$\det(\mathbf{1} - c \mathbf{J}_n)$, where $\mathbf{J}_n$ is an $n\times n$ matrix of ones

Question

Let $\mathbf{J}_n$ be an $n\times n$ matrix of ones, and let $c$ be a real number. Compute the following determinant:

$$\det(\mathbf{1} - c \mathbf{J}_n)$$

where $\mathbf{1}$ denotes the $n\times n$ identity matrix.

$n$ is a positive integer, of course.

hint: eigenvalues if you are allow to, otherwise induction over $n$. — user251257, Nov 23 '16 at 21:26
$1 - c J_n$ is symmetric, thus diagonalizable. The eigenvalues of $J_n$ are $0$ and $n$ (why? $0$ only if $n>1$). — user251257, Nov 23 '16 at 21:32

score 0 · Accepted Answer · answered Nov 23 '16 at 21:32

0

Hint: Show that $\det(id-cJ_n)=1-nc$ by induction. For $n=1$, $id-cJ_1=(1-c)$, so the determinant is $1-c$.

Another way is to see that $id-cJ_n$ is a symmetric matrix of a very special kind, where more general determinant formulas are known.

answered Nov 23 '16 at 21:32

Dietrich Burde

Can you expand on the known "more general determinant formulas"? – a06e Nov 23 '16 at 21:35
See for (symmetric) Toeplitz matrices and its determinants. – Dietrich Burde Nov 23 '16 at 21:36

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