Positive Definite Linear Algebra

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mahjk17

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Prove that the inner product associated with a positive definite quadratic form [FONT=MathJax_Math]q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT] is given by the polarization formula [FONT=MathJax_Main]⟨[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]⟩[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/2[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]][/FONT].


How will I be able to do this problem. I know in order to be positive definite we need that [FONT=MathJax_Main]⟨[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]⟩[/FONT][FONT=MathJax_Main]=([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]^T[/FONT][FONT=MathJax_Math])K[/FONT][FONT=MathJax_Math]y[/FONT] for [FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Math]^n[/FONT] where K has positive pivots hence detK > 0 , but how do I go on to apply that here? I don't even know what the question is asking for.
 
The crucial thing to think about here is, what does it mean for an inner product to be "associated" with a quadratic form?
 
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