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$A=\begin{pmatrix}6&1&1&1\\ 2&7&2&2\\ 3&3&8&3\\ 4&4&4&9\end{pmatrix}$

Find all the eigenvalues of this matrix, without computing it's characteristic polynomial.

Since all the entries of each column add up to 15, one eigenvalue is $\lambda = 15$. The trace gives $tr(A) = 30$ thus the sum of the other eigenvalues is $15$. and their product is $125$ so my guess is that the other eigenvalues are $\lambda = 5,5,5$, which when calculated turns out to be correct, however. This was just a lucky guess, is there a way to tell for sure?

Question 2 Does this matrix belong to a specific type?

Question 3 Are there more matrices who's eigenvalues can be found without calculating $p(\lambda)$? (For the exception of singular matrices).

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Hint: Subtract $5I$. (more stuff to get to 30 characters)

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  • $\begingroup$ Ah I see, so you saw that $5I$ is an eigenvector. However can you tell me does this matrice belong to a specific type and are there other types who's eigenvalues can be guessed? $\endgroup$
    – gvidoje
    Jul 29 '16 at 15:20
  • $\begingroup$ I don't know of such types. I just noticed that this particular matrix looked suspicious when I ignored the diagonal. I suppose you could make up other trick examples that yield to the same observation. $\endgroup$ Jul 29 '16 at 15:24

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