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Linear Algebra, 2020a

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332 Chapter Four. Determinants<br />

each n, the function that such a formula calculates. We will define this function<br />

by a list of properties. We will then prove that a function with these properties<br />

exists and is unique, and also describe how to compute it. (Because we will<br />

eventually prove this, from the start we will just say ‘det(T)’ instead of ‘if there<br />

is a unique determinant function then det(T)’.)<br />

2.1 Definition A n×n determinant is a function det: M n×n → R such that<br />

(1) det(⃗ρ 1 ,...,k· ⃗ρ i + ⃗ρ j ,...,⃗ρ n )=det(⃗ρ 1 ,...,⃗ρ j ,...,⃗ρ n ) for i ≠ j<br />

(2) det(⃗ρ 1 ,...,⃗ρ j ,...,⃗ρ i ,...,⃗ρ n )=−det(⃗ρ 1 ,...,⃗ρ i ,...,⃗ρ j ,...,⃗ρ n ) for i ≠ j<br />

(3) det(⃗ρ 1 ,...,k⃗ρ i ,...,⃗ρ n )=k · det(⃗ρ 1 ,...,⃗ρ i ,...,⃗ρ n ) for any scalar k<br />

(4) det(I) =1 where I is an identity matrix<br />

(the ⃗ρ ’s are the rows of the matrix). We often write |T| for det(T).<br />

2.2 Remark Condition (2) is redundant since<br />

T<br />

ρ i +ρ j<br />

−→<br />

−ρ j +ρ i<br />

−→<br />

ρ i +ρ j<br />

−→<br />

−ρ i<br />

−→<br />

ˆT<br />

swaps rows i and j. We have listed it for consistency with the Gauss’s Method<br />

presentation in earlier chapters.<br />

2.3 Remark Condition (3) does not have a k ≠ 0 restriction, although the Gauss’s<br />

Method operation of multiplying a row by k does have it. The next result shows<br />

that we do not need that restriction here.<br />

2.4 Lemma A matrix with two identical rows has a determinant of zero. A matrix<br />

with a zero row has a determinant of zero. A matrix is nonsingular if and only<br />

if its determinant is nonzero. The determinant of an echelon form matrix is the<br />

product down its diagonal.<br />

Proof To verify the first sentence swap the two equal rows. The sign of the<br />

determinant changes but the matrix is the same and so its determinant is the<br />

same. Thus the determinant is zero.<br />

For the second sentence multiply the zero row by two. That doubles the determinant<br />

but it also leaves the row unchanged, and hence leaves the determinant<br />

unchanged. Thus the determinant must be zero.<br />

Do Gauss-Jordan reduction for the third sentence, T →···→ ˆT. By the first<br />

three properties the determinant of T is zero if and only if the determinant of ˆT is<br />

zero (although the two could differ in sign or magnitude). A nonsingular matrix<br />

T Gauss-Jordan reduces to an identity matrix and so has a nonzero determinant.

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