Linear Algebra trouble

[79wjd]

Am obviously missing something really stupid, but how does w = [-3,1,-1,1]^T?

 

y=[-1,1,-1,1]^T

w = y - ||y||e₁ = [-1,1,-1,1]^T - 2e₁.

So why wouldn't w = [-3,-1,-3,-1]^T?

[Guest]

I'm no math wiz, but this website should help.

http://www.wolframalpha.com/

[79wjd]

Just now, MagnesiumPC said:

I'm no math wiz, but this website should help.

http://www.wolframalpha.com/

No.

[Guest]

Just now, djdwosk97 said:

No.

? I can't help you personally, but resources like Woflram & Khan Academy can really help. Unless, this is an equation you already know the answer to, and in that case, no thanks.

[Dash Lambda]

What is e₁? From this it looks to just be [1 0 0 0]^T, is it a basis?

[79wjd]

51 minutes ago, Dash Lambda said:

What is e₁? From this it looks to just be [1 0 0 0]^T, is it a basis?

Yeah, that's what it is. I realized that last night after playing around with it a bit. 

[ono]

4 hours ago, djdwosk97 said:

Yeah, that's what it is. I realized that last night after playing around with it a bit. 

I take it that means you know why w is equal that now and this topic is closed? Just checking in case you're still confused on something.

[79wjd]

29 minutes ago, ono said:

I take it that means you know why w is equal that now and this topic is closed? Just checking in case you're still confused on something.

Yeah, don't go too far though, I might have another question related to something slightly different in a little bit.

[KieuVanQuan]

On 9/26/2017 at 4:39 AM, djdwosk97 said:

 

y=[-1,1,-1,1]^T

w = y - ||y||e₁ = [-1,1,-1,1]^T - 2e₁.

So why wouldn't w = [-3,-1,-3,-1]^T?

 

Yep, I completely agree with that.

[79wjd]

On 9/27/2017 at 3:46 PM, ono said:

I take it that means you know why w is equal that now and this topic is closed? Just checking in case you're still confused on something.

Any thoughts on this one: (the problem is in the first slide)

 

I looked at the lecture slides, and I'm pretty sure I want to do this to solve for y (to approximate b): (this is called from the second code snippet)

find_y(A,z,W){
  	T = A.T #A transpose
	x = (T*W*A)^(-1)*T*z
	y = A*x
	return y
}

And then I was thinking something like this, but I'm not really sure: 

#A is a given 4x2 matrix
#z is a given 4x1 matrix
#W is a matrix that I pass in that 
###is one of the three options from 
###the first screenshot -- so, let's say 
###W is a 4x4 identity matrix. 

T = A transpose
Wg = T*W*A
y = find_y(W,z,Wg)

 

[Teddy07]

sick shit man, lol. I feel like an amateur reading that

[ono]

3 hours ago, djdwosk97 said:

Any thoughts on this one: (the problem is in the first slide)

 

I looked at the lecture slides, and I'm pretty sure I want to do this to solve for y (to approximate b): (this is called from the second code snippet)

find_y(A,z,W){
  	T = A.T #A transpose
	x = (T*W*A)^(-1)*T*z
	y = A*x
	return y
}

And then I was thinking something like this, but I'm not really sure: 

#A is a given 4x2 matrix
#z is a given 4x1 matrix
#W is a matrix that I pass in that 
###is one of the three options from 
###the first screenshot -- so, let's say 
###W is a 4x4 identity matrix. 

T = A transpose
Wg = T*W*A
y = find_y(W,z,Wg)

 

I'm not familiar with this method and I don't have the time to dig into this (I just visit on small breaks at work), but it looks strange to me that you calculate Wg=T*W*A in your second snippet when in your first snippet, you already have T*W*A. Though I'm not sure what the purpose of and where A comes from quickly glancing at the material. EDIT: Oh nevermind, I missed the part that the solution is constrained to be in the column space of some chosen A. So yeah, doesn't your first snippet do all the work? Doesn't seem like you need  the second snippet.

 

Neat your course uses Python, at least that's the only language I'm aware of that uses numpy. I used matlab/octave back in my schooling days.

Edited September 29, 2017 by ono