# Lengths and distrances (inner products), angles, cauchy-schwartz inequality, triangle inequality

by OC6291

Department

MathematicsCourse Code

MAT224H1Professor

Martin, BurdaThis

**preview**shows page 1. to view the full**4 pages of the document.**Lecture 14 notes byY. Burda

1An exercise to beused later

Exercise:Provethat the determinantof apositive-deﬁnite matrix is apos-

itivenumber.

Solution 1(due to Kirby): Let Abeapositive-deﬁnite matrix.Since

it is Hermitian (and every complex positivedeﬁnite matrix bydeﬁnition is

Hermitian), it is diagonalizable. Hence det Ais the product of A’s eigenval-

ues. Since the eigenvalues of apositive-deﬁnite matrix are real and positive,

their product is also positive.

Solution 2(due to Kristie): Since Ais positive-deﬁnite, there exists an

invertible matrix Bsuchthat A=B∗B.Hence det A=det B∗det B=

det BTdet B=det Bdet B=det Bdet B=|det B|2>0(in the last step

weused the fact that Bis invertible and hence its determinantis non-zero).

2Lengths and distances in inner product spaces

If Vis an inner product space wecan deﬁne the length of avector bythe

formula

kvk=p(v,v)

This length is also called the “norm” of the vector vand weuse the

symbol kvkinstead of |v|to distinguish itfrom the length computed using

the standard dot-product on Rn.

Distance between twopoints u, v∈Vis deﬁned as the length of the

vector from vto u:

d(u, v)=ku−vk

Example:If (−,−)is the usual dot product on C2,then k(i

1)k=pii +11 =

√2and d((i

1),(1

i)) =ki−1

1−ik=p|i−1|2+|1−i|2=2

Example:If (−,−)is theinner product on R2given by((x1

x2),(y1

y2)) =

2x1y1+x2y2then the unit circle centered at origin is the set

{v∈R2|d(v,0) =1}={v∈R2|kvk=1}=

={(x

y)|p2x2+y2=1}={(x

y)|2x2+y2=1}

1

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