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Who wants to explain linear algebra to me?

Fenixgoon

Fenixgoon

Lifer
yes it's my homework. no i'm not looking for just the answer. i want to understand how the math works.

the problem is:

show that:

(A dyadic B)(C dyadic D) = (B dot C)(A dyadic D)

skype, vent, PM, AIM, whatever works if you're willing to explain the math.

TIA, flamers can go to hell since i'm not looking for ATOT to do the work for me.
 
Yeah, this makes absolutely no sense. What are the dyadics, what are the tensors, and what are the operators? I have no idea what the line means.
 
Originally posted by: Born2bwire
Yeah, this makes absolutely no sense. What are the dyadics, what are the tensors, and what are the operators? I have no idea what the line means.
Click to expand...

as far as i know, ABCD are all vectors.
 
Originally posted by: Quintox
what the hell is a dyadic, sounds like medicine or something.

Good luck!!
Click to expand...

I googled it and found the wiki page. I always learned that operation as an outer product.

A = [a1 a2 a3]
B = [b1 b2 b3]

Inner product = A * B = a1b1 + a2b2 + a3b3

Outer product =

a1b1 a1b2 a1b3
a2b1 a2b2 a2b3
a3b1 a3b2 a3b3

That's a matrix btw.

It seems like the best way to do it is to just write out a general vector the way I have, do the multiplication, and see if you can get the product on one side to be the same as on the other.
 
Originally posted by: silverpig
Originally posted by: Quintox
what the hell is a dyadic, sounds like medicine or something.

Good luck!!
Click to expand...

I googled it and found the wiki page. I always learned that operation as an outer product.

A = [a1 a2 a3]
B = [b1 b2 b3]

Inner product = A * B = a1b1 + a2b2 + a3b3

Outer product =

a1b1 a1b2 a1b3
a2b1 a2b2 a2b3
a3b1 a3b2 a3b3

That's a matrix btw.

It seems like the best way to do it is to just write out a general vector the way I have, do the multiplication, and see if you can get the product on one side to be the same as on the other.
Click to expand...

Weird......

I was taught (University of Manitoba) the names of those products as:

Inner Product -> Dot (Scalar) Product
Outer Product -> Cross (Vector) Product

I've never heard them referred to as Inner and Outer products.
 
Originally posted by: futuristicmonkey
Originally posted by: silverpig
Originally posted by: Quintox
what the hell is a dyadic, sounds like medicine or something.

Good luck!!
Click to expand...

I googled it and found the wiki page. I always learned that operation as an outer product.

A = [a1 a2 a3]
B = [b1 b2 b3]

Inner product = A * B = a1b1 + a2b2 + a3b3

Outer product =

a1b1 a1b2 a1b3
a2b1 a2b2 a2b3
a3b1 a3b2 a3b3

That's a matrix btw.

It seems like the best way to do it is to just write out a general vector the way I have, do the multiplication, and see if you can get the product on one side to be the same as on the other.
Click to expand...

Weird......

I was taught (University of Manitoba) the names of those products as:

Inner Product -> Dot (Scalar) Product
Outer Product -> Cross (Vector) Product

I've never heard them referred to as Inner and Outer products.
Click to expand...

A cross product is definitely not an outer product. A cross product is only valid for 2 3-d vectors and gives a vector perpendicular to the plane of both input vectors. The result of an outer product is a matrix. An inner product is a dot product though, so that's right.
 
I got A's in Calc I through III but couldn't quite grasp Linear Algebra. I did end up getting a B in it but I have no clue how to do any of the math, and of course have not used it since the last day of class (I'm a software developer).
 
Originally posted by: Eeezee
Simple answer - linear algebra is stupid and useless, and no one should ever take it for any reason.
Click to expand...

well according to my mechanical behavior of solids class, and probably material modelling as well, it's not.

also, the whole idea behind my class right now is to use indicial notation.

apparently A dyadic B = AiBj (Ei dyadic Ej) where Ei and Ej are the respective unit vectors in an orthonormal basis.

going with that, (A dyadic B) (C dyadic D) gives Ai Bj Ci Dj (Ei dyadic Ej)^2

the other side, (B dot C) (A dyadic D) gives Bi Ci Ai Dj (Ei dyadic Ej)

the confusing part is that I don't know which indices to use for each vector. With (A dyadic B)*(C dyadic D) does the answer come of the form Ai Bj Ck Dl (that's an L next to D)? hell if i know, and i don't know how to reduce it further.

grr :|
 
Originally posted by: Fenixgoon
yes it's my homework. no i'm not looking for just the answer. i want to understand how the math works.

the problem is:

show that:

(A dyadic B)(C dyadic D) = (B dot C)(A dyadic D)

skype, vent, PM, AIM, whatever works if you're willing to explain the math.

