# Determinant of an n x n matrix

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 :

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

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

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

• 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) – Paul Vinur Dec 27 '18 at 17:07
• @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. – 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.