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Below is the question I want to answer, but how do I even work out a spectrum of something? Please explain.

Consider the complete graph $K_6$ (with $6$ vertices). Show that its spectrum is given by: $\sigma(K_6)=\{[-1]^5,[5]\}.$

My final solution after doing the row reduction https://i.stack.imgur.com/pI0dd.jpg

lauren
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    Your book should have a definition of the term, and you should also type up any work you've done on it. – Adrian Keister Feb 07 '20 at 19:39
  • My booklet litterally has nothing on spectrums this is the confusion I’m having I don’t even know where to start I just have the drawing of K6 – lauren Feb 07 '20 at 19:45
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    The spectrum of a graph is the set of eigenvalues of its adjacency matrix. So, you are meant to show that the adjacency matrix of $K_6$ has eigenvalues $-1$ with multiplicity $5$ and $5$ with multiplicity $1$. – Ben Grossmann Feb 07 '20 at 19:48
  • Right ok so I have a 6x6 matrix and have to do row reduction? – lauren Feb 07 '20 at 19:49
  • That being said: I strongly suspect that if you check the index for the word "spectrum", you will find a definition exactly along the lines of what I have said. – Ben Grossmann Feb 07 '20 at 19:50
  • @lauren what does row-reduction have to do with finding eigenvalues? – Ben Grossmann Feb 07 '20 at 19:50
  • That’s what I was told previous in order to find the Eigenvalues I’d have to do row operations – lauren Feb 07 '20 at 19:51
  • You could find the eigenvalues of $A$ by row-reducing $A - \lambda I$ with $\lambda$ as a variable. You could find the eigenvector associated with an eigenvalue by row-reducing $A - \lambda I$, with an eigenvalue $\lambda$ plugged in, in order to find the eigenvectors. This same idea allows you to test wheter a suspected value is an eigenvalue. However, row-reducing $A$ directly doesn't lead you anywhere unless $0$ happens to be an eigenvalue. – Ben Grossmann Feb 07 '20 at 19:55
  • I think that the nicest approach for your problem, however, is to apply the ideas given in this post or the ideas from any of the posts linked therein. – Ben Grossmann Feb 07 '20 at 19:57
  • I think I’ve completed it however can you check my algebra is correct as idk if what I’ve done is allowed – lauren Feb 07 '20 at 23:18
  • @lauren I didn't see this until now; typically to make sure that someone sees your comment, you have to put an "@" with the name of who you're talking to (if you type the first letter, the site usually fills in the rest so you can click on it). In any case the work you've shown seems fine, but I'm not sure how you got it "after doing row-reduction" since I'm not sure what matrix you ended up with. – Ben Grossmann Feb 09 '20 at 21:23
  • @lauren according to some, you technically shouldn't have divided by $\lambda$ while row-reducing. – Ben Grossmann Feb 09 '20 at 21:25

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