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Linear Algebra Inner Product Spaces-1, guaranteed and verified 100% Pass

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Linear Algebra Inner Product Spaces-1, guaranteed and verified 100% PassLinear Algebra Inner Product Spaces-1, guaranteed and verified 100% PassLinear Algebra Inner Product Spaces-1, guaranteed and verified 100% PassLinear Algebra Inner Product Spaces-1, guaranteed and verified 100% PassLinear Algebra Inner Product Spaces-1, guaranteed and verified 100% PassLinear Algebra Inner Product Spaces-1, guaranteed and verified 100% PassLinear Algebra Inner Product Spaces-1, guaranteed and verified 100% Pass

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Institution
Math
Course
Math

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1


Inner Product Spaces


Def. Let 𝑉 be a vector space (over ℝ). An inner product on 𝑉 is a function that
assigns to every pair of vectors 𝑣, 𝑤 ∈ 𝑉 a real number < 𝑣, 𝑤 > such that for all
𝑢, 𝑣, 𝑤 ∈ 𝑉 and 𝑐 ∈ ℝ the following hold:
a. < 𝑣 + 𝑢, 𝑤 > =< 𝑣, 𝑤 > +< 𝑢, 𝑤 >
b. < 𝑐𝑣, 𝑤 > = 𝑐 < 𝑣, 𝑤 >
c. < 𝑣, 𝑤 > =< 𝑤, 𝑣 >
d. < 𝑣, 𝑣 > > 0 if 𝑣 ≠ 0.
A vector space 𝑉 with an inner product, < , > , is called an inner product space.


Ex. Let 𝑣, 𝑤 ∈ ℝ𝑛 be given by 𝑣 =< 𝑎1 , … , 𝑎𝑛 > , 𝑤 =< 𝑏1 , … , 𝑏𝑛 > in the
standard ordered basis for ℝ𝑛 . Then define
< 𝑣, 𝑤 >= ∑𝑛𝑖=1 𝑎𝑖 𝑏𝑖 = 𝑎1 𝑏1 + ⋯ + 𝑎𝑛 𝑏𝑛 .
This is the standard inner product on ℝ𝒏 .
Notice that this inner product satisfies conditions a-d above. For example:
Let 𝑢 =< 𝑑1 , … , 𝑑𝑛 > then
< 𝑣 + 𝑢, 𝑤 > =≪ (𝑎1 + 𝑑1 ), … , (𝑎𝑛 + 𝑑𝑛 ) >, < 𝑏1 , … , 𝑏𝑛 >>
= ∑𝑛𝑖=1(𝑎𝑖 + 𝑑𝑖 )𝑏𝑖
= ∑𝑛𝑖=1(𝑎𝑖 𝑏𝑖 + 𝑑𝑖 𝑏𝑖 )
= ∑𝑛𝑖=1 𝑎𝑖 𝑏𝑖 + ∑𝑛𝑖=1 𝑑𝑖 𝑏𝑖
=< 𝑣, 𝑤 > +< 𝑢, 𝑤 >.

, 2


Ex. Let 𝑉 = 𝐶[0,1] = {Continuous real valued functions on [0,1]}. We can define
an inner product on 𝐶[0,1] by
1
< 𝑓, 𝑔 > = ∫0 𝑓 (𝑥 )𝑔(𝑥 )𝑑𝑥 .
Standard properties of Riemann integrals allow us to verify conditions a-d.


Theorem: Let 𝑉 be an inner product space. For 𝑢, 𝑣, 𝑤 ∈ 𝑉 and 𝑐 ∈ ℝ we have
i. < 𝑢, 𝑣 + 𝑤 > =< 𝑢, 𝑣 > +< 𝑢, 𝑤 >
ii. < 𝑢, 𝑐𝑣 > = 𝑐 < 𝑢, 𝑣 >
iii. < 𝑢, 0 > =< 0, 𝑢 > = 0
iv. < 𝑢, 𝑢 > = 0 if and only if 𝑢 = 0.
v. If < 𝑢, 𝑣 > =< 𝑢, 𝑤 > for all 𝑢 ∈ 𝑉 then 𝑣 = 𝑤.


Proof of i. and v.:
i. < 𝑢, 𝑣 + 𝑤 >=< 𝑣 + 𝑤, 𝑢 > by property c.
=< 𝑣, 𝑢 > +< 𝑤, 𝑢 > by property a.
=< 𝑢, 𝑣 > +< 𝑢, 𝑤 > by property c.


v. Suppose < 𝑢, 𝑣 > =< 𝑢, 𝑤 > for all 𝑢 ∈ 𝑉 then
< 𝑢, 𝑣 > −< 𝑢, 𝑤 > = 0
< 𝑢, 𝑣 > +< 𝑢, − 𝑤 > = 0 by property ii.
< 𝑢, 𝑣 − 𝑤 > = 0 by property i.
The last line is true for all 𝑢 ∈ 𝑉, so in particular 𝑢 = 𝑣 − 𝑤 .
< 𝑣 − 𝑤, 𝑣 − 𝑤 > = 0 ⟹ 𝑣 − 𝑤 = 0 by iv.
So 𝑣 = 𝑤.

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