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I have a question where I am asked suppose we have an $(N \times N)$ matrix with one's on the diagonal, and constant, $c$ everywhere else. What would you expect the eigenvalues to be ?

I tried calculating it using a $(2\times 2)$ matrix, but I then got stuck for $(3\times 3)$ and so forth. Is there some sort of trick that can be used here ?

Hunter Batley
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PutsandCalls
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    The matrix of all ones can be written as $e e^T$ where $e$ is a vector of ones and it is straightforward to compute the eigenvalues. Multiply by $c$ and add an appropriate multiply of $I$ to get the matrix above. What would that do to the eigenvalues? – copper.hat Apr 15 '20 at 00:42

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Hint Let $A$ be this matrix. Then $$A-(1-c)I$$ is the matrix with $c$ in all entries. This matrix has rank 1.

Therefore $(1-c)$ is an eigenvalue of $A$ of geometric multiplicity $N-1$. It follows that $1-c$ is an eigenvalue of $A$ of algebraic multiplicity at least $N-1$.

Therefore, $N-1$ eigenvalue are equal to $1-c$. Use the fact that $\tr(A)$ is the sum of all eigenvalues to find the last eigenvalue.

N. S.
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  • wait sorry, because $A - (1-c)I$ is a rank 1 matrix, how were you able to conclude that $(1-c)$ is an eigenvalue of $A$ with multiplicity $N-1$ ? Would you be able to refer me to this theorem or help explain why ? – PutsandCalls Apr 15 '20 at 00:51
  • @PutsandCalls: What dimension would the kernel of a rank one operator be? – copper.hat Apr 15 '20 at 01:03
  • @PutsandCalls The rank nullity theorem says that the dimension of the eigenspace plus the rank of $(A-\lambda I)$ equals $N$. – N. S. Apr 15 '20 at 01:48
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It is well known that all one matrix ($n\times n$) given as $J$ has $n-1$ zero eigenvalues and $n$ as single positive eigenvalue. Multiplying by a scalar $cJ$ has $n-1$ zeros and eigenvalue $cn$ Adding: $cJ+(1-c)I$ you add to each eigenvalue of $cJ$, $1-c$ to get all of them.

Toni Mhax
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