How to show this equality? This is a equation from a book of control theory:
$$c(sI-A)^{-1}b=\frac{\det(sI-A)-\det(sI-A-bc)}{\det(sI-A)}$$
$I$ is identity matrix, $A$ is $n\times n$, $b$ is $n \times 1$ vector, $c$ is $1 \times n$ vector.
I was trying to use induction, but seems not work. I would very appreciate some advice.
 A: Note that $bc$ is a rank-1-matrix. We will find a formula for $(B - bc)^{-1}$, where $B$ is any $n\times n$-matrix ($sI - A$ in your case). We will prove 
$$ \det (B - bc) = (1 - cB^{-1}b)\det(B) $$
To prove that, note 
\begin{align*}
   \det(B- bc) &= \det B\det(I - B^{-1}bc)\\
   &= \det B \det\begin{pmatrix} I- B^{-1}bc & -B^{-1}b\\ 0  & 1 \end{pmatrix}\\
   &= \det B \det \begin{pmatrix} I & 0\\ c & 1\end{pmatrix}
 \det\begin{pmatrix} I- B^{-1}bc & -B^{-1}b\\ 0  & 1 \end{pmatrix}\begin{pmatrix} I & 0\\ -c & 1\end{pmatrix}\\
   &= \det B \cdot \det\begin{pmatrix} I & -B^{-1}b\\ 0 & 1 - cB^{-1}b\end{pmatrix}\\
   &= \det B \cdot (1 - cB^{-1}b) 
\end{align*}
Hence 
$$ -\det(sI - A - bc) = -\det(sI - A) \cdot \bigl(1 - c(sI-A)^{-1}b\bigr) $$
Now solve for $c(sI- A)^{-1}b$.
A: It's equivalent to an interesting identity for $n$-dimensional vectors $u,v$:

$$\det(I+uv^t)=1+v^tu$$  

The identity can be proved, of course, via direrctly computing the determinant, but a more elegant way is to consider the eigenvalues of $uv^t$. Suppose $u,v\not=0$, otherwise the identity is trivial. Note that $uv^t$, when considered as a linear transform, maps $\mathbb{R}^n$(or some $n$-dimensional vector space, whatever) into the subspace spanned by $u$, since $v^tw$ is a scalar and thus $uv^t(w)=u(v^tw)=(v^tw)u$. So the matrix is of rank $1$. Now it's easy to see that $uv^t$ has $n-1$ eigenvalues $0$(since the rank is $1$), and one eigenvalue $v^tu\not=0$(since $uv^tu=(v^tu)u$; in case that $v^tu=0$ we perturb $u$ a little by considering $u'=u+\epsilon v$ and letting $\epsilon\to 0$ to make the discussion consistent). Thus we have shown
$$\det(I+uv^t)=\prod_{i=1}^n(\lambda_i+1)=1+v^tu$$

Now I briefly explain the equivalence of this identity and your question. On the one hand, taking $A=(s-1)I$ and $b=-u,c=v^t$ yields the identity. On the other hand, take $u=(sI-A)^{-1}b$ and $v=-c^t$ to see
$$1+v^tu=1-c(sI-A)^{-1}b$$
\begin{align}
\det(I+uv^t)&=\det\bigg((sI-A)^{-1}(sI-A)-(sI-A)^{-1}bc\bigg)\\
&=\det(sI-A)^{-1}\det(sI-A-bc)=\frac{\det(sI-A-bc)}{\det(sI-A)}
\end{align}
and hence the result.
