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".
 A: 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.
A: The value of a  determinant as well of a trace is independent of the choice of basis. So suppose that the image of $A$ is generated by a vector $v_1$. Complement this vector with $v_2,...,v_n$ to form a base. In this base the matrix of $A$ takes the form:
  $$ \underline{A} = 
\left( \begin{matrix} a_{11} & a_{12} & ... & a_{1n}
\\  0 & 0 & ... & 0 \\
\\  . & . & ... & . \\
\\ 0 & 0 & ... & 0 
\end{matrix}
\right)
$$
and that of $E+A$:
$$ \underline{1+A} = 
\left( \begin{matrix} 1+a_{11} & a_{12} & ... & a_{1n}
\\  0 & 1 & ... & 0 \\
\\  . & . & ... & . \\
\\ 0 & 0 & ... & 1 
\end{matrix}
\right)
$$
Then clearly ${\rm tr} \ A=a_{11}\ $  and $\ \det (1+A)=1+a_{11}$.
In less abstract terms, as $A$ has rank 1, we may write  $A= u v^T$ where $u$ and $v$ are column vectors. Suppose that $e_1^T u\neq 0$. Then $u,e_2,...,e_n$ forms a basis. Carrying out the products one verifies:
 $$ (1+ A) 
\left[ \begin{matrix} u & e_2 & ... & e_n \end{matrix} \right] =
\left[ \begin{matrix} u & e_2 & ... & e_n \end{matrix} \right]
\left[ \begin{matrix} 1+v^T u  & v_2 & ... & v_n\\
  0 & 1 & ... & 0 \\
  . & . & ... & . \\
  0 & 0 & ... & 1 \end{matrix} \right]$$ 
so $1+A$ is conjugated to the matrix on the right which verifies the claimed identity. And determinant and trace is invariant under conjugation.
A: Suppose $\;A=(a_{ij})\;$ with rank$\,A=1\;$ and assume that the only row in $\;A\;$ which is linearly independent (= non all-zeros row) is the first one, so that the other ones are scalar multiples of this first one. Say the $\;i\,-$ th row is $\;k_i\;$ times the first one
$$|A+E|=\begin{vmatrix}1+a_{11}&a_{12}&\ldots&a_{1n}\\a_{21}&1+a_{22}&\ldots&a_{2n}\\\ldots&\ldots&\ldots&\ldots\\a_{n1}&a_{n_2}&\ldots&1+a_{nn}\end{vmatrix}=$$
We now apply elementary operations on the above which do not change the determinant: from row $\;i\;$ substract $\;k_i\;$ times the first row:
$$=\begin{vmatrix}1+a_{11}&a_{12}&a_{13}&\ldots&a_{1n}\\-k_2&1&0&\ldots&0\\-k_3&0&1&0\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\-k_n&0&0&\ldots&1\end{vmatrix}\stackrel{\text{develop 2nd row}}=k_2\begin{vmatrix}a_{12}&a_{13}&\ldots&a_{1(n-1)}&a_{1n}\\0&1&0&\ldots&0\\0&0&1&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\0&0&0&\ldots&1\end{vmatrix}+$$$${}$$
$$\begin{vmatrix}1+a_{11}&a_{13}&\ldots&a_{1(n-1)}&a_{1n}\\-k_3&1&0&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\-k_n&0&0&\ldots&1\end{vmatrix}\stackrel{\text{Inductively}}=$$$${}$$
$$=k_2a_{12}+k_3a_{13}+\ldots+k_{n-1}a_{1(n-1)}+k_na_{1n}+1+a_{11}=$$
$$=1+a_{11}+a_{22}+\ldots+a_{nn}=1+\text{Tr.}\,A$$
Try to fill in details. After some thought, if you still have doubts write back.
