42

Also....look at yahoo groups -- there's SEVERAL groups for linear algebra....scary, ain't it?

Quoted: Quoted: Never mind, I quit at subspaces.  You ain't seen nothin' yet, though, wait until you get to double dual vector spaces. Oh yeah, and wait till you get to double dual secret vector probation spaces!

The answer is 42.

Quoted: Quoted: Yes I can do them however Im really busy w school this week 1. When is it due 2. What AR parts 3. Can you send me a scan because I can't read that red and yellow BS yeah, what he said, i think i still have my linear algebra book. what's "Nul()"? everything else i think i remember, but i don't have a clue what Nul is. is it like elements which are zero (like null)?

i assume solve for the nullspace (null vector space) ie the trivial solution, but i havent looked at the problems yet

Quoted: i assume solve for the nullspace (null vector space) ie the trivial solution, but i havent looked at the problems yet

I think that is right.  My book says the nullspace of an m x n matrix A, written as Nul A, is a set of all solutions to the homogeneous equation Ax=0.

Quoted: Yes I can do them however Im really busy w school this week 1. When is it due 2. What AR parts 3. Can you send me a scan because I can't read that red and yellow BS

It is due on Friday.  Please give me you email so I can send a scan.



thanks
Scott

Quoted: Quoted: Take home test, huh? Be back in a while, to see if i can figure this one out. Yeah, and my teacher is from China and I can't understand a thing he says.  Everyone is lost in the class. Scott

Do the rules of the exam forbid you from getting help from other people?

Number 6 makes sense intuitively. If W and V didn't have a nonzero vector in common, and they were both three-dimensional subspaces, they would have to be in R6 to fit together. Given that they're in R5, they have to have at least one nonzero vector in common.

I know that's not a rigorous proof, but it should help you understand a bit.

Edited to add: I hate it when you're taking some technical class and the professor is from China, the TAs are from China, and most of the other students are from China. I'm even Chinese myself, and it still drives me crazy.

Quoted: Number 6 makes sense intuitively. If W and V didn't have a nonzero vector in common, and they were both three-dimensional subspaces, they would have to be in R6 to fit together. Given that they're in R5, they have to have at least one nonzero vector in common. I know that's not a rigorous proof, but it should help you understand a bit. Edited to add: I hate it when you're taking some technical class and the professor is from China, the TAs are from China, and most of the other students are from China. I'm even Chinese myself, and it still drives me crazy.

This is correct, you just need to word it more rigorously.

HINTS

I will fill them in as I have time


#1:
The kernel of a linear transformation between two vector spaces is its nullspace (ie ker(A) = null(A) )

#2

#3

#4
Recall that dimension of column space = dimension of row space = rank of A (probably theres a proof for this in the book).
Specfically consider: if you know what the rank of C is, what is the rank of C(transpose) ?  (Hint, what is the rank of the identity matrix Im?  What about its transpose?)  That should help you get started at least.

Also consider this:  under what circumstances could the rank of a Matrix go UP?  Wouldn't that require making the matrix itself of larger dimensions?   Is it even possible for the rank to increase doing multiplication?

#5
P is invertible => P is nonsingular => P has linearly independent columns => P has rank m (by def'n of rank)
Same for Q, which has rank n
PA is mxm
AQ is mxn
Try to work from here

#6
Someone else posted an outline earlier.  A strategy here might be to start out by saying "assume V and W do not have a nonzero vector in common" then prove a contradiction.

#7
Look up LU factorization in your book

#8
Think about how the rank of A and the rank of a reduced (eg Gaussian reduced) matrix A are the same thing.

#9

Tb is your first transformation matrix
Tc is your second transformation matrix
Tbc is your transformation matrix from b to c

[e1 e2]*Tb = [b1 b2]
[e1 e2]*Tc = [c1 c2]

if
[b1 b2] = [e1 e2]*Tb   and
[c1 c2] = [e1 e2]*Tc   and
[e1 e2] = [b1 b2]*inv(Tb)   then
[c1 c2] = [b1 b2]*inv(Tb)*Tc   therefore

Tbc = inv(Tb)*Tc

I made that up in 20 seconds please double check it

#10

Not sure what the teach means here but you could just do the multiplication out by hand and show that the transformed b1 =
[a*b11+ b*b11]
[        d*b21     ]

where b11 is the first entry of b1 and b12 is the second entry of b1

WoW

I actually sat here and tried to follow what was going on with your math and the possibilities for the solve.

My brain hurts now, thanks.

Quoted: HINTS #5 You must've mistyped this one because if P is mxn and A is mxn you cannot multiply them unless m=n.  Also, what is Q? #10

P is m x m

A is m x n

Q is n x n

bump

updated 5 and 10