TITLE: About Sylvester's determinant
QUESTION [8 upvotes]: 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?)

REPLY [8 votes]: The common ground of those two formulas is related to the Woodbury matrix identity. This relation is a useful statement that shows what happens to the inverse when one "updates" a matrix $A\in\mathbb{C}^{n\times n}$ with a term $UCV$, with $U,V\in\mathbb{C}^{n\times m}$, $C\in\mathbb{C}^{m\times m}$. The typical situation is when $m<n$ and this update is a small-rank term. The rank-1 version, often useful in its own, is called Sherman-Morrison formula. 
A statement on determinants can be derived with a small modification to the proof: take Equations (1) and (2) here; compute determinants of the first and third member of those chains of equalities, and equate them:
$$
\det A^{-1}\det(C-VA^{-1}U)^{-1} = \det\begin{bmatrix}A & U\\V&C\end{bmatrix}^{-1} = \det(A-UC^{-1}V)^{-1}\det C^{-1};
$$
invert everything and you get
$$
\det A \det(C-VA^{-1}U) = \det(A-UC^{-1}V)\det C, \tag{*}
$$
which by continuity holds also when $\begin{bmatrix}A & U\\V&C\end{bmatrix}$ isn't invertible. Your two formulas can be derived easily from this one ($C=I$ in both cases).
Equation (*) is also known as (the two versions of) the Schur complement formula; an equivalent statement appears for instance here (last equation of the section titled Background).
People in applied linear algebra often prefer talking about inverses than determinants, but it's essentially the same result.