that third term in h should be h3 and not he.
Great balls of fire. It’s too early in the morning —
these equations
a[sup]T[/sup]h = theta*a[sup]T[/sup]V[sup]-1[/sup]h
and
1 = theta*a[sup]T[/sup]V[sup]-1[/sup]h
should be
a[sup]T[/sup]h = theta*a[sup]T[/sup]V[sup]-1[/sup]a
and
1 = theta*a[sup]T[/sup]V[sup]-1[/sup]a
I’m not clear on what this means. Can you elaborate?
I’m actually beginning to think that S[sub]i[/sub] was your notation for “sum over i”. I’d originally parsed it as a diagonal matrix of coefficients. The rest of the post assumes that’s what you meant (which may well be false)
In general, an inner product is a symmetric bilinear form, which can be given in matrices by
u·v = u[sup]T[/sup]Sv
= \Sigma[sub]i,j=1[/sub][sup]n[/sup] u[sub]i[/sub]S[sub]ij[/sub]v[sub]j[/sub]
Where S is a symmetric n-by-n matrix. In general over the real numbers (and many other situations) we can choose a basis so that the matrix S is diagonal, which could be written (suppressing the summation sign)
S[sub]i[/sub]u[sub]i[/sub]v[sub]i[/sub]
Yes, that is what I meant. I was hoping the bounds would be clear from context.
What we need is a bulletin-board that recognizes LaTeX.
Thanks David: you nailed it.
a’h = h’a : Yikes, why didn’t I see that?
…and permits inserted images (a graph every now and then can be diverting).
You didn’t see it? Who did? We don’t do these things every day. If we did we would immediately see that h[sup]T[/sup]a is the dot product h·a and the dot product is commutative so it’s equal to a·h or a[sup]T[/sup]h.