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

29 views4 pages 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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## Document Summary

Exercise: prove that the determinant of a positive-de nite matrix is a pos- itive number. Solution 1(due to kirby): let a be a positive-de nite matrix. Since it is hermitian (and every complex positive de nite matrix by de nition is. Hence det a is the product of a"s eigenval- ues. Since the eigenvalues of a positive-de nite matrix are real and positive, their product is also positive. Solution 2(due to kristie): since a is positive-de nite, there exists an invertible matrix b such that a = b b. 2 lengths and distances in inner product spaces. If v is an inner product space we can de ne the length of a vector by the formula kvk = p(v, v) This length is also called the norm of the vector v and we use the symbol kvk instead of |v| to distinguish it from the length computed using the standard dot-product on rn.