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Let $S$ and $A$ be a symmetric and a skew-symmetric $n \times n$ matrix over $\mathbb{R}$, respectively. When calculating (numerically) the product $S^{-1} A S^{-1}$ I keep getting the factor $\det S$ in the denominator, while I would expect to get the square $$S^{-1} A S^{-1} = \frac{(\text{adj }S) A (\text{adj }S)}{(\det S)^2},$$ where $\text{adj }S$ is the adjugate of $S$.

Is there a way to prove that the combination $(\text{adj }S) A (\text{adj }S)$ already contains a factor of $\det S$?

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  • $\begingroup$ Not true for $n=3$ (where having a factor of $\det S$ would mean being $0$). That said, we do (for $n=3$) have $\left(\operatorname{adj} S\right) A \left(\operatorname{adj} S\right) = h \left(\operatorname{adj} S\right)$, where $h$ is a certain scalar depeding on $A$ and $S$. $\endgroup$ – darij grinberg Aug 13 '15 at 18:12
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    $\begingroup$ The conjecture does hold for even $n$, though. Better yet: $\operatorname{adj} S$ is divisible by the Pfaffian $\operatorname{Pf} S$ (and as you know, we have $\left(\operatorname{Pf} S\right)^2 = \det S$). I'll post this as an answer once I've figured out a nice proof that doesn't use Hilbert's Nullstellensatz. $\endgroup$ – darij grinberg Aug 13 '15 at 18:21
  • $\begingroup$ @darijgrinberg I'd be interested in the proof that does use it, if it can be summarized $\endgroup$ – Ben Grossmann Aug 13 '15 at 18:27
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    $\begingroup$ Summary of the proof using the Nullstellensatz: Let us work over an algebraically closed field. Then, $\operatorname{Pf} S$ is an irreducible polynomial in the entries over $S$ (I think this is clear, though I'm not 100% sure), and whenever it vanishes, so does every entry of $\operatorname{adj} S$ (since the vanishing of $\operatorname{Pf} S$ shows that $\operatorname{rank} S < 2n$ and thus $\operatorname{rank} S \leq 2n-2$, since the rank of a skew-symmetric matrix must be even). Gauss' lemma can then be used to pull back the divisibility from the algebraically closed field to any ring. $\endgroup$ – darij grinberg Aug 13 '15 at 18:28
  • $\begingroup$ Anyway, it seems to me that we have something explicit: If $S$ is a skew-symmetric $n\times n$-matrix with $n$ even, then $\operatorname{adj} S = \operatorname{Pf} S \cdot \operatorname{Pdj} S$, where $\operatorname{Pdj} S$ is the Pfaffian adjoint of $S$ (that is, the $n\times n$-matrix whose $\left(i,j\right)$-th entry is $\left(-1\right)^{i+j+\left[i>j\right]} p_{i,j}\left(S\right)$, where $p_{i,j}\left(S\right)$ is the Pfaffian of the matrix obtained by removing the $i$-th and $j$-th rows and the $i$-th and $j$-th columns from $S$). Here, $\left[i>j\right]$ means $1$ if $i>j$ ... $\endgroup$ – darij grinberg Aug 13 '15 at 18:31
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Assume that $\det(S)=0$. Then $adj(S)$ has rank $1$ and is symmetric; then $adj(S)=avv^T$ where $a\in \mathbb{R}$ and $v$ is a vector. Thus $adj(S)Aadj(S)=a^2v(v^TAv)v^T$. Since $A$ is skew-symmetric, $v^TAv=0$ and $adj(S)Aadj(S)=0$. We use the Darij's method; here, the condition is that $\det(S)$ is an irreducible polynomial when $S$ is a generic symmetric matrix; if it is true, then $\det(S)$ is a factor of every entry of $adj(S)Aadj(S)$.

EDIT 1. For the proof that $\det(S)$ is an irreducible polynomial when $S$ is a generic symmetric matrix, cf. https://mathoverflow.net/questions/50362/irreducibility-of-determinant-of-symmetric-matrix and we are done !

