Vector in equation

[Sauraj]

Hello, $\displaystyle w, w^T, w_x, w_y$ are vectors of same dimensions
$\displaystyle (w^Tw_x-w^Tw_y)^2=w^T(w_x-w_y)(w_x-w_y)^Tw$
I dont understand where the right $\displaystyle w$ comes from.
I do:
\(\displaystyle
(w^Tw_x-w^Tw_y)^2 = (w^Tw_x-w^Tw_y)(w^Tw_x-w^Tw_y)^T = ?\) here I would do $\displaystyle w^T(w_x-w_y)w^T(w_x-w_y)^T$ but its not correct
thx

 

[yoscar04]

Hi.
I think you missed that (AB)^T=B^TA^T. That's the reason why the [MATH]\omega[/MATH] gets to the right at the end.

 

[Steven G]

(w^Twx−w^Twy)²=(w^Twx−w^Twy)(w^Twx−w^Twy)^T

Where does that last T come from. Doesn't square mean to multiply the base by itself two times?

 

[yoscar04]

(w^Twx−w^Twy)²=(w^Twx−w^Twy)(w^Twx−w^Twy)^T

Where does that last T come from. Doesn't square mean to multiply the base by itself two times?

Not in this case. There is no meaning to a vector square per se. We know how to compute A², given a square matrix. But if you take, let's say u as having components x,y,z (column vector): how would you square it? There is no meaning to the multiplication of a 3X1 matrix by itself (or a 1X3 if you want to put it the other way). When we multiply matrices, the number of columns of the 1st matrix must equal the number of rows of the 2nd matrix. So, in general when people talk about the square of a vector they mean u^Tu.

 

[yoscar04]

(w^Twx−w^Twy)²=(w^Twx−w^Twy)(w^Twx−w^Twy)^T

Where does that last T come from. Doesn't square mean to multiply the base by itself two times?

By the way, Jomo, you are right here. I didn't pay attention to the beginning of his answer, in the question. So I am not sure of what he did. It would be nice to see the original question.

 

[Sauraj]

(w^Twx−w^Twy)²=(w^Twx−w^Twy)(w^Twx−w^Twy)^T

Where does that last T come from. Doesn't square mean to multiply the base by itself two times?

I think the norm is missing here, but in the lecture it was also without norm (||.||)

By the way, Jomo, you are right here. I didn't pay attention to the beginning of his answer, in the question. So I am not sure of what he did. It would be nice to see the original question.

There is no original question, this equation refers to Linear discriminant analysis (Maximizes mean class dierence).
Yes I forgot about the transpose rule.
$\displaystyle (w^Tw_x-w^Tw_y)^2 = (w^Tw_x-w^Tw_y)(w^Tw_x-w^Tw_y)^T = w^T(w_x-w_y)(w^T(w_x-w_y))^T = w^T(w_x-w_y)(w_x-w_y)^Tw$
thx yoscar04 !

 

[yoscar04]

I think the norm is missing here, but in the lecture it was also without norm (||.||)


There is no original question, this equation refers to Linear discriminant analysis (Maximizes mean class dierence).
Yes I forgot about the transpose rule.
$\displaystyle (w^Tw_x-w^Tw_y)^2 = (w^Tw_x-w^Tw_y)(w^Tw_x-w^Tw_y)^T = w^T(w_x-w_y)(w^T(w_x-w_y))^T = w^T(w_x-w_y)(w_x-w_y)^Tw$
thx yoscar04 !

What is funny is that each term inside the parenthesis is a number. I got confused after writing my observation, because you wrote (..)²=(..)(..)^T instead of (..)²= (..)^T(..). But since the objects inside the parenthesis are numbers, it doesn't care about the order.