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Let $M$ denote the real $6×6$ matrix all of whose off-diagonal entries are $−1$ and all of whose diagonal entries are $5$. List out the eigenvalues of $M$ (each eigenvalue must be written as many times as its multiplicity):

Attempt:- I could prove that determinant($M$)=$0$. So, It has $0$ as an eigenvalue. Also, we know that the sum of all eigenvalues equals $30$. How do I find other eigenvalues? Solving $\det(M-tI)=0$ is very lengthy. Is there any shortcut?

Truth_searcher
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2 Answers2

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Hints: First, try to find the eigenvalues of $M - 6I$, which is a $6 \times 6$ matrix where all the entries are $-1$. This matrix has rank $1$, so you know that $5$ of the eigenvalues of $M - 6I$ are zero. The other one should be easy to find.

If $v$ is an eigenvector of $M-6I$ with eigenvalue $\lambda$, then $v$ is also an eigenvector of $M$ with eigenvalue $\lambda+6$. Do you see why?

JimmyK4542
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Define $A=M-6I$, observe that $A$ is the matrix with all of its entries as $-1$. Therefore, $A$ has rank $1$. To know anything about the eigenvalues of $M$ we need to get information about $\det(M-\lambda I)$. $A$ is easy to observe, $$\det (A-\beta I) = (\beta-0)^5(\beta+6)^1$$ As,

  • $\det(A-0I) = \det(A) = 0 $ so $0$ with multiplicity $5$ ($A$ has rank $1$. It is real symmetric, therefore # non-zero eigenvalues = rank) ref
  • $\det(A+6I) = \det(M) = 0$ so $-6$ with multiplicity $1$

Now we can say $$ \det (M-\lambda I ) = \det (A-(\lambda -6 )I) = (\lambda -6 )^5(\lambda -0)^1 $$

sougata
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