Determinant of complex matrix with almost constant lines

Question

Let $0\neq c\in\mathbb{C}$. Take the matrix $$A_C=\begin{pmatrix} n&c&\dots&c&c \\ c&n&c &\dots & c\\ c &c & n &c &\dots\\ \vdots &\vdots&\vdots&\ddots & \vdots \\ c & \dots &\dots &c& n \end{pmatrix}\in \mathbb{C}^{n \times n}.$$

I want to show that $A_C$ invertible. It seems like one can compute the determinant explicitly but I messed with it a bit.

How can I prove $\det A_C\neq 0$?

EDIT: We assume $c\neq n$.

Answer 1

This is a circulant matrix and as such has normalized eigenvectors $$v_j=\frac{1}{\sqrt{n}}(1,\omega_j,\omega_j^2,\ldots,\omega_j^{n-1})$$ where $$\omega_j=exp\left(\frac{2\pi i j}{n}\right)$$ The eigenvalues are $$\lambda_j=n+c\omega_j+c\omega_j^2+\cdots+c\omega_j^{n-1}$$ Taking the product of the eigen values gives the determinant. Now since the $\omega_j$ are roots of unity we have $$\sum_{k=1}^{n-1}\omega_j^k=-1$$ which gives $\lambda_j=n-c$. Since we have assumed that $n\neq c$ no eigenvalue is equal to zero. Hence the determinant is non-zero.

Answer 2

See if its so take it in eigenvalues. We know that sum of eigenvalues is the trace of matrix as $n \neq c $ is not equal to $0$ so we can easily conclude that $det(A) \neq 0$ you can also do that as $det(A)=π \lambda_i$ and as $n\neq 0$ so the determinant is also non-zero and its done.