If [b[sup]ij[/sup]] = [a[sub]ij[/sub]][sup]-1[/sup] solve the n x n system
(I) y[sub]i[/sub] = a[sub]ij[/sub] x[sub]j[/sub] for x[sub]j[/sub] in terms of y[sub]i[/sub]
It seems to me that x[sub]j[/sub] just equals [b[sup]ij[/sup]] y[sub]i[/sub]
But the text says to multiply both sides of (I) by b[sup]ki[/sup] and sum on i.
And then x[sub]j[/sub] = b[sup]ji[/sup] y[sub]i[/sub]
Could someone give me some idea of what’s going on here?
Yes, I was wondering the very same thing this morning. Anyone?
Do you understand the notation? x[sub]j[/sub]=b[sub]ij[/sub]y[sub]i[/sub] can’t be right because the subscripts don’t match up. b[sub]ij[/sub]y[sub]i[/sub] is shorthand for b[sub]1j[/sub]y[sub]1[/sub]+b[sub]2j[/sub]y[sub]2[/sub]+…+b[sub]nj[/sub]y[sub]n[/sub] and something[sub]j[/sub]=something[sub]j[/sub] means it’s true for j=1,2…n, but your answer has a sub i on the left side and a sub j on the right (ignoring the two 'i’s , which are shorthand for a sum).
As a shortcut, think of A and B as matrices and X and Y as vectors.
You have Y=B[sup]-1[/sup]X so BY=X and put in the subscripts to get b[sub]st[/sub]y[sub]t[/sub]=x[sub]s[/sub].
When you multiply matrices, you have to match up the inner indices, just like you have to match up the inner dimensions. Do you recall that you can only multiply an n × m matrix by an m × p matrix?
You should have a textbook that explains it better than I can, with some examples. Make sure you understand ‘(einsteinian) summation convention.’ You might want to try doing it their way, and then write all the steps out in full in terms of 1,2,3 (assume n=3, makes it easier to write).
Jpeg: I can’t tell. Are you joking? You don’t know, or you’d say so, but you bothered to read a LA thread…
:smack: Yeah, I meant to say what Achernar said as well.
You get a lot of mathematicians on SDMB 
No I don’t understand the notation unless it’s clearly explained—heavy indices notation looks like gobbledygook to me. I have had no trouble with calculus, differential equations, partial differential equations or “Elementary Linear Algebra.” The last because the author went easy on indicial stuff and stayed with matrix notation. But because he did I don’t see where I’m ever going to be able to pick this stuff up.
Now that you and Achernar have explained it, I sort of get it, but otherwise I think I’m dead. I just can’t see this stuff.
Can I always substitute matrix notation and put in the subscripts later?
The very next problem says (Using caps so I don’t have to sup/sub so many times)
Show that under a change of coordinates Xi(barred) = AijXj the quadric hypersurface CijXiXj = 1 transforms to CijXiXj(all barred) =1 where Cij(barred) = CrsBriBsj with (Bij) = (Aij)[sup]-1[/sup].
I can’t even tell what they’re asking. Can this be translated into matrix notation?
Also inre my first post. If y = Ax then x = A[sup]-1[/sup] y.
But A[sup]-1[/sup] = [a[sub]ij[/sub]][sup]-1[/sup] = [b[sup]ij[/sup]]
But the solution is x[sub]j[/sub] = [b[sup]ji[/sup]] y[sub]i[/sub]
Can you just switch indices around without changing the matrix? Are these just dummy indices?
You need to find a good book or website and I don’t really know. Is this at university? Is there anyone to help you with this sort of stuff?
Yes, these represent matrices, so you can switch notations, but there are some things only subscript notation can do.
Remember what the subscripts mean. Aij means the (i,j) entry in matrix A. Yj means the j’th entry in vector Y. AijYj means Ai1Y1 + Ai2Y2 + Ai3Y3, which is the same vector as AY. It might make you feel better if you check this is true from the definition of multiplication of matrices. Would it help to write X=(X11,X12…) A=(Ai1|Ai2|Ai3) where this means replace Ai1 with the column vector (A11,A21,A31) and do some examples? Notice the terms can be numbers so AB is represented by AijBjk=BjkAij but not equal BkjAij (which is BA).
