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Is such a matrix possible?


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I want a matrix W of size nxk (where n>>k) with the following two properties:

1. Sum of all elements of each row is equal to one, i.e. Sumj wij = 1 for all i.
2. Sum of squares of all elements of each column is equal to one, i.e. Sumi wij2 = 1 for all j.
Is such a matrix possible? Any hint at how to prove one way or the other would be appreciated. Thanks in advance.
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Sure. A trivial example would be [math]w_{i1}=1[/math], and [math]w_{ij}=0[/math] for all [math]j\neq 1[/math].

How does that satisfy the second property?

 

Are negative numbers allowed?

=Uncool-

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How does that satisfy the second property?

 

Are negative numbers allowed?

=Uncool-

 

Because 12=1, and the rest of the elements in the row are zero. So [math]\sum_j (w_{ij})^2=1^2+0^2+0^2+...+0^2=1[/math].

 

EDIT: You're right, I read the question wrong. The simplest matrix I can think of which satisfies 1. and 2. is the identity matrix, but that doesn't satisfy the condition that n>>k.

Edited by elfmotat
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It's a bit cheesy, but if you let [math]n=k^2[/math] and set every element to [math]\frac{1}{k}[/math], then for sufficiently large [math]k[/math] you'll have such a matrix.

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It's a bit cheesy, but if you let [math]n=k^2[/math] and set every element to [math]\frac{1}{k}[/math], then for sufficiently large [math]k[/math] you'll have such a matrix.

 

Condition 1. is satisfied by setting every element to [math]1/k[/math], but condition 2. is only satisfied if [math]n/k=1[/math], i.e [math]n=k[/math] in which case the condition that [math]n\gg k[/math] is not satisfied.

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Condition 1. is satisfied by setting every element to [math]1/k[/math], but condition 2. is only satisfied if [math]n/k=1[/math], i.e [math]n=k[/math] in which case the condition that [math]n\gg k[/math] is not satisfied.

 

I'm not sure I follow. If we have an [math]n \times k[/math] matrix with [math]n=k^2[/math] and each element set to [math]\frac{1}{k}[/math], then each column will consist of [math]k^2[/math] elements of [math]\frac{1}{k}[/math] each. Since condition 2 deals with the sum of the squares along each column, we end up with [math]k^2 \left(\frac{1}{k^2}\right) =1[/math], as condition 2 requires.

Edited by John
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I'm not sure I follow. If we have an [math]n \times k[/math] matrix with [math]n=k^2[/math] and each element set to [math]\frac{1}{k}[/math], then each column will consist of [math]k^2[/math] elements of [math]\frac{1}{k}[/math] each. Since condition two deals with the sum of the squares along each column, we end up with [math]k^2 \left(\frac{1}{k^2}\right) =1[/math], as condition 2 requires.

 

Nevermind, you're right. Fortunately/unfortunately I'm a bit intoxicated at the moment.

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