determinant of the symmetric matrix $8\times8$

Question

How to compute the determinant of the following matrix:

$ \left( \begin{array}{cccccccc} 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 \end{array} \right)$

Answer 1

you can write your matrix as $$I + uu^T, \det(I + uu^T) = n+1.$$ we have $u = (1,1,1, \ldots, 1)$ and the eigenvalues of $uu^T$ are $n, 0, 0, \ldots , 0.$ that means the eigenvalues of $I + uu^T$ are $n+1, 1, 1, \cdots, 1.$ so the determinant is $n+1.$

Answer 2

Write this matrix as a linear combination of two matrices:

$Q$ is the $8\times 8$ matrix in which every entry is $1/8$. Show that the rank of $Q$ is $1$.

$P=I-Q$, where $I$ is the $8\times 8$ identity matrix. Show that the rank of $P$ is $7$.

Show that $Q$ is the matrix of the orthogonal projection onto the space $A=\{ x(1,1,1,1,1,1,1,1)^\top : x\in\mathbb R\}$ and $P$ is the matrix of the complementary orthogonal projection onto the space $B=\{(x_1,\ldots,x_8)^\top : x_1+\cdots+x_8=0\}$.

It follows that there is an orthonormal basis in which $P$ and $Q$ are both diagonal matrices with no scalars on the diagonal except $0$ and $1$: One member of that basis spans the $1$-dimensional subspace $A$ above; the other seven space the $7$-dimensional orthogonal complement $B$.

It follows that $\det (aP+bQ) = a^7 b$.

You just need to find $a$ and $b$, and that's the easiest part of all this.

Answer 3

There is another trick available here which works for any such $n \times n$ matrix). Call your matrix $A$. For $1 \leq i \leq n-1,$ let $v_{i}$ be the columnn vector with $1$ in entry $i, -1$ in entry $i+1$, and $0$ elsewhere. Then $Av_{i} = v_{i}$ for each $i$, and $\{v_{i}: 1 \leq i \leq n-1 \}$ is linearly independent. Hence $A$ has $n-1$ linearly independent eigenvectors with eigenvalue $1$, so $1$ occurs as an eigenvalue of $A$ with algebraic multiplicity at least $n-1.$ Since trace($A$) $=2n,$ the other eigenvalue of $A$ must be $n+1,$ and det($A$)$=n+1.$