Easy way to multiplying standard basis

It seems to be easy, but I am not good at it.
I would like to generate the following matrix that is multiplying of two standard basis:
E_{ij}:= (e_i - e_j)(e_i - e_j)^T
Could any body knows how to write it in few lines?
Thanks

6 comentarios

Azzi Abdelmalek
Azzi Abdelmalek el 22 de Feb. de 2013
What is T?
Saber
Saber el 22 de Feb. de 2013
It shows exactly the transpose of the first term or in MATLAB like a'.
Azzi Abdelmalek
Azzi Abdelmalek el 22 de Feb. de 2013
Editada: Azzi Abdelmalek el 22 de Feb. de 2013
What is the size of e. Can you provide an example?
Saber
Saber el 22 de Feb. de 2013
@Azzi Abdelmalek: Please look at below
Azzi Abdelmalek
Azzi Abdelmalek el 22 de Feb. de 2013
Post an example with expected result
Azzi Abdelmalek
Azzi Abdelmalek el 22 de Feb. de 2013
Editada: Azzi Abdelmalek el 22 de Feb. de 2013
It seems that second Wayne's answer is what you are looking for

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Respuestas (3)

Thiru jeyan
Thiru jeyan el 2 de Mayo de 2017

You can use this method

e=@(k,n) [zeros(k-1,1);1;zeros(n-k,1)]

For an example

>> e(1,3)

ans =

     1
     0
     0

>> e(2,5)

ans =

     0
     1
     0
     0
     0
Wayne King
Wayne King el 22 de Feb. de 2013
I agree with the comment above that you should provide an example because it's not clear what the size of your problem is or how scalable you want it. For example, are you looking for something as simple as:
E1 = [1 0 0]';
E2 = [0 1 0]';
matrx = (E1-E2)*(E1-E2)';

1 comentario

Saber
Saber el 22 de Feb. de 2013
Editada: Saber el 22 de Feb. de 2013
Let's our dimension is n=5. Now in this dimension we have five standard basis which all together build identity matrix I up. e_i shows this unit vectors(columns of I) that in i'th position is 1 and other entries are all zero. for example
e_3= 0
0
1
0
0
now I want to build up a routine way to create this vectors and label them as the position of 1's.
I hope that now it is clear a bit.

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Wayne King
Wayne King el 22 de Feb. de 2013
You can just do
I = eye(5);
Each column of I is a standard basis vector
Then,
idx = [3 4];
Is = I(:,idx(1))-I(:,idx(2));
X = Is*Is';
gives you what you want for an example where i=3 and j= 4

2 comentarios

Saber
Saber el 22 de Feb. de 2013
The out put is too strange:
X =
0 0 0 0 0
0 0 0 0 0
0 0 1 -1 0
0 0 -1 1 0
0 0 0 0 0
I need to separate matrix
I =
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
as unit indexed vectors
1 0 0 0 0
0 1 0 0 0
e_1= 0 e_2= 0 e_3= 1 e_4= 0 e_5= 0
0 0 0 1 0
0 0 0 0 1
Wayne King
Wayne King el 22 de Feb. de 2013
so just extract the columns, what is so difficult about that?
Did you look at the Is vector? that is just a column vector.

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