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Lecture

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

Department
Mathematics
Course Code
MAT224H1
Professor
Martin, Burda

This 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=BB.Hence det A=det Bdet 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, vVis deﬁned as the length of the
vector from vto u:
d(u, v)=kuvk
Example:If (,)is the usual dot product on C2,then k(i
1)k=pii +11 =
2and d((i
1),(1
i)) =ki1
1ik=p|i1|2+|1i|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
{vR2|d(v,0) =1}={vR2|kvk=1}=
={(x
y)|p2x2+y2=1}={(x
y)|2x2+y2=1}
1
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