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Linear Algebra, 2020a

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Section II. Geometry of Determinants 355<br />

II<br />

Geometry of Determinants<br />

The prior section develops the determinant algebraically, by considering formulas<br />

satisfying certain conditions. This section complements that with a geometric<br />

approach. Beyond its intuitive appeal, an advantage of this approach is that<br />

while we have so far only considered whether or not a determinant is zero, here<br />

we shall give a meaning to the value of the determinant. (The prior section<br />

treats the determinant as a function of the rows but this section focuses on<br />

columns.)<br />

II.1<br />

Determinants as Size Functions<br />

This parallelogram picture is familiar from the construction of the sum of the<br />

two vectors.<br />

(<br />

x2<br />

y 2<br />

)<br />

(<br />

x1<br />

y 1<br />

)<br />

1.1 Definition In R n the box (or parallelepiped) formed by 〈⃗v 1 ,...,⃗v n 〉 is the<br />

set {t 1 ⃗v 1 + ···+ t n ⃗v n | t 1 ,...,t n ∈ [0 ... 1]}.<br />

Thus the parallelogram above is the box formed by 〈 ( x 1<br />

y 1<br />

)<br />

,<br />

( x2<br />

y 2<br />

)<br />

〉. A three-space<br />

box is shown in Example 1.4.<br />

We can find the area of the above box by drawing an enclosing rectangle and<br />

subtracting away areas not in the box.<br />

B<br />

A<br />

y 2 D<br />

y 1 C<br />

F<br />

E<br />

x 2 x 1<br />

area of parallelogram<br />

= area of rectangle − area of A − area of B<br />

− ···− area of F<br />

=(x 1 + x 2 )(y 1 + y 2 )−x 2 y 1 − x 1 y 1 /2<br />

− x 2 y 2 /2 − x 2 y 2 /2 − x 1 y 1 /2 − x 2 y 1<br />

= x 1 y 2 − x 2 y 1<br />

That the area equals the value of the determinant<br />

∣ x 1 x ∣∣∣∣<br />

2<br />

= x<br />

∣<br />

1 y 2 − x 2 y 1<br />

y 1 y 2

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