A matrix defined inner product of R^n generated by the invertible nxn matrix A, < u,v >= Au dot Av. An orthogonal matrix is an invertible matrix where A^T=A^-1
The question asks to prove that if A is an orthogonal nxn matrix then it generates the Euclidean Inner Product in R^n.
My approach to this proof was A^T*A=AA^T=I(identity matrix) which leads to the equivalent characterization that a matrix A is orthogonal if its transpose is equal to its inverse A^T=A^-1 so to be orthogonal it must generate the inner product in R^n and in regards to < u, v >= Au dot Av , i know that this could be expressed as < u ,v > = (Av)^TAu which is equivalent to < u, v > = v^TA^TAu. I guess what I'm stuck on is the part where you have to combine all of this into a formal proof, Any ideas?
Best Answer
The confusion, I guess, is about the formulation of the question itself.
Take these steps.
I start with $R^n$ as an (abstract) vector space, and pretend that I am not aware of its inner product (its geometry).
I am also given this matrix $A$ with the property that $AA^T=1$.
Now I am super clever (!) and define, out of blue, this real-values function on pairs of vectors $v, w \in R^n$: $$ v.w:= (Av)^T(Aw)$$ where I understand the right hand side as a matrix multiplication of one $1\times n$ vector by an $n\times 1$ vector -- the result of which is a $1\times 1$ matrix, a real number.
Then, I prove that $v.w$ define as above is
a) linear in both $v$ and $w,$ b) $v.v$ is always positive, except when $v=0,$ c) $v.w = w.v$ d) [anything else that we need for an inner product]
That's how we "discover" the inner product in $R^n$ only using an orthogonal matrix. And I think that was what the question had asked for.