EDIT 2. @ darij grinberg , hi Darij, I read quickly your Theorem 1 (for $K$, a commutative ring with unity) and I think that your proof works; yet it is complicated! I think (as you wrote in your comment above) that it suffices to prove the result when $K$ is a field; yet I do not kwow how to write it rigorously...

STEP 1. $K$ is a field. If $\det(S)=0$, then $adj(S)=vw^T$ and $adj(S).A.adj(S)=v(w^TAw)v^T=0$ (even if $char(K)=2$). Since $\det(.)$ is irreducible over $M_n(K)$, we conclude as above.

STEP 2. Let $S=[s_{ij}],A=[a_{i,j}]$. We work in the ring of polynomials $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i<j}]$ in the indeterminates $(s_{i,j}),(a_{i,j})$. This ring has no zero-divisors, is factorial and its characteristic is $0$ and even is integrally closed. Clearly the entries of $adj(S).A.adj(S)$ are in $\mathbb{Z}[(s_{i,j})_{i,j},(a_{i,j})_{i<j}]$; moreover they formally have $\det(S)$ as a factor.

Now, if $K$ is a commutative ring with unity, we must use an argument using a variant of Gauss lemma showing that the factor $\det(S)$ is preserved over $K$. What form of the lemma can be used and how to write it correctly ?

I just see that the OP takes for himself the green chevron; we are our own best advocates

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  • $\begingroup$ Why is the rank of adj S = 1 for det S = 0? $\endgroup$ – user54031 Aug 13 '15 at 23:06
  • $\begingroup$ It is true for any matrix ! $\endgroup$ – user91684 Aug 14 '15 at 9:59
  • $\begingroup$ In fact the rank is 0 or 1 $\endgroup$ – user91684 Aug 14 '15 at 10:00
  • $\begingroup$ +1. Nice trick in here! But you can spare yourself the algebraic geometry orgy by generalizing: For any $n\times n$-matrix $S$ and any alternating $n\times n$-matrix $A$, the entries of $\left(\operatorname{adj} S\right) \cdot A \cdot \left(\operatorname{adj}\left(S^T\right)\right)$ are divisible by $\det S$. The irreducibility of $\det S$ is much more elementary here, and instead of $\operatorname{adj} S = avv^T$ you now need $\operatorname{adj} S = vw^T$ (which is also easier to prove and less reliant on the base field). $\endgroup$ – darij grinberg Aug 14 '15 at 21:08
  • $\begingroup$ Yes Darij, you are right. $\endgroup$ – user91684 Aug 15 '15 at 4:22
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I have managed to prove the claim in a pedestrian way. I'll just present the final result because the proof is quite involved and probably of interest only to me.

In the following, I will use the (abstract) index notation and Einstein summation convention.

First of all, the determinant of a square $n \times n$ matrix $S$ is given by

$$\det S = \frac{1}{n!} \epsilon_{a_1 \ldots a_n} \epsilon_{b_1 \ldots b_n} S_{b_1 a_1} \ldots S_{b_n a_n}.$$

Second, the adjugate of $S$ is given by a similar expression $$(\operatorname{ajd} S)_{a_1 b_1} = \frac{1}{(n-1)!} \epsilon_{a_1 a_2 \ldots a_n} \epsilon_{b_1 b_2 \ldots b_n} S_{b_2 a_2} \ldots S_{b_n a_n}.$$

Finally, we will need the another tensor of order $4$, defined as $$(\operatorname{ajd}_2 S)_{a_1 a_2,b_1 b_2} = \frac{1}{(n-2)!} \epsilon_{a_1 a_2 a_3 \ldots a_n} \epsilon_{b_1 b_2 b_3 \ldots b_n} S_{b_3 a_3} \ldots S_{b_n a_n}.$$

Then the following identity holds $$((S^{-1}) A (S^{-1})^{T})_{ab} = \frac{1}{2} \frac{(\operatorname{ajd}_2 S)_{ab,cd} A_{cd}}{\det S}$$ for $A$ skew-symmetric.

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Here is a proof.

