bb700092 Posted March 4, 2013 Share Posted March 4, 2013 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. Sum_{j} w_{ij} = 1 for all i. 2. Sum of squares of all elements of each column is equal to one, i.e. Sum_{i} w_{ij}^{2} = 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. Link to comment Share on other sites More sharing options...

elfmotat Posted March 4, 2013 Share Posted March 4, 2013 Sure. A trivial example would be [math]w_{i1}=1[/math], and [math]w_{ij}=0[/math] for all [math]j\neq 1[/math]. Link to comment Share on other sites More sharing options...

uncool Posted March 4, 2013 Share Posted March 4, 2013 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- Link to comment Share on other sites More sharing options...

elfmotat Posted March 4, 2013 Share Posted March 4, 2013 (edited) How does that satisfy the second property? Are negative numbers allowed? =Uncool- Because 1^{2}=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 March 4, 2013 by elfmotat Link to comment Share on other sites More sharing options...

John Posted March 4, 2013 Share Posted March 4, 2013 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. Link to comment Share on other sites More sharing options...

elfmotat Posted March 4, 2013 Share Posted March 4, 2013 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. Link to comment Share on other sites More sharing options...

John Posted March 4, 2013 Share Posted March 4, 2013 (edited) 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 March 4, 2013 by John Link to comment Share on other sites More sharing options...

elfmotat Posted March 4, 2013 Share Posted March 4, 2013 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. 1 Link to comment Share on other sites More sharing options...

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