Proof of Cauchy's determinant identity I am reading Matrix Analysis 2nd. Page 66 gives the proof of Caucuh's determinant identity as follow:

where $(1.3.23)$ is as follow:

I am very confused about these questions:

*

*How does second identity make sense? What does $\lambda_i$ stand for?

*The $\lambda_i$ should be mulitplied into $I$ in the third identity. But it did not happen. Why?

Could anyone help me out? Thanks in advance! I am very very confused...
 A: *

*if $B$, $C$ are $n\times n$ matrices then

$$\det (\lambda I_n + B C) = \det(\lambda I_n + C B)$$


*If $k \le n$, $B$ is $n\times k$, $C$ is $k\times n$ then

$$\det( \lambda I_n + B C) = \lambda^{n-k} \det ( \lambda I_k + C B)$$
How to prove 2. from 1.: Consider the $n\times n$ matrices $\tilde B$, $\tilde C$ obtained as follows: complete $B$ with $n-k$ columns of $0$ to the right, and $C$ with $n-k$ rows of zero at the bottom. Now check that $\tilde B \cdot \tilde C = B\cdot C$, while $\tilde C \cdot \tilde B$ is $n\times n$ with the $ k \times k$ block $C\cdot B$ and $0$ in rest. The conclusions follows now.
To prove 1. We have
$$\det C \cdot \det (\lambda I_n + B C) = \det ( C( \lambda I_n + B C)) = \det ( ( \lambda I_n + C B) C) = \det ( \lambda I_n + C B) \cdot \det C$$
and if this is considered as a polynomial identity, we can simplify with $\det C\ne 0$.
Form 2.  taking $\lambda=1$ we get


*If  $B$ is $n \times k$ and $C$ is $k\times n$ then

$$\det ( I_n + B C) = \det( I_k + C B)$$
Now assume that $A$ is $n\times n$, $B$ is $n\times k$, and $C$ is $k\times n$. We get from 3. (assume for now $A$ invertible)
$$\det (A + B C) = \det ( A( I_n + A^{-1} B C) = \det A \det (I_n + (A^{-1} B) C) =\\= \det A \cdot \det(I_k + C A^{-1} B)$$
Now assume that $k=1$. Then $\det (I_k + C A^{-1} B) = 1 + C A^{-1} B$, so we get
$$\det A ( 1 + C A^{-1} B) = \det A + C (\det A \cdot A^{-1}) B$$
Note that $\det A \cdot A^{-1} = \operatorname{adj} A$. Therefore we get


*For $A$ $n\times n$, $B$ $\ n\times 1$, $C$ $\ 1\times n$
$$\det( A + B C) = \det A + C \cdot \operatorname{adj}(A)\cdot B$$
Note that by density the property holds for all $A$ $\ n\times n$.
$\bf{Added:}$ The result above is very similar to the formula for the determinant of a bordant matrix. So let $A$ be $n\times n$, $x$ an $n\times 1$, $y$ $\ 1\times n$, $\alpha$ a constant. Consider the $(n+1)\times (n+1)$ matrix $\tilde A$ obtained from $A$ by bordering the row $n+1$ with $x$, column $n+1$ with $y$, and element $(n+1, n+1)$ to be $\alpha$. Then we have
$$\det \tilde A = \alpha \cdot \det A - y \operatorname{adj} A x$$
(notice the $-$ sign) (the Cauchy expansion)
May assume $\alpha\ne 0$, then $\alpha=1$. Then, we can reduce the last row to $0$ ( except the entry $\alpha$) Then the $n\times n$ block becomes $A - x \cdot y$, and we apply the known result.
