lemma Let A be a square matrix, then A(AdjA) = (AdjA)A = |A|I. Further A−1 = 1 D(A)AdjA.
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Find the matrices of minors, cofactors and the adjoint of the following matrix: 2 √ 3 √ 2 −1 1 √ 2 √ 3 2 . Verify that (AdjA)A = A(AdjA) = |A|I. Hence compute the inverse if it exists.
Minors: M(A) = − √ 3 −(2 + √ 2) − √ 3 (2 − √ 2) √ 3 2 (2 − √ 2) √ 3 √ 3 (2 + √ 2) √ 3 . Cofactors: C(A) = − √ 3 (2 + √ 2) − √ 3 ( √ 2 − 2) √ 3 2 ( √ 2 − 2) √ 3 √ 3 −(2 + √ 2) √ 3 .
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AdjA = C(A)T = − √ 3 ( √ 2 − 2) √ 3 √ 3 (2 + √ 2) 2 −(2 + √ 2) − √ 3 ( √ 2 − 2) √ 3 √ 3 . A(AdjA) = 2 √ 3 √ 2 −1 1 √ 2 √ 3 2 − √ 3 ( √ 2 − 2) √ 3 √ 3 (2 + √ 2) 2 −(2 + √ 2) − √ 3 ( √ 2 − 2) √ 3 √ 3 = Further, expanding by the second row |A| =
√ 3 √ 2 −1 1 √ 2 √ 3 2
√ 3 − √ 6) − 1(2 √ 3 − √ 6) = 0 = ⇒ A(AdjA) = O = |A|I and the inverse does not exist.
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Theorem (Rank via determinants) Let A be any square matrix. Then rank(A) = max
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Find the rank by determinants. Verify by row reduction. (i) 2 −3 2 5 −3 5 (ii) 4 3 −8 −6 16 12 .
(i)
−3 2 5 −3 5
⇒ rank(A) ≤ 2 since there is no 3 × 3 "submatrix" of non-zero determinant. Also
2
(ii)
⇒ rank(A) < 2 = ⇒ rank(A) = 1 because A = [0].
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Let Ax = b be a system of n equations in the same number n of
call such a system as a square system. Theorem (Cramer’s Rule) Let Ax = b be a square system and |A| = 0. Then the solution vector x exists and is unique. Further, the components of x are expressible as xk =
k[b]
, where Aˆ
k[b] = [A1, ..., b, ..., An] is the n × n matrix obtained from A by
replacing its kth column by b. Proof: Omitted.
Decode the column x = A−1b = (AdjA)b |A| entry-wise.
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Find the values of β for which Cramer’s rule is applicable. For the remaining value(s) of β, find the number of solutions. x + 2y + 3z = 20 x + 3y + z = 13 x + 6y + βz = β. The coefficient matrix has determinant β + 5, hence for β = −5 Cramer’s Rule is applicable. For β = −5, the augmented matrix A+ = 1 2 3
1 3 1
1 6 −5
.
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1 2 3
1 3 1
1 6 −5
→ 1 2 3
1 −2
4 −8
→ 1 2 3
1 −2
. The pivot in the last column means that ρ(A) = 2 < 3 = ρ(A+). Hence the system is not solvable for β = −5.
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