In the following, we fix a commutative ring $\mathbb{K}$. All matrices are over $\mathbb{K}$.

Theorem 1. Let $n\in\mathbb{N}$. Let $S$ be an $n\times n$-matrix. Let $A$ be an alternating $n\times n$-matrix. (This means that $A^{T}=-A$ and that the diagonal entries of $A$ are $0$.) Then, each entry of the matrix $\left( \operatorname*{adj}S\right) \cdot A\cdot\left( \operatorname*{adj}\left( S^{T}\right) \right) $ is divisible by $\det S$ (in $\mathbb{K}$).

[UPDATE: A slight modification of the below proof of Theorem 1 can be found in the solution to Exercise 6.42 in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019. More precisely, said Exercise 6.42 claims that each entry of the matrix $\left(\operatorname{adj} S\right)^T \cdot A \cdot \left(\operatorname{adj} S\right)$ is divisible by $\det S$; now it remains to substitute $S^T$ for $S$ and recall that $\left(\operatorname{adj} S\right)^T = \operatorname{adj} \left(S^T\right)$, and this immediately yields Theorem 1 above. Still, the following (shorter) version of this proof might be useful as well.]

The main workhorse of the proof of Theorem 1 is the following result, which is essentially (up to some annoying switching of rows and columns) the Desnanot-Jacobi identity used in Dodgson condensation:

Theorem 2. Let $n\in\mathbb{N}$. Let $S$ be an $n\times n$-matrix. For every $u\in\left\{ 1,2,\ldots,n\right\} $ and $v\in\left\{ 1,2,\ldots ,n\right\} $, we let $S_{\sim u,\sim v}$ be the $\left( n-1\right) \times\left( n-1\right) $-matrix obtained by crossing out the $u$-th row and the $v$-th column in $S$. (Thus, $\operatorname*{adj}S=\left( \left( -1\right) ^{i+j}S_{\sim j,\sim i}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$.) For every four elements $u$, $u^{\prime}$, $v$ and $v^{\prime}$ of $\left\{ 1,2,\ldots,n\right\} $ with $u\neq u^{\prime}$ and $v\neq v^{\prime}$, we let $S_{\left( \sim u,\sim u^{\prime}\right) ,\left( \sim v,\sim v^{\prime }\right) }$ be the $\left( n-2\right) \times\left( n-2\right) $-matrix obtained by crossing out the $u$-th and $u^{\prime}$-th rows and the $v$-th and $v^{\prime}$-th columns in $S$. Let $u$, $i$, $v$ and $j$ be four elements of $\left\{ 1,2,\ldots,n\right\} $ with $u\neq v$ and $i\neq j$. Then, \begin{align} & \det\left( S_{\sim i,\sim u}\right) \cdot\det\left( S_{\sim j,\sim v}\right) -\det\left( S_{\sim i,\sim v}\right) \cdot\det\left( S_{\sim j,\sim u}\right) \\ & = \left( -1\right) ^{\left[ i<j\right] +\left[ u<v\right] }\det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) . \end{align} Here, we use the Iverson bracket notation (that is, we write $\left[ \mathcal{A}\right] $ for the truth value of a statement $\mathcal{A}$; this is defined by $\left[ \mathcal{A}\right] = \begin{cases} 1, & \text{if }\mathcal{A}\text{ is true;}\\ 0, & \text{if }\mathcal{A}\text{ is false} \end{cases} $).