TIA, flamers can go to hell since i'm not looking for ATOT to do the work for me.
Click to expand...


Okay this turns out to be pretty straight forward with the right notation.

First, A, B, C and D are vectors and we will denote the i th component as A_i ect.

(Typically we use both up and down indicees, but here it won't be needed.)

Using this notation, the dot product between A and B is the sum over i of A_i*B_i.

Using this notation, the dyadic product of C and Dgives a matrix whose jk th component is C_j*D_k

Now we can show this relationship holds:

The components of A dyadic B are A_i*B_j and C dyadic D C_j*D_k

Taking the product of these matrices we sum over all the j components

Sum( A_i*B_j*C_j*D_k)

now the inner term of the sum is B_j*C_j which is exactly the dot product and we are left with A_i*D_k the components of the dyadic of A and D.

Hope this helps in some way! Best of luck.

 
Originally posted by: Fenixgoon
Originally posted by: Eeezee
Simple answer - linear algebra is stupid and useless, and no one should ever take it for any reason.
Click to expand...

well according to my mechanical behavior of solids class, and probably material modelling as well, it's not.

also, the whole idea behind my class right now is to use indicial notation.

apparently A dyadic B = AiBj (Ei dyadic Ej) where Ei and Ej are the respective unit vectors in an orthonormal basis.

going with that, (A dyadic B) (C dyadic D) gives Ai Bj Ci Dj (Ei dyadic Ej)^2

the other side, (B dot C) (A dyadic D) gives Bi Ci Ai Dj (Ei dyadic Ej)

the confusing part is that I don't know which indices to use for each vector. With (A dyadic B)*(C dyadic D) does the answer come of the form Ai Bj Ck Dl (that's an L next to D)? hell if i know, and i don't know how to reduce it further.

grr :|
Click to expand...

Dyadic is not an operator, a dyad is a second rank tensor. Dyadic describes an operator that is a dyad, like the Dyadic Green's Function. I think you mean to say the dyad AB, you can represent this in introductory matrix notation as AB' where ' is the transpose of B and both A and B are single rank tensors. All of this is just a fancy way of talking about vectors and matrices. So if I have the tensor A = [a1 a2 a2]' and B = [b1 b2 b3]' (note the transpose operator) then the dyad AB is,

a1b1 a1b2 a1b3
a2b1 a2b2 a2b3
a3b1 a3b2 a3b3

What is the difference between a matrix and a dyad you ask? Well Timmy, a dyad generally has a coordinate system associated with it. So we have a tensor A = [a1x a2y a3z]' where x, y, and z are all the hatted directional vectors. So the first row of the dyad would be [a1b1xx a1b2xy a1b3xz]. This can be used to describe an action by a tensor in one set of coordinates that result in another tensor. So the Dyadic Green's Function will have these xx, xy, xz, etc. terms that stand for the x-directed electric field produced by an x-directed current, a y-directed current, and a z-directed current respectively.
 
Originally posted by: silverpig
Originally posted by: futuristicmonkey
Originally posted by: silverpig
Originally posted by: Quintox
what the hell is a dyadic, sounds like medicine or something.

Good luck!!
Click to expand...

I googled it and found the wiki page. I always learned that operation as an outer product.

A = [a1 a2 a3]
B = [b1 b2 b3]

Inner product = A * B = a1b1 + a2b2 + a3b3

Outer product =

a1b1 a1b2 a1b3
a2b1 a2b2 a2b3
a3b1 a3b2 a3b3

That's a matrix btw.

It seems like the best way to do it is to just write out a general vector the way I have, do the multiplication, and see if you can get the product on one side to be the same as on the other.
Click to expand...

Weird......

I was taught (University of Manitoba) the names of those products as:

Inner Product -> Dot (Scalar) Product
Outer Product -> Cross (Vector) Product

I've never heard them referred to as Inner and Outer products.
Click to expand...

A cross product is definitely not an outer product. A cross product is only valid for 2 3-d vectors and gives a vector perpendicular to the plane of both input vectors. The result of an outer product is a matrix. An inner product is a dot product though, so that's right.
Click to expand...

No kidding! I guess my mind went into autopilot there; I guess, for me, discussions involving the dot product always seem to at least mention the cross product. That'll teach me to type without looking first 😱
 
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