Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$

Assume $a\ne 0$, We know that

$$\det(A+uv^T)=(1+v^TA^{-1}u)\det(A)$$

Here $A=aI, u = be, v=e$

\begin{align}\det(aI+bee^T)&=(1+be^T(aI)^{-1}e)\det(aI) \\ &=\left(1+\frac{bu}a\right)a^u\\ &=a^{u}+bua^{u-1}\end{align}

If $a=0$ and $u>1$, then the determinant is $0$.

If $a=0$ and $u=1$, then the determinant is $b$.

Note that $G$ is a circulant matrix, so we know the eigenvectors are $(1,\zeta,\zeta^2,\dots,\zeta^{u-1})$ where $\zeta^u=1$. Its eigenvalue is $$ a+b\sum_{j=0}^{u-1} \zeta^j= \begin{cases} a & \text{if }\zeta\neq 1\\ a+bu & \text{if }\zeta=1 \end{cases} $$ and so the determinant is $a^{u-1}(a+bu)$.

Use eigenvalues!

The matrix has $a$ as eigenvalue and the corresponding eigenspace is the space that is orthogonal to $e$: If $v^Te = 0$, then $$ Gv = (aI + bee^T)v = av + bee^Tv = av. $$ Hence the eigenvalue $a$ has multiplicity $n-1$ (if the matrix is $n\times n$). The other eigenvalue is $a+nb$ and the eigenspace is one dimensional an spanned by $e$: $$ Ge = (aI + bee^T)e = ae + bee^Te = (a+nb)e. $$ Since the determinant is the product of eigenvalues, we get $$ \det(G) = a^{n-1}(a+nb). $$