There are several ways to prove Theorem 2: I am aware of one argument that derives it from the Plücker relations (the simplest ones, where just one column is being shuffled around). There is at least one combinatorial argument that proves Theorem 2 in the case when $i = 1$, $j = n$, $u = 1$ and $v = n$ (see Zeilberger's paper); the general case can be reduced to this case by permuting rows and columns (although it is quite painful to track how the signs change under these permutations). (See also a paper by Berliner and Brualdi for a generalization of Theorem 2, with a combinatorial proof too.) There is at least one short algebraic proof of Theorem 2 (again in the case when $i = 1$, $j = n$, $u = 1$ and $v = n$ only) which relies on "formal" division by $\det S$ (that is, it proves that \begin{align} & \det S \cdot \left(\det\left( S_{\sim 1,\sim 1}\right) \cdot\det\left( S_{\sim 2,\sim 2}\right) -\det\left( S_{\sim 1,\sim 2}\right) \cdot\det\left( S_{\sim 2,\sim 1}\right) \right) \\ & = \left(\det S\right)^2 \cdot\det\left( S_{\left( \sim 1,\sim 2\right) ,\left( \sim 1,\sim 2\right) }\right) , \end{align} and then argues that $\det S$ can be cancelled because the determinant of a "generic" square matrix is invertible). (This proof appears in Bressoud's Proofs and Confirmations; a French version can also be found in lecture notes by Yoann Gelineau.) Unfortunately, none of these proofs seems to release the reader from the annoyance of dealing with the signs. Maybe exterior powers are the best thing to use here, but I do not see how. I have written up a division-free (but laborious and annoying) proof of Theorem 2 in my determinant notes; more precisely, I have written up the proof of the $i < j$ and $u < v$ case, but the general case can easily be obtained from it as follows:

Proof of Theorem 2. We need to prove the equality \begin{align} & \det\left( S_{\sim i,\sim u}\right) \cdot\det\left( S_{\sim j,\sim v}\right) -\det\left( S_{\sim i,\sim v}\right) \cdot\det\left( S_{\sim j,\sim u}\right) \\ & = \left( -1\right) ^{\left[ i<j\right] +\left[ u<v\right] }\det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) . \label{darij.eq.1} \tag{1} \end{align} If we interchange $u$ with $v$, then the left hand side of this equality gets multiplied by $-1$ (because its subtrahend and its minuend switch places), whereas the right hand side also gets multiplied by $-1$ (since $S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right)}$ does not change, but $\left[ u<v\right]$ either changes from $0$ to $1$ or changes from $1$ to $0$). Hence, if we interchange $u$ with $v$, then the equality \eqref{darij.eq.1} does not change its truth value. Thus, we can WLOG assume that $u \leq v$ (since otherwise we can just interchange $u$ with $v$). Assume this. For similar reasons, we can WLOG assume that $i \leq j$; assume this too. From $u \leq v$ and $u \neq v$, we obtain $u < v$. From $i \leq j$ and $i \neq j$, we obtain $i < j$. Thus, Theorem 6.126 in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019 (applied to $A=S$, $p=i$ and $q=j$) shows that \begin{align} & \det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) \\ & = \det\left( S_{\sim i,\sim u}\right) \cdot\det\left( S_{\sim j,\sim v}\right) -\det\left( S_{\sim i,\sim v}\right) \cdot\det\left( S_{\sim j,\sim u}\right) \label{darij.eq.2} \tag{2} \end{align} (indeed, what I am calling $S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }$ here is what I am calling $\operatorname{sub}^{1,2,\ldots,\widehat{u},\ldots,\widehat{v},\ldots,n}_{1,2,\ldots,\widehat{i},\ldots,\widehat{j},\ldots,n} A$ in my notes).

But both $\left[i < j\right]$ and $\left[u < v\right]$ equal $1$ (since $i < j$ and $u < v$). Thus, $\left( -1\right) ^{\left[ i<j\right] +\left[ u<v\right] } = \left(-1\right)^{1+1} = 1$. Therefore \begin{align} & \underbrace{\left( -1\right) ^{\left[ i<j\right] +\left[ u<v\right] }}_{=1}\det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) \\ & = \det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) \\ & = \det\left( S_{\sim i,\sim u}\right) \cdot\det\left( S_{\sim j,\sim v}\right) -\det\left( S_{\sim i,\sim v}\right) \cdot\det\left( S_{\sim j,\sim u}\right) \end{align} (by \eqref{darij.eq.2}). This proves Theorem 2. $\blacksquare$

Finally, here is an obvious lemma:

Lemma 3. Let $n\in\mathbb{N}$. For every $i\in\left\{ 1,2,\ldots ,n\right\} $ and $j\in\left\{ 1,2,\ldots,n\right\} $, let $E_{i,j}$ be the $n\times n$-matrix whose $\left( i,j\right) $-th entry is $1$ and whose all other entries are $0$. (This is called an elementary matrix.) Then, every alternating $n\times n$-matrix is a $\mathbb{K}$-linear combination of the matrices $E_{i,j}-E_{j,i}$ for pairs $\left( i,j\right) $ of integers satisfying $1\leq i<j\leq n$.

