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.
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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
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