Another proof for Sherman Morrison Formula? The proof of Sherman Morrison Formula is on wikipedia as well as this question Proof of the Sherman-Morrison Formula. 
Isn't there a proof which does not uses multiplication of the inverse and the matrix?
I mean, it definitely arises from some equalities that wind up to this. 
$$(A + \mathbf{u}\mathbf{v}^T)^{-1}=A^{-1} - \frac{A^{-1}\mathbf{u} \mathbf{v}^T A^{-1}}{(1+\mathbf{v}^TA^{-1}\mathbf{u})}$$
 A: Here's a proof adapted from wikipedia's proof for the (more general) Woodbury matrix identity.
We would like to find a matrix $X$ such that
$$
(A + uv^T)X = I \implies
AX + uv^TX = I
$$
Now, if we define $Y = (v^TX)$, then we can rewrite this as a system of equations:
$$
A X + uY = I\\
v^TX - Y = 0
$$
That is,
$$
\pmatrix{A & u\\v^T&-1} \pmatrix{X\\Y} = \pmatrix{I\\0}
$$
We can solve this system using an augmented matrix and block-matrix operations.  In particular, we have
$$
\left[
\begin{array}{cc|c}
A & u & I\\
v^T & -1&0
\end{array}
\right] \to 
\left[
\begin{array}{cc|c}
I & A^{-1}u & A^{-1}\\
v^T & -1&0
\end{array}
\right] \to 
\left[
\begin{array}{cc|c}
I & A^{-1}u & A^{-1}\\
0 & -1 - v^TA^{-1}u & -v^TA^{-1}
\end{array}
\right] \to\\
\left[\begin{array}{cc|c}
I & A^{-1}u & A^{-1}\\
0 & 1 & \frac{1}{1 + v^TA^{-1}u}v^TA^{-1}
\end{array}
\right] \implies
\begin{cases}
X + A^{-1}uY = A^{-1}\\
Y = \frac{1}{1 + v^TA^{-1}u}v^TA^{-1}
\end{cases}
$$
All that remains is substitution.  That is, we have
$$
X = A^{-1} - A^{-1}uY = A^{-1} - A^{-1}u\left( \frac{1}{1 + v^TA^{-1}u}v^TA^{-1}\right) = 
A^{-1} - \frac{A^{-1}uv^TA^{-1}}{1 + v^TA^{-1}u}
$$
A: Write $A+uv^T=A(I+A^{-1}uv^T)$; we are to find an inverse of $I+A^{-1}uv^T$. It's a bit simpler if we set $w=-u$, so instead we look for an inverse of $I-A^{-1}wv^T$; the idea that comes to mind is to consider, formally,
$$
(I-A^{-1}wv^T)^{-1}=I+A^{-1}wv^T+(A^{-1}wv^T)^2+(A^{-1}wv^T)^3+\dotsb \tag{*}
$$
taking from $\frac{1}{1-x}=1+x+x^2+\dotsb$
Now
$$
(A^{-1}wv^T)^2=A^{-1}wv^TA^{-1}wv^T=(v^TA^{-1}w)A^{-1}wv^T
$$
and
$$
(A^{-1}wv^T)^3=
A^{-1}wv^TA^{-1}wv^TA^{-1}wv^TA^{-1}wv^T=
(v^TA^{-1}w)^2A^{-1}wv^T
$$
and, by induction,
$$
(A^{-1}wv^T)^n=(v^TA^{-1}w)^{n-1}A^{-1}wv^T
$$
so the formal sum (*) becomes
$$
I+A^{-1}wv^T+(v^TA^{-1}w)A^{-1}wv^T+(v^TA^{-1}w)^2A^{-1}wv^T+(v^TA^{-1}w)^3A^{-1}wv^T+\dotsb
$$
and therefore
$$
I+\biggl(\,\sum_{n\ge0}(v^TA^{-1}w)^n\biggr)A^{-1}wv^T
$$
The term in parentheses is the inverse of $1-v^TA^{-1}w$. Returning to $u$, we find that the inverse should be
$$
I-\frac{1}{1+v^TA^{-1}u}A^{-1}uv^T
$$
Multiplying on the right by $A^{-1}$ we see that the inverse of $A+uv^T$ should be
$$
A^{-1}-\frac{1}{1+v^TA^{-1}u}A^{-1}uv^TA^{-1}
$$
Now we can do the multiplication and verify that the intuition is correct.
A: Let $w=-A^{-1}u$. Then the problem boils down to proving the equivalent identity that
$$
(I-wv^T)^{-1} = I+\frac{wv^T}{1-v^Tw}.\tag{1}
$$
Let us abuse the symbol $v$ and denote by $v(\cdot)$ the linear functional $x\mapsto v^Tx$. Then $I-wv^T$ is a matrix representation of the linear function
$$
y = f(x) = x - v(x)w.
$$
The inverse of this mapping is clearly
$$
x = f^{-1}(y) = y+v(x)w\tag{2}
$$
but we wish to express $v(x)$ in terms of $y$. Now, since $v$ is a linear functional,
$$
v(y)=v\left(x-v(x)w\right)=v(x)-v(x)v(w).
$$
Therefore $v(x)=\frac{v(y)}{1-v(w)}$ and $(2)$ gives
$$
f^{-1}(y) = y+\frac{v(y)w}{1-v(w)}
$$
and $(1)$ follows immediately.
One may argue that the above proof is not what you want because it "uses multiplication of the inverse and the matrix" implicitly, but I think it is worthwhile to prove the identity from an alternative perspective.
