Page 22 - Vector Analysis

18 CHAPTER 1. Linear Algebra

Proposition 1.62. Let E be an elementary matrix. Then

1. det(E) ‰ 0.
2. det(E) = det(ET).

3. If A is an n ˆ n matrix, then det(EA) = det(E) det(A).

The proof of the proposition above is not difficult, and is left as an exercise.

Corollary 1.63. Let v1, ¨ ¨ ¨ , vn P Rn be (column) vectors, c P R, and

A = [ ... ¨ ¨ ¨ ... ] ,
v1 vn

B = [ ... ¨ ¨ ¨ ...vj´1 ... λvj ...vj+1 ... ¨ ¨ ¨ ... ] ,
v1 vn

C = [ ... ¨ ¨ ¨ ...vj´1 ... vj + µvi ...vj+1 ... ¨ ¨ ¨ ... ] for some i ‰ j .
v1 vn

Then det(B) = λ det(A), and det(C) = det(A).

Proof. The corollary is easily concluded since B = E1A and C = E2A for some elementary

matrices E1 and E2 with det(E1) = c and det(E2) = 1. ˝

Corollary 1.64. Let A be an n ˆ n matrix. Then A is invertible if and only if det(A) ‰ 0.

Proof. (ñ) Since A is invertible, Theorem 1.55 implies that

A = EkEk´1 ¨ ¨ ¨ E2E1

for some elementary matrices E1, ¨ ¨ ¨ , Ek, and this corollary follows from Proposition
1.62.

(ð) Note that A is invertible if and only if rank(A) = rank(AT) = n. Therefore, if A is not
invertible, the row vectors of A are linearly dependent; thus there exists a non-zero
vectors (α1, ¨ ¨ ¨ , αn) P Fn such that

α1v1 + α2v2 + ¨ ¨ ¨ αnvn = 0 ,

where AT = [ ... ¨ ¨ ¨ ... ] Suppose that αj ‰ 0. Then
v1 vn .

vj = β1v1 + ¨ ¨ ¨ βj´1vj´1 + βj+1vj+1 + ¨ ¨ ¨ βnvn ;