* If $A$ is any $n \times m$ matrix and $B$ is any $m \times n$ matrix then one familiar form of the Sylvester's identity is $\det(I + AB) = \det(I + BA)$. 


Now somehow curiously this above identity is often enough used along side another statement which says that for any vector $v$ and any invertible square matrix $A$, it is true that $\det(A + v v^T) = \det(A)\det(I + v^TA^{-1}v)$.


Is there some relationship between these two identities? Like can one be gotten from the other or vice versa?
* A particularly useful (at least in some recent big breakthrough researches!) form of the Sylvester's identity is when $A= tu, B = v^\dagger $ where ($u$ and $v$ are complex vectors and $t$ is some complex number). This then shows that $\det(I + t u v^\dagger) = 1 + tv^\dagger u$


Now this simple statement above apparently implies this more powerful formulation which is not clear to me - If $C$ is any $n\times n$ (necessarily invertible?) matrix and $A$ is a rank-$1$ matrix then is $\det( I + tCA)$ is a degree $1$ polynomial in $t$? Why? (and this is the same as saying that this is "affine-linear" in t?) 



About Sylvester's determinant
For the first point, note that by Sylvester's identity


$$\det(I\_n + v^TA^{-1}v) = \det(I\_n + (v^T)(A^{-1}v)) = \det(I\_n+(A^{-1}v)(v^T)) = \det(I\_n + A^{-1}vv^T),$$


so 


$$\det(A)\det(I\_n+v^TA^{-1}v) = \det(A)\det(I\_n + A^{-1}vv^T) = \det(A + vv^T).$$


Given $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$, their *outer product* is the $m\times n$ matrix $uv^T$. If $u$ and $v$ are non-zero, then their outer product has rank one. Conversely, a rank one $m\times n$ matrix can be written as the outer product of some non-zero $u$ and $v$.


Now suppose $C$ is an $n\times n$ matrix (not necessarily invertible) and $A$ is a rank one $n\times n$ matrix. By the above discussion, there are $u, v \in \mathbb{R}^n$ such that $A = uv^T$. So


\begin{align\*}
\det(I\_n + tCA) &= \det(I\_n + tCuv^T)\\ 
&= \det(I\_n + (tCu)(v^T))\\ 
&= \det(I\_1 + (v^T)(tCu))\\ 
&= \det(I\_1 + tv^TCu)\\ 
&= 1 + tv^TCu
\end{align\*}


where the last equality follows because $I\_1 + tv^TCu$ is a $1\times 1$ matrix.


A function of the form $f(t) = at + b$ is called *affine*. I'm guessing this is what is meant by affine-linear.


**Added Later:** Let me add a theorem which further demonstrates the relationship between the rank of a matrix and outer products. In the following statement, I am taking the definition of the rank of a matrix to be the dimension of its column space.



> 
> **Theorem:** An $m\times n$ matrix $A$ has rank $k$ if and only if the minimal number of outer products needed to express $A$ as a sum is $k$ (i.e. $A$ can be written as the sum of $k$ outer products, but not $k - 1$).
> 
> 
> 


Proving this is a really nice exercise in elementary linear algebra. I would hate to rob you of the experience by posting the solution here.


