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Show that $v_1,...,v_n$ form an orthonormal basis of $\mathbb{R^n}$ for the inner product $\langle v,w\rangle = v^TKw$ for $K > 0$ iff $A^TKA = I$ where $A= (v_1v_2...v_n)$.

How will I be able to do this problem? I know that in order to be an orthonormal basis it must have a unit vector equal to one and must be orthogonal, but how will I be able to show that here?

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Hint: what is $(v_1 0 ... 0)^TK(v_1 0 ... 0)$? – anonymous Oct 20 '12 at 23:28
@anonymous it is a positive definite matrix. – diimension Oct 20 '12 at 23:33
up vote 1 down vote accepted

Recall two ways of multiplying matrices. If $A$ is given and $B = (b_1,b_2,\ldots,b_n)$, then $JK = (Jk_1,Jk_2,\ldots,Jk_n)$. In particular, if the transpose of $A$ is $A^T = (a_1,a_2,\ldots,a_n)$, then the product has entries $(AB)_{ij} = a_i^Tb_j$ (for $i,j = 1,\cdots,n)$.

Observe then that $KA = (Ka_1, Ka_2,\ldots, Ka_n)$, so the entries of $A^TKA$ are $(A^TKA)_{ij} = a_i^TKa_j$. Since $\langle a_i,a_j\rangle = a_i^TKa_j$, $i,j = 1,\ldots,n$. I think you can take it from here.

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I understood what you have given me but I can't find a way to complete the proof. What I got to complete your proof is since $\langle a_i,a_j\rangle = a_i^TKa_j$ and we know that $A^TKA = I$ we have that the entries are 0 when i does not equal j and when i = j we have that it equals 1 which will make it orthonormal ? – diimension Oct 20 '12 at 23:44
Well, is $\langle a_i,a_j\rangle = \delta_{ij}$ equivalent to $(a_i)$ forming an orthonormal basis? – Neal Oct 20 '12 at 23:47
Yes, since we have already established that $a_i,a_j$ to be = I which will make its entries equal to 1 ? – diimension Oct 20 '12 at 23:49
@diimension I'm not sure what you mean by "$a_i,a_j$ to be = I", and the identity matrix's entries are not all equal to 1. I would guess you've probably proven earlier in your course that $\langle a_i,a_j\rangle = \delta_{ij}$ iff the $(a_i)$ form an orthonormal basis of your vector space. – Neal Oct 21 '12 at 2:29
No, no need to apologize! If we have a collection of $n$ vectors $(a_i)$ which have $\langle a_i, a_j\rangle = \delta_{ij}$, then each vector has unit length and all the vectors are orthogonal, hence linearly independent. By dimension count, they must span the space, hence they form a basis. But the condition that $\langle a_i,a_j\rangle = \delta_{ij}$ is exactly orthonormality. – Neal Oct 22 '12 at 12:02

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