Is {V1-Vn} an Orthonormal Basis in R^n for a Positive Definite Inner Product?

[blue2004STi]

Homework Statement

Show that {V1--Vn} form an orthonormal basis of R^n for the inner product
<v,w>= (v^T)Kw for K>0(positive definite) if and only if (A^T)KA=I where A={v1,v2---vn}

Homework Equations

I don't know what to do in terms of do I write it out using actual matrices or are there some simple properties that I should use like inverse or transposes? Am I missing something?

The Attempt at a Solution

(v^T)Kw=0 for K>0 iff (A^T)KA=I <v,w>=(v^T)Kw=0 and ||v||=||w||=1

[v1--vn][-k1-][v1] = [v1--vn][-k1*v-] = v^2(k1+k2+ --- +kn) = I
[-k2-][v2] [-k2*v-]
[ | ][ | ] [ | ]
[-kn-][vn] [-kn*v-]

-k1-, -k2-, -kn- are the rows of K
v1, vn, are the elements of V

--- represents through (ie. v1--vn means 'V' one through 'V' 'N')

||*|| represents the norm(any norm)

(v^T)Kw represents the quadratic form of the inner product



I tried doing it by expanding but I don't know where to go from here. I thought about using the inverse rules to get somewhere but I don't think that'll help. Any thoughts are appreciated.

 

[morphism]

Let K=[k₁ | k₂ | ... | k_n]. The (i,j)th entry of A^TK is v_i^T k_j. Now what happens when we multiply A^TK by A?

 

[blue2004STi]

It becomes (v^T)Kv which when the "subs" i & j for i=j it equals 1 and for i not equal to j it equals 0. Which produces the identity. THANKS


Matt

Correct me if I'm wrong.