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### math-2011-note11

Course: ECON 205, Spring 2011
School: Korea University
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53 Theorem (Thm 11.2, p.243). Let A = (a1 , a2 , , am). If a1 , a2 , , am are linearly independent if and only if det A = 0. Proof. a) only if: If a1 , a2 , , am are linearly independent, then A is one-to-one and onto, hence has an inverse function. Call it A1 . Then A1 A = I . Hence det A = 0. b) if: If a1 , a2 , , am are linearly dependent, then det A = 0. Theorem 54 (Variant of Thm 11.8, p.248). The...

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53 Theorem (Thm 11.2, p.243). Let A = (a1 , a2 , , am). If a1 , a2 , , am are linearly independent if and only if det A = 0. Proof. a) only if: If a1 , a2 , , am are linearly independent, then A is one-to-one and onto, hence has an inverse function. Call it A1 . Then A1 A = I . Hence det A = 0. b) if: If a1 , a2 , , am are linearly dependent, then det A = 0. Theorem 54 (Variant of Thm 11.8, p.248). The following statements are equivalent (a) An m m matrix A = (a1 , a2 , , am) is a one-to-one and onto linear function from Rm to Rm . (b) column vectors a1 , a2 , , am are linearly independent. (c) det A = 0. (d) A is invertible. Proof. We only need to show that (c) implies (d). (c) implies that A is one-to-one and onto, hence that it has an inverse function. Call it A1 . Given y1 and y2 , let y1 = Ax1 and y2 = Ax2 . Then sy1 + ry2 = A(sx1 + rx2 ), and hence A1 (sy1 + ry2 ) = sx1 + rx2 = sA1 (y1 ) + rA1(y2 ). Thus A1 is a linear function, which can be regarded as an m m matrix. And AA1 = A1 A = I since AA1 (y) = y and A1 A(x) = x for all x, y Rm . From the denition of the determinant one can see that det A = det AT . Hence, the row vectors of A are linearly independent if and only if the column vectors are. Finding the eigenvalues Given an m m matrix A, r is an eigenvalue of A if there exists a non-zero vector v such that Av = rv. Av = rv Av = rI v (A rI )v = 0 det(A rI ) = 0 3 1 1 Example 11. Consider B = 1 3 1 . 1 1 5 3r 1 1 det(B ) rI = det 1 3 r 1 = (r 2)(r 3)(r 6) 1 1 5 r Hence, 2, 3, and 6 are the eigenvalues of B. 47 16.3 Calculating the Determinant Lemma 5. det A = m=1 (1)1+ ja1 j det A1 j , where A1 j is a matrix obtained from A by j deleting row 1 and column j. Proof. det A = 1 jr m s( j1 , j2, , jm)a1 j1 a2 j2 am jm = a11 2 jr m s(1, j2, , jm)a2 j2 am jm +a12 jr =2 s(2, j2, , jm)a2 j2 am jm + + a1m 1 jr m1 s(m, j2, , jm )a2 j2 am jm = a11 2 jr m s( j2, , jm)a2 j2 am jm +a12 (1) jr =2 s( j2, , jm)a2 j2 am jm + + a1m (1)m1 1 jr m1 s( j2, , jm)a2 j2 am jm = m=1 (1)1+ j a1 j det A1 j j Theorem 55 (Thm 26.1, p.724). det A = m=1 (1)i+ j ai j det Ai j , where Ai j is a matrix j obtained from A by deleting row i and column j. Proof. Denote by A(h, k) the matrix obtained from A by switching h-th and k-th row. Also denote by A(h) the matrix obtained from A by by bringing the h-th row to the top, without changing the order of the other rows. Then det A(i) = (1)i1 det A (since det A(h 1, h) = det A). By the previous Lemma, we have m det A = (1)i1 det A(i) = (1)i1 (1)1+ j ai j det Ai j j =1 m = (1)i+ jai j det Ai j . j =1 Example 12. a11 a12 a13 det a21 a22 a23 = a11 det a31 a32 a33 a12 det a22 a23 a32 a33 a21 a23 a31 a33 a21 a22 a31 a32 = a11 (a22 a33 a23 a32 ) +a13 det a12 (a21 a33 a23 a31 ) +a13 (a21 a32 a22 a31 ) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 a13 a22 a31 a12 a21 a33 a11 a23 a32 48
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