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bugatti79
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I have just realized that I accidently put it in wrong sub forum. This should be in 'calculus and beyond'.
Prove the function <x,y>=x_1y_1+x_2y_2+x_3y_3 defines an inner product space on the real vector space R^3 where x=(x1,x2,x3) and y=(y1,y2,y3)
Axiom 1 <x,y> >=0 since we have that x_n and y_n for n=1,2,3 are in R
Axiom 2a <x,y> =x1y1+x2y2+x3y3, then <x,y>=0 iff x_n and y_n for n=1,2,3 both =0
Axiom 2b <ax,y>=a<x,y>
<ax,y> = ax1y1+ax2y2+ax3y3
= a(x1y1+x2y2+x3y3)
=a<x,y>
Axiom 3 <y,x>= complex of <x,y>
<y,x>=(y1x1+y2x2+y3x3) but y complex =y and x complex=x in R therefore
= (y1x1 complex+y2x2 complex +x3y3 complex)
=<y,x> complex
=<x,y> complex
Axiom 4 <x+y,z>=<x,z>+<y,z>, let z=(z1,z2,z3) in R^3
<x+y,z>=(x1+y1+x2+y2+x3+y3)(z1+z2+z3)
=x1z1 +x2z2+x3z3+y1z1+y2zy3z3
=<x,z>+<y,z>
...?
Homework Statement
Prove the function <x,y>=x_1y_1+x_2y_2+x_3y_3 defines an inner product space on the real vector space R^3 where x=(x1,x2,x3) and y=(y1,y2,y3)
The Attempt at a Solution
Axiom 1 <x,y> >=0 since we have that x_n and y_n for n=1,2,3 are in R
Axiom 2a <x,y> =x1y1+x2y2+x3y3, then <x,y>=0 iff x_n and y_n for n=1,2,3 both =0
Axiom 2b <ax,y>=a<x,y>
<ax,y> = ax1y1+ax2y2+ax3y3
= a(x1y1+x2y2+x3y3)
=a<x,y>
Axiom 3 <y,x>= complex of <x,y>
<y,x>=(y1x1+y2x2+y3x3) but y complex =y and x complex=x in R therefore
= (y1x1 complex+y2x2 complex +x3y3 complex)
=<y,x> complex
=<x,y> complex
Axiom 4 <x+y,z>=<x,z>+<y,z>, let z=(z1,z2,z3) in R^3
<x+y,z>=(x1+y1+x2+y2+x3+y3)(z1+z2+z3)
=x1z1 +x2z2+x3z3+y1z1+y2zy3z3
=<x,z>+<y,z>
...?
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