The second example was fairly easy, once I ignored the hard words. I use X* to mean X bar and Xi* for X bar sub i . It’s really asking "Given Xi*=AijXj and CijXiXj=1 eliminate X to find an equation for X’ in terms of C and A.
To explain what I think it’s getting at, you know how you can represent a line in 2d by x=2y ? This can be written X=(x,y) and giving a matrix equation for X. Now draw some new axes at 45 degrees to the old ones. Coordinates with respect to these axes can be (x’,y’)=X’. And X’=AX where A is a rotation matrix. This is the same thing, but in 4d or something.
Let’s try matrix notaion, using Xt for X as a row vector and X for X as a column vector and . for multiply.
X*=A.X
and
A=invB
so
B.X*=X and Xt=X*t.Bt
Now
Xt.C.X =1
so
(Xt.Bt).C.(B.X)=1
or
Xt.(Bt.C.B).X=1
or
Xt.C.X*=1 where C*=(Bt.C.B)
Do you want to try that with subscript notation and see how far you get?
What do you mean by ‘switch the indices’? It doesn’t matter which letters you use, so Aij=Bji means the same as Amn=Bnm or, indeed, Aji=Bij. If it occurs twice it can be changed to anything, so for instance AijXj-AikXk=AilXl-AilXl=0. Is that what you asked?
Thanks Shade, I’m going to print this out and give the subscript notation a try.
In the first problem I thought b[sub]ji[/sub] would equal (b[sub]ij[/sub])[sup]T[/sup]
And no there’s no one I can ask for help. And I haven’t been able to find a website that covers this stuff so You guys are it.
Thanks again.
It does.
You’re sunk 
Are you doing this just for fun then, or what?
I had a google, but couldn’t find anything helpful. this covers what I said above, I think, and hopefully won’t have any typos in.
PS. Goodnight.
In your link it asks the following:
Exercise: Given
[a[sub]ij[/sub]] = [1 0 2][0 1 2][3 0 3]
Evaluate (a) a[sub]ii[/sub] – no problem, (b) a[sub]ij[/sub] a[sub]ij[/sub] – no problem
© a[sub]jk[/sub]a[sub]kj[/sub] – problem
Does this mean j still represents the column and k is just a substitute index for i? IOW is a[sub]jk[/sub]a[sub]kj[/sub] the same as a[sub]ji[/sub]a[sub]ij[/sub]?
I presume it is but I don’t know that for sure.
Where’s ultrafilter when you need him?
It is indeed the same, Sacrociliac. I’d point out that a[sub]ij[/sub]a[sub]ij[/sub] and a[sub]ij[/sub]a[sub]ji[/sub] are not the same, however, except perhaps by numerical coincidence.
Where the hell did that extra “c” come from? Oops.
Damn. I wish I would have named myself Sacro Ciliac or, maybe Sacro Iliac. Either one sounds/looks better than plain old Sacroiliac.
Thanks Gator Guy. How did the Gators do this year? Damn if I can remember. But I did win a pool. NC state won it for me.
Eh, let’s just say that the calls for Ron Zook’s head started early in the season and picked up steam as the season progressed. Not that I think it was an awful season by most standards, but still…
One final question and then I’ll drop this. In the first problem I posted
y[sub]i[/sub] = a[sub]ij[/sub] x[sub]j[/sub] for x[sub]j[/sub] in terms of y[sub]i[/sub]
In matrix notation this is just y = Ax, so x = A[sup]-1[/sup]y
But the answer in index notation is x[sub]j[/sub] = [b[sub]ji[/sub]]y[sub]i[/sub]
But Shade said that [b[sub]ji[/sub]] = (b[sub]ij[/sub])[sup]T[/sup]
So given 1 wouldn’t this mean that x = [A[sup]-1[/sup]][sup]T[/sup]y – Vs. x = A[sup]-1[/sup]y?
Reading the thread, marveling at how much linear algebra I’ve forgotten. 
More help if I can think of it.