Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words.
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}n_{22} + n_{11} + n_{22} + 1 - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{11} + n_{22} + 1 = n_{11} + n_{22} + 1$
Which shows that it's true for a 2*2 matrix.
Looking at other 2*2 matrices
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = n_{11}n_{22} + n_{22} - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{22} = n_{22}$
Similarly:
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}$
$\begin{vmatrix}
n_{11} & n_{12} + 1 \\
n_{21} & n_{22}
\end{vmatrix} = -n_{21}$
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} + 1 & n_{22}
\end{vmatrix} = -n_{12}$
Positive if on the main diagonal, negative if on the side diagonal.
Using that, we can show that
$\begin{vmatrix}
n_{11} + 1 & n_{12} & n_{13} \\
n_{21} & n_{22} + 1 & n_{23} \\
n_{31} & n_{32} & n_{33} + 1
\end{vmatrix} = (n_{11} + 1)(n_{22} + n_{33} + 1) - n_{12}n_{21} - n_{13}n_{31} = n_{11}n_{22} - n_{12}n_{21} + n_{11}n_{33} - n_{13}n_{31} + n_{11} + n_{22} + n_{33} + 1 = n_{11} + n_{22} + n_{33} + 1$
And in a similar fashion we can apply this to higher order matrices.
But I haven't the slightest clue how to word or show this "similar fashion".

Since $\texttt{rank}(A) = 1$ $A = u v^T$ for some $n \times 1$ matrices $u$ and $v$ and consequently $\texttt{trace}(A) = v^Tu.$
We have $$\begin{pmatrix} I & 0 \\ v^T & 1 \end{pmatrix} \begin{pmatrix}I & u \\ -v^T & 1 \end{pmatrix}  = \begin{pmatrix} I & u \\ 0 & 1 + v^Tu \end{pmatrix}. \tag{1}$$
And we also have $$\begin{pmatrix} I & -u \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I & u \\ -v^T & 1 \end{pmatrix} = \begin{pmatrix} I + uv^T & 0 \\ -v^T & 1 \end{pmatrix}. \tag{2}$$
Taking determinant of $(1)$ and $(2)$ the result follows.
