Show that $f(t) = \det(A + tB)$ is a line 
Suppose $A,B \in \mathbb{R}^{n \times n}$ where rank$(B) = 1$. Show that the function $f: \mathbb{R} \to \mathbb{R}$ where $$f(t) = \det(A + tB)$$ is a line, i.e., has the form $f = mt + b$.


So far, I really only have that $f(0) = \det(A) = b$. I thought about explicitly calculating the derivative as in,
$$ \lim_{t\to0} \frac{\det(A + tB) - b}{t}$$
but I'm not sure where I would go next. Also, I am not sure how to use that rank$(B) = 1$. Any help would be appreciated.
 A: Theorem: (Matrix Determinant Lemma)
$$
\det \left(\mathbf {A} +\mathbf {uv} ^{\textsf {T}}\right)
=\det \left(\mathbf {A} \right)+\mathbf {v} ^{\textsf {T}}\mathrm {adj} \left(\mathbf {A} \right)\mathbf {u}
\underset{\text{if $\exists A^{-1}$}}{=}\left(1+\mathbf {v} ^{\textsf {T}}\mathbf {A} ^{-1}\mathbf {u} \right)\,\det \left(\mathbf {A} \right)$$
Thus, since $\operatorname{rank}(B)=1\implies\exists u,v : B=uv^{\textsf {T}}$, we have
$$
f(t) 
= \det(\mathbf {A} + t\mathbf {B})
=\det \left(\mathbf {A} +t\mathbf {uv} ^{\textsf {T}}\right)
=\det \left(\mathbf {A} \right)+t\cdot\mathbf {v} ^{\textsf {T}}\mathrm {adj} \left(\mathbf {A} \right)\mathbf {u}
= a+bt
$$
A: You can find $S\in Gl_n$ such that $SB=C$ with only nonzero entries in the first row, i.e.
$$ C= 
\begin{pmatrix}
- 
& x^t & -\\
-&0&-\\
&\vdots &\\
-&0&-
\end{pmatrix}
$$
Then $\det(A+tB)=\det(S^{-1})\det(SA+tC)$ and you can use the row-wise linearity of the determinant, i.e. if
$$
SA= 
\begin{pmatrix}
- 
& y_1^t & -\\
-&y_2^t&-\\
&\vdots &\\
-&y_n^t&-
\end{pmatrix},
$$
then
$$
\det(SA+tC)=\det(SA)+t\det 
\begin{pmatrix}
- 
& x^t & -\\
-&y_2^t&-\\
&\vdots &\\
-&y_n^t&-
\end{pmatrix}
$$
A: Since $\mathrm{rank}(B)=1$, there exists $P,Q \in GL_n(\mathbb{R})$ such that
$$B=PJQ, \quad \text{where } J=\begin{pmatrix}
1 & 0 & \dots & \dots & \dots & 0  \\
0 & 0 & \ddots & & \dots  & 0 \\
\vdots&\ddots&\ddots&&\dots&\vdots\\
0 & 0 & \dots & \dots & \dots & 0
\end{pmatrix}$$
So
$$\det(A+tB) = \det(P^{-1}AQ^{-1}+tJ)\det(P)\det(Q)$$
Denoting by $(m_{ij})$ the coefficients of $P^{-1}AQ^{-1}$, you have
$$P^{-1}AQ^{-1}+tJ = \begin{pmatrix}
m_{11}+t & m_{12} & \dots & \dots & \dots & m_{1n}  \\
m_{21} & m_{22} & \ddots & & \dots  & m_{2n} \\
\vdots&\ddots&\ddots&&\dots&\vdots\\
m_{n1} & m_{n2} & \dots & \dots & \dots & m_{nn}
\end{pmatrix}$$
so its determinant is clearly a polynomial of degree $1$ in $t$, so finally $\det(A+tB)$ is also a polynomial of degree $1$ in $t$.
A: Generalizing, suppose that we have square matrices ${\bf A}, {\bf B} \in {\Bbb R}^{n \times n}$, where matrix $\bf B$ is rank-$r$ and $r \leq n$. Hence, there exist matrices ${\bf U}, {\bf V} \in {\Bbb R}^{n \times r}$ such that ${\bf B} = {\bf U} {\bf V}^\top$. Assuming that matrix $\bf A$ is invertible and using the generalized matrix determinant lemma,
$$f (t) := \det \left( {\bf A} + t \, {\bf B} \right) = \det \left( {\bf A} + t \, {\bf U} {\bf V}^\top\right) = \cdots = \det \left( {\bf A} \right) \underbrace{\det \left( {\bf I}_r + t \, {\bf V}^\top {\bf A}^{-1} {\bf U} \right)}_{=: g (t)}$$
If $\color{blue}{r = 1}$, then $g (t)$ is affine in $t$. What if $r > 1$?
