I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?

This is the matrix :


  • 1
    $\begingroup$ It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question. $\endgroup$ – Sigur Dec 27 '18 at 13:37
  • $\begingroup$ Do you have a particular matrix in mind, or just an arbitrary $n \times n$ matrix? $\endgroup$ – Clive Newstead Dec 27 '18 at 13:37
  • $\begingroup$ I am sorry, the link to the picture was not included. I added it now. $\endgroup$ – Paul Vinur Dec 27 '18 at 13:41
  • $\begingroup$ Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$? $\endgroup$ – Test123 Dec 27 '18 at 14:08

Let $M_n$ be your matrix.

Let $\eta_n$ be the $n\times n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you

  1. Subtract row $k+1$ from row $k$ for $k = 1,2,\ldots,n-1$.
    This is equivalent to multiply $M_n$ by $I_n - \eta_n$ from the left

  2. Subtract column $k-1$ from column $k$ for $k = n,n-1,\ldots,2$ (notice the order of $k$).
    This is equivalent to multiply $(I_n-\eta_n)M_n$ by $I_n - \eta_n$ from the right.

After you do this, your matrix simplifies to $$(I_n - \eta_n) M_n (I_n - \eta_n) = \begin{bmatrix} n-1&-n&0&\cdots&0&0&0\\ 0&n-1&-n&\cdots&0&0&0\\ 0&0&n-1&\ddots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&n-1&-n&0\\ 0&0&0&\cdots&0&n-1&-\lambda\\ 1&0&0&\cdots&0&0&\lambda-1 \end{bmatrix}$$

From this, you can deduce

$$\det[M_n] = \det[(I_n - \eta_n)M_n(I_n - \eta_n)] = (n-1)^{n-1}(\lambda-1) + n^{n-2}\lambda$$

  • $\begingroup$ This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ) $\endgroup$ – Paul Vinur Dec 27 '18 at 17:07
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    $\begingroup$ @PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i \text{ or } i+1 \pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $\prod_{i} a_{ii}$ and $\prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms. $\endgroup$ – achille hui Dec 27 '18 at 17:12

This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.

Then $\det(A) = \det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.

This is defined if $\lambda \ne 1$. For $\lambda = 1$, take the limit.


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