TITLE: Determinant of a rank $1$ update of a scalar matrix, or characteristic polynomial of a rank $1$ matrix
QUESTION [67 upvotes]: This question aims to create an "abstract duplicate"  of numerous questions that ask about determinants of specific matrices (I may have missed a few):

Characteristic polynomial of a matrix of $1$'s
Eigenvalues of the rank one matrix $uv^T$
Calculating $\det(A+I)$ for matrix $A$ defined by products
How to calculate the determinant of all-ones matrix minus the identity?
Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)
Determinant of a special $n\times n$ matrix
Find the eigenvalues of a matrix with ones in the diagonal, and all the other elements equal
Determinant of a matrix with $t$ in all off-diagonal entries.
Characteristic polynomial - using rank?
Caclulate $X_A(x) $ and $m_A(x) $ of a matrix $A\in \mathbb{C}^{n\times n}:a_{ij}=i\cdot j$
Determinant of rank-one perturbations of (invertible) matrices

The general question of this type is

Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?

A clearly very closely related question is

What is the characteristic polynomial of a matrix $A$ of rank$~1$?

REPLY [2 votes]: Here’s an answer without using eigenvalues: the rank of $A$ is $1$ so its image is spanned by some nonzero vector $v$. Let $\mu$ be such that
$$Av=\mu v.$$
We can extend this vector $v$ to a basis of $\mathbb{C}^n$. With respect to this basis now, we have that the matrix of $A$ has all rows except the first one equal to $0$. Since determinant and trace are basis-independent it follows by expanding the first column of $A$ wrt to this basis that
$$\det(A-\lambda I)= (-1)^n(\lambda -\mu)\lambda^{n-1}.$$
Using this same basis as above we also see that $\text{Tr}(A) =\mu$, so the characteristic polynomial of $A$ turns out to be
$$(-1)^n(\lambda -\text{Tr}(A))\lambda^{n-1}.$$