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Show that the determinant of

$$M=\begin{pmatrix}0&1&1&1&\cdots&1\\1&0&1&1&\cdots&1\\1&1&0&1&\cdots&1\\1&1&1&0&\cdots&1\\\vdots&\vdots&\vdots&\vdots&&\vdots\\1&1&1&1&\cdots&0\end{pmatrix}$$

is nonzero.

Solution I was provided: If the above matrix is $n\times n$ then

$$\begin{pmatrix}0&1&1&\cdots&1\\1&0&1&\cdots&1\\1&1&0&\cdots&1\\\vdots&\vdots&\vdots&&\vdots\\1&1&1&\cdots&0\end{pmatrix}\begin{pmatrix}n-2&-1&-1&\cdots&-1\\-1&n-2&-1&\cdots&-1\\-1&-1&n-2&\cdots&-1\\\vdots&\vdots&\vdots&&\vdots\\-1&-1&-1&\cdots&n-2\end{pmatrix}=\begin{pmatrix}-(n-1)&0&0&\cdots&0\\0&-(n-1)&0&\cdots&0\\0&0&-(n-1)&\cdots&0\\\vdots&\vdots&\vdots&&\vdots\\0&0&0&\cdots&-(n-1)\end{pmatrix}$$

Clearly the determinant of the right hand matrix is just the product of the diagonal entries, which is nonzero $(n$ is assumed to be at least 3). Since for any square matrices $A$ and $B,\det(AB)=$ $\det(A)\det(B)$, we must have that the determinant of both matrices on the left hand side are nonzero as well.

Can someone explain how does the solution work, I don't get why we multiply M by this specific matrix? And why does it only work for n greater than 3, not 2?

RobPratt
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Jarek
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    What makes you say that it doesn't work for $n = 2$? (It should be clear why it doesn't work for $n=1$, esp if you write it out.) – Calvin Lin Nov 25 '24 at 19:14
  • But why does it say "n is assumed to be at least 3". Also why the choice of this matrix in particular, if the goal is to get a diagonal matrix? Can it be done with a different one or is this uinque? I don't think this proof is rigorous though – Jarek Nov 25 '24 at 19:16
  • 1/ Maybe because that's what the original problem says? EG The problem could have asked us to prove this for (say) $n = 100$. 2/ Clearly if $ AB = D$ diagonal matrix, then $ A(kB) = kD$ still a diagonal matrix for any real $k$. Note that if such a matrix exists, then $B = k' A^{-1}$ is uniquely determined up to a scalar. 3/ The goal (of this approach) isn't necessarily to get to a diagonal matrix, but simply to get to a matrix where we can show that the determinant is non-zero. Then, $ \det(A)\det(B) = \det(D)$, so if $\det(D) \neq 0$, then $ \det(A) \neq 0$. – Calvin Lin Nov 25 '24 at 19:19
  • Observe that $(M+I)^2=n(M+I)$ and deduce that $M$ is inversible, so its determinant is nonzero. – hamam_Abdallah Nov 25 '24 at 20:06
  • A similar question has been asked some days ago here. In fact, up to a complete reversal of the columns (which is an invertible operation), it is the same problem... – Jean Marie Nov 25 '24 at 21:35

1 Answers1

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The matrix is of the form $M=N-I,$ where $N$ is the matrix with all terms equal $1.$ The eigenvalues of $N$ are equal $n$ and $0.$ The eigenspace corresponding to $n$ is one dimensional and contains $(1,\ldots,1)^t.$ The eigenspace corresponding to $0$ has dimension $n-1$ and consists of vectors whose sum of the terms is equal $0.$ Therefore the eigenvalues of $M=N-I$ are equal $n-1$ and $-1,$ with eigenspaces described above. Hence $M$ is invertible.

The determinant is equal to the product of the eigenvalues. Thus $\det M=(-1)^{n-1}(n-1).$

Remark The determinant can be calculated directly. First add all rows $2,\ldots,n$ to the first row. In the new matrix subtract the first column from all other columns. The resulting matrix is lower triangular with $n-1$ at the upper left corner and with $-1$ on remaining part of the diagonal.

Ryszard Szwarc
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  • More generally, if $f(x)$ is a polynomial with eigenvalues $\lambda_i$, then the eigenvalues of $f(M)$ are $f(\lambda_i)$. You are establishing this result for $f(x)=x-1$. – Robert Shore Nov 25 '24 at 21:43
  • @RobertShore I am not establishing this, nearly obvious, result, but applying it. – Ryszard Szwarc Nov 26 '24 at 01:07