Proof of Theorem 1. We shall use the notation $E_{i,j}$ defined in Lemma 3.

We need to prove that every entry of the matrix $\left( \operatorname*{adj} S\right) \cdot A\cdot\left( \operatorname*{adj}\left( S^{T}\right) \right) $ is divisible by $\det S$. In other words, we need to prove that, for every $\left( u,v\right) \in\left\{ 1,2,\ldots,n\right\} ^{2}$, the $\left( u,v\right) $-th entry of the matrix $\left( \operatorname*{adj} S\right) \cdot A\cdot\left( \operatorname*{adj}\left( S^{T}\right) \right) $ is divisible by $\det S$. So, fix $\left( u,v\right) \in\left\{ 1,2,\ldots,n\right\} ^{2}$.

We need to show that the $\left( u,v\right) $-th entry of the matrix $\left( \operatorname*{adj}S\right) \cdot A\cdot\left( \operatorname*{adj} \left( S^{T}\right) \right) $ is divisible by $\det S$. This statement is clearly $\mathbb{K}$-linear in $A$ (in the sense that if $A_{1}$ and $A_{2}$ are two alternating $n\times n$-matrices such that this statement holds both for $A=A_{1}$ and for $A=A_{2}$, and if $\lambda_{1}$ and $\lambda_{2}$ are two elements of $\mathbb{K}$, then this statement also holds for $A=\lambda_{1}A_{1}+\lambda_{2}A_{2}$). Thus, we can WLOG assume that $A$ has the form $E_{i,j}-E_{j,i}$ for a pair $\left( i,j\right) $ of integers satisfying $1\leq i<j\leq n$ (according to Lemma 3). Assume this, and consider this pair $\left( i,j\right) $.

