Need help solving this in matlab!
Eigenvalue problems 3) The popularity problem The google search algorithm, in its infancy, solved a problem in linear algebra called the popularity problem. It attempts to answer the following question in a straightforward yet sophisticated way: If I ask a group of n websites which among the other n - 1 sites are the most popular, how can rank each website's popularity? The simplest method is just count links. Every time a website is linked to, it's popularity increases by 1. But this is problematic, and doesn't always give results that people agree with Maybe one site is linked to many times, but it is exclusively from sites that are unpopular themselves. Also, maybe certain sites contain a ton of links, so their individual links shouldn't carry as much weight. We would like to find the popularity of each website, where instead of simply counting links, we sum the popularity of each website that linked to each other website. Read the following paragraphs slowly and several times. Let Rijbe a 1 ora 0 depending on whether or not the 'h website linked to the website, and let p, be the popularity of the website. In that case, a sensible expression for the popularity of the 1" website would be P: = R12P2 + ... RIP We could write a similar equation for every single website, i 1... n. This creates a system of equations of the form Rp = p where Ris a matrix containing all of the R entries, and p is an unknown column vector containing the popularities of each website. This should be immediately recognizable as an eigenvalue problem. Remember, the eigenvalue equation is Az = 2x It is not at all clear that the above equation (Rp = p) would, in general, have a solution. That's because it isn't a general eigenvalue equation. It specifically has an eigenvalue of 1. An arbitrary matrix R will not necessarily have that eigenvalue. However, we can fix this problem if we enforce the following rule:every column of R must sum to 1. If every column adds up to one, the matrix is guaranteed to have 1 as an eigenvalue, and the above equation will have a solution. Essentially, each website is only allowed to give one total"link", evenly distributed amongst all the different websites it links to. The above fact is not obvious or simple to prove. The jh column of the R matrix, again, is the number of times the website linked to the others. If we take each column and divide it by its sum, it will then meet our requirement. That is to say, we normalize each column, individually, so they all sum to 1. If the 5 website linked to you, but it also linked to 9 other websites, you only get an Rig value of 0.1 Links from Website 3 Website 4 | Website 5 Website 1 Website 2 Website 1 10 Website 2 1 Website 3 11 11 Website 4 11 1 Website 5 0 0 10 1 O For example, from the table we see: From column 1, website 1 links to websites 2, 3, and 4 From row 2, we see that website 2 was linked from website 1,3, and 5. For the table above, solve for the popularities if the websites in question. List, in a comment, the website in order from most popular to least popular Eigenvalue problems 3) The popularity problem The google search algorithm, in its infancy, solved a problem in linear algebra called the popularity problem. It attempts to answer the following question in a straightforward yet sophisticated way: If I ask a group of n websites which among the other n - 1 sites are the most popular, how can rank each website's popularity? The simplest method is just count links. Every time a website is linked to, it's popularity increases by 1. But this is problematic, and doesn't always give results that people agree with Maybe one site is linked to many times, but it is exclusively from sites that are unpopular themselves. Also, maybe certain sites contain a ton of links, so their individual links shouldn't carry as much weight. We would like to find the popularity of each website, where instead of simply counting links, we sum the popularity of each website that linked to each other website. Read the following paragraphs slowly and several times. Let Rijbe a 1 ora 0 depending on whether or not the 'h website linked to the website, and let p, be the popularity of the website. In that case, a sensible expression for the popularity of the 1" website would be P: = R12P2 + ... RIP We could write a similar equation for every single website, i 1... n. This creates a system of equations of the form Rp = p where Ris a matrix containing all of the R entries, and p is an unknown column vector containing the popularities of each website. This should be immediately recognizable as an eigenvalue problem. Remember, the eigenvalue equation is Az = 2x It is not at all clear that the above equation (Rp = p) would, in general, have a solution. That's because it isn't a general eigenvalue equation. It specifically has an eigenvalue of 1. An arbitrary matrix R will not necessarily have that eigenvalue. However, we can fix this problem if we enforce the following rule:every column of R must sum to 1. If every column adds up to one, the matrix is guaranteed to have 1 as an eigenvalue, and the above equation will have a solution. Essentially, each website is only allowed to give one total"link", evenly distributed amongst all the different websites it links to. The above fact is not obvious or simple to prove. The jh column of the R matrix, again, is the number of times the website linked to the others. If we take each column and divide it by its sum, it will then meet our requirement. That is to say, we normalize each column, individually, so they all sum to 1. If the 5 website linked to you, but it also linked to 9 other websites, you only get an Rig value of 0.1 Links from Website 3 Website 4 | Website 5 Website 1 Website 2 Website 1 10 Website 2 1 Website 3 11 11 Website 4 11 1 Website 5 0 0 10 1 O For example, from the table we see: From column 1, website 1 links to websites 2, 3, and 4 From row 2, we see that website 2 was linked from website 1,3, and 5. For the table above, solve for the popularities if the websites in question. List, in a comment, the website in order from most popular to least popular
