6
$\begingroup$

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 :

$$\begin{bmatrix}n&n-1&n-2&\cdots&2&1\\1&n&n-1&\cdots&3&2\\1&1&n&\cdots&4&3\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\1&1&1&\cdots&n&n-1\\1&1&1&\cdots&1&\lambda\end{bmatrix}$$

$\endgroup$
  • 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
6
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.