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Let $M$ be a real $n\times n$ matrix with all diagonal entries equal to $r$ and all non-diagonal entries equal to $s$. Compute the determinant of $M$. This was a question from ISI M.Math test $2021$.

My attempt: I tried directly computing for small $n$ to see if there are any patterns. For $n=1,2,3$ the determinants are respectively $r, r^2-s^2, r^3+2s^3-3rs^2$. I couldn't find any pattern other than the obvious fact that there has to an entry $r^n$ for each $n$ so that the identity matrix satisfies the formula. And the sum of the coefficients must be zero for $n>1$ (by having a matrix with all entries equal to $1$). But I can't proceed any further.

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    See Andrea's answer in https://math.stackexchange.com/questions/84206/how-to-calculate-the-determinant-of-all-ones-matrix-minus-the-identity – leslie townes Aug 15 '23 at 05:23
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    $M=(r-s)I_n+see^T$ where $e$ denotes the vector of ones. It follows that you can determine the eigenvalues of $M$ if you can determine the eigenvalues of $A=ee^T$. Since $A$ is a rank-one matrix, it has $n-1$ zero eigenvalues and the remaining eigenvalue is equal to the trace of $A$. – user1551 Aug 15 '23 at 05:25
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    The decomposition just given by @user1551 is a "Rank-one update of a scalar matrix". A Google search with this type of query will bring you many answers such as this one. – Jean Marie Aug 15 '23 at 05:29
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    https://en.wikipedia.org/wiki/Circulant_matrix#Determinant – Anne Bauval Aug 15 '23 at 06:27
  • Thank you everyone –  Aug 15 '23 at 06:41

1 Answers1

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Let $M$ be a real $n\times n$ matrix with all diagonal entries equal to $r$ and all non-diagonal entries equal to $s$. Compute the determinant of $M$.

I would rewrite $M=(r-s)I+A$ where $A$ is a matrix whose entries are identically $s$.

Now, it is easy to check that eigenvalues of $A$ are

  • $0$ (by linear dependence of rows) and
  • $ns$ (by constant row sum)

with algebraic multiplicities $n-1$ and $1$ respectively.

The eigenvalues of $(r-s)I+A$ will be $(r-s)+0$ and $(r-s)+ns$ with same multiplicities.

Determinant being product of eigenvalues, we have $$\det(M)=(r-s)^{n-1}(r-s+ns)$$

The question in this post is a special case.

Nothing special
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