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

where $(1.3.23)$ is as follow: 1.3.23

I am very confused about these questions:

  1. How does second identity make sense? What does $\lambda_i$ stand for?
  2. 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...

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stander Qiu
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  • The identity itself is fairly well-known: see https://math.stackexchange.com/questions/1514408/matrix-determinant-lemma-with-adjugate-matrix?rq=1 . Not sure about that proof. I guess $\lambda_i\left(K\right)$ means the $i$-th eigenvalue of $K$ in some order. – darij grinberg Nov 23 '17 at 01:31
  • @darijgrinberg I think the OP actually meant to ask why the second and the third equality signs are correct. – user1551 Nov 23 '17 at 01:33
  • The second equality sign follows because the determinant of a matrix is the product of its eigenvalues. The third holds because the eigenvalues of $I + B$ are $1 + \mu$ where $\mu$ ranges over the eigenvalues of $B$. – darij grinberg Nov 23 '17 at 01:34
  • @darijgrinberg I know the det of a matrix is the product of its eigenvalues, but it seems that $I + A^{-1}xy^T$ is still a matrix... – stander Qiu Nov 23 '17 at 01:37
  • @darijgrinberg I think I misunderstand the notation. Could you tell me which matrix does $\lambda_i$ correspond to? – stander Qiu Nov 23 '17 at 01:43
  • Yes, but its eigenvalues are $1 + y^TA^{-1}x$ (one time) and $1$ (many times), so the product of these eigenvalues is $1 + y^TA^{-1}x$. – darij grinberg Nov 23 '17 at 01:43
  • @darijgrinberg I see. $\lambda_i$ is not specific value here, it is a function... Thanks for your help – stander Qiu Nov 23 '17 at 01:46

1 Answers1

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  1. if $B$, $C$ are $n\times n$ matrices then

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

  1. 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

  1. 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

  1. 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.

orangeskid
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