We have $\operatorname*{adj}S=\left( \left( -1\right) ^{x+y}\det\left( S_{\sim y,\sim x}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n}$ and \begin{align} \operatorname*{adj}\left( S^{T}\right) & = \left( \left( -1\right) ^{x+y}\det\left( \underbrace{\left( S^{T}\right) _{\sim y,\sim x} }_{=\left( S_{\sim x,\sim y}\right) ^{T}}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n} \\ & = \left( \left( -1\right) ^{x+y}\underbrace{\det\left( \left( S_{\sim x,\sim y}\right) ^{T}\right) }_{=\det\left( S_{\sim x,\sim y}\right) }\right) _{1\leq x\leq n,\ 1\leq y\leq n} \\ & = \left( \left( -1\right) ^{x+y}\det\left( S_{\sim x,\sim y}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n} . \end{align} Hence, \begin{align} & \underbrace{\left( \operatorname*{adj}S\right) }_{=\left( \left( -1\right) ^{x+y}\det\left( S_{\sim y,\sim x}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n}}\cdot\underbrace{A}_{=E_{i,j}-E_{j,i}}\cdot \underbrace{\left( \operatorname*{adj}\left( S^{T}\right) \right) }_{=\left( \left( -1\right) ^{x+y}\det\left( S_{\sim x,\sim y}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n}} \\ & =\left( \left( -1\right) ^{x+y}\det\left( S_{\sim y,\sim x}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n}\cdot\left( E_{i,j}-E_{j,i}\right) \\ & \qquad \qquad \cdot\left( \left( -1\right) ^{x+y}\det\left( S_{\sim x,\sim y}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n} \\ & = \left( \left( -1\right) ^{x+i}\det\left( S_{\sim i,\sim x}\right) \cdot\left( -1\right) ^{j+y}\det\left( S_{\sim j,\sim y}\right) \right. \\ & \qquad \qquad \left. -\left( -1\right) ^{x+j}\det\left( S_{\sim j,\sim x}\right) \cdot\left( -1\right) ^{i+y}\det\left( S_{\sim i,\sim y}\right) \right) _{1\leq x\leq n,\ 1\leq y\leq n} . \end{align} Hence, the $\left( u,v\right) $-th entry of the matrix $\left( \operatorname*{adj}S\right) \cdot A\cdot\left( \operatorname*{adj}\left( S^{T}\right) \right) $ is \begin{align} & \left( -1\right) ^{u+i}\det\left( S_{\sim i,\sim u}\right) \cdot\left( -1\right) ^{j+v}\det\left( S_{\sim j,\sim v}\right) -\left( -1\right) ^{u+j}\det\left( S_{\sim j,\sim u}\right) \cdot\left( -1\right) ^{i+v} \det\left( S_{\sim i,\sim v}\right) \\ & = \left( -1\right) ^{i+j+u+v}\left( \det\left( S_{\sim i,\sim u}\right) \cdot\det\left( S_{\sim j,\sim v}\right) -\det\left( S_{\sim i,\sim v}\right) \cdot\det\left( S_{\sim j,\sim u}\right) \right) . \label{darij.eq.3} \tag{3} \end{align} We need to prove that this is divisible by $\det S$. If $u=v$, then this is obvious (because if $u=v$, then the right hand side of \eqref{darij.eq.3} is $0$). Hence, we WLOG assume that $u\neq v$. Thus, \eqref{darij.eq.3} shows that the $\left( u,v\right) $-th entry of the matrix $\left( \operatorname*{adj}S\right) \cdot A\cdot\left( \operatorname*{adj}\left( S^{T}\right) \right) $ is \begin{align} & \left( -1\right) ^{i+j+u+v}\underbrace{\left( \det\left( S_{\sim i,\sim u}\right) \cdot\det\left( S_{\sim j,\sim v}\right) -\det\left( S_{\sim i,\sim v}\right) \cdot\det\left( S_{\sim j,\sim u}\right) \right) }_{\substack{=\left( -1\right) ^{\left[ i<j\right] +\left[ u<v\right] }\det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) \\\text{(by Theorem 2)}}} \\ & = \left( -1\right) ^{i+j+u+v}\left( -1\right) ^{\left[ i<j\right] +\left[ u<v\right] }\det S\cdot\det\left( S_{\left( \sim i,\sim j\right) ,\left( \sim u,\sim v\right) }\right) , \end{align} which is clearly divisible by $\det S$. Theorem 1 is thus proven. $\blacksquare$

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  • $\begingroup$ Please see my answer. Does it agree with your answer? $\endgroup$ – user54031 Aug 18 '15 at 8:19
  • $\begingroup$ @LBO: I fear I cannot tell, as I am not acquainted with the tensor notation you are using. $\endgroup$ – darij grinberg Aug 18 '15 at 11:52
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Because $S\mbox{adj}(S)=\mbox{adj}(S)S=\mbox{det}(S)I$, then $S$ is invertible iff $\mbox{det}(S)\ne 0$, and, in that case $$ S^{-1} = \frac{1}{\mbox{det}(S)}\mbox{adj}(S). $$ Therefore, $$ S^{-1}AS^{-1} = \frac{1}{\mbox{det}(S)^{2}}\mbox{adj}(S)\,A\,\mbox{adj}(S). $$

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  • $\begingroup$ Yes, thank you, I'm aware of that. I would like to prove that the numerator of your last expressio contains det S. $\endgroup$ – user54031 Aug 13 '15 at 20:32
  • $\begingroup$ I don't really understand what you mean by contains $\mbox{det}(S)$. There are determinants buried in the inverses on the left that match the determinants on the right. Is there something in particular bothering you about the expression? $\endgroup$ – Disintegrating By Parts Aug 13 '15 at 20:44
  • $\begingroup$ Ok, to word it differently, I want to show that, whenever det S = 0, (adj S) A (adj S) also vanishes. $\endgroup$ – user54031 Aug 13 '15 at 21:00

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