Lengths and distrances (inner products), angles, cauchy-schwartz inequality, triangle inequality
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Lecture 14 notes byY. Burda
1An exercise to beused later
Exercise:Provethat the determinantof apositive-deﬁnite matrix is apos-
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
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:
Example:If (−,−)is the usual dot product on C2,then k(i
1)k=pii +11 =
Example:If (−,−)is theinner product on R2given by((x1
2x1y1+x2y2then the unit circle centered at origin is the set
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