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In this question, we just consider square matrices.

The cofactor matrix $\mathrm{C}(\mathrm{A}) = (c_{ij})$ of a $n$-by-$n$ matrix $\mathrm{A} = (a_{ij})$ is a $n$-by-$n$ matrix ($n > 0$) with

$$c_{ij} = (-1)^{i+j} \cdot \det{(\mathrm{A}_{ij})} \; \forall i,j,$$

where $\mathrm{A}_{ij}$ is the matrix remaining after removing the $i$-th row and the $j$-th column from $\mathrm{A}$.

Problem: Let $\mathrm{A},\mathrm{B}$ be $n$-by-$n$ matrices. Prove that $\mathrm{C}(\mathrm{AB}) = \mathrm{C}(\mathrm{A})\cdot\mathrm{C}(\mathrm{B})$.

I start learning Linear Algebra recently, and this property is one of the problems giving me hard time proving. After giving up, I begin searching for a solution but not succeeding (my effort might not be enough). Therefore, I hope I can find a solution or a link to a solution here, and it's better that this solution uses no linear space and mapping knowledge. For further information about cofactor matrices and its relatives, you can read the Wikipedia article linked.

Thanks in advance!

egreg
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Vincent J. Ruan
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2 Answers2

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Let $C(A)$ be matrix with entries $\alpha_{ij}$, and likewise $B(A)=(\beta_{ij})$. Then the entries $\gamma_{ij}$ of $C(A)C(B)$ can be written as $$ \gamma_{ij} = \sum_{k=1}^n \alpha_{ik} \beta_{kj} = \sum_{k=1}^n(-1)^{i+j+2k} \det(A_{ik}) \det(B_{kj}) = (-1)^{i+j}\sum_{k=1}^n\det(A_{ik}) \det(B_{kj}). $$ One can check that $$ (AB)_{ij} = A_{i\cdot}B_{\cdot j}, $$ where $A_{i\cdot}$ and $B_{\cdot j}$ are submatrices of $A$ and $B$ without the $i$th row and $j$th columns respectively, $A_{i\cdot}\in\mathbb R^{n-1,n}$, $B_{\cdot j}\in\mathbb R^{n,n-1}$ (or any other field instead of $\mathbb R$).

By the Cauchy-Binet formula, $$ \det(A_{i\cdot}B_{\cdot j})= \sum_{k=1}^n\det(A_{ik}) \det(B_{kj}). $$ This proves $$ \gamma_{ij} = (-1)^{i+j}\sum_{k=1}^n\det(A_{ik}) \det(B_{kj}) =(-1)^{i+j}\det((AB)_{ij} ), $$ hence the entries of $C(AB)$ and $C(A)C(B)$ conincide.

darij grinberg
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daw
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  • Thank you. However, I wonder whether the Cauchy-Binet formula - the most important part of your proof - would be easily proved, at least in your specific case. – Vincent J. Ruan Jun 17 '14 at 09:41
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    There is a proof on the wikipedia page, which can be simpliflied in this case ofc. I am not aware of any more elementary proof. – daw Jun 17 '14 at 09:53
  • I've found a straightforward proof (having the same idea of the proof in the Wikipedia article, I think so) for the Cauchy-Binet formula (in general cases): http://goo.gl/iUOBQM. P.S: The third equation in the proof is implied (in my thought) from using the Leibnitz formula of determinants. – Vincent J. Ruan Jun 21 '14 at 01:37
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Since you are just beginning to learn linear algebra, I'm afraid the following proofs are not appropriate. Yet, I'll leave them here for future references.

Proof 1. (Assume you are talking about real or complex matrices.) For any real or complex matrix $X$ and $t\in\mathbb R$, note that $\det(X+tI)$ and the entries of $C(X+tI)$ are polynomials (and hence continuous functions) in $t$. Since every polynomial in $t$ has only finitely many roots, it follows that when $t>0$ is sufficiently small, $X+tI$ is invertible.

Now, from $X^{-1}=\frac1{\det(X)}\operatorname{adj}(X)$ one obtains $C(X) = \det(X)(X^{-1})^T$. Therefore, if $X$ and $Y$ are invertible matrices, we have \begin{align} C(XY)&=\det(XY)((XY)^{-1})^T\\ &=\det(XY)(Y^{-1}X^{-1})^T\\ &=\det(XY)(X^{-1})^T(Y^{-1})^T\\ &=\det(X)\det(Y)(X^{-1})^T(Y^{-1})^T\\ &=C(X)C(Y). \end{align} In particular, $C((A+tI)(B+tI))=C(A+tI)C(B+tI)$ when $t>0$ is sufficiently small. Let $t\to0$, the result follows.

The above proof is an example of the so-called continuity argument. Often, if some property is known to be preserved in the limit, and we have difficulties in proving the property at a certain point, we may construct a sequence that converges to that point, and prove that the property holds for the sequence instead.

Proof 2. (Assume you are talking about matrices over, say, an integral domain or a field.) As $C(X) = \det(X)(X^{-1})^T$ when $X$ is invertible, it is straightforward to show that the target equality holds for invertible matrices. In particular, the target equality holds when $A,B$ are two matrices whose entries are $2n^2$ different indeterminates $a_{11},a_{12},\ldots,a_{nn},b_{11},b_{12},\ldots,b_{nn}$. Hence the equality also holds when $A$ and $B$ are specialised to any particular values.

One more impressive use of the above technique of considering "generic" matrices is to prove the Cayley-Hamilton theorem (which is usually introduced in the middle of an introductory linear algebra course). See q313284.

Community
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user1551
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  • +1 for Proof 1. I do not understand Proof 2: How to proceed from invertible matrices to non-invertible matrices. – daw Jun 17 '14 at 10:46
  • @daw We don't proceed from invertible matrices to noninvertible matrices. Every generic matrix is invertible. – user1551 Jun 17 '14 at 12:45
  • I still don't understand. "The target equality holds when $A,B$ are two matrices whose entries are $2n^2$ different indeterminates $a_{11},…,a_{nn},b_{11},…,b_{nn}$"- it's true but at that point in Proof 2, it has just been proved in the case of which consists of a condition of $f(a_{11},…,a_{nn})f(b_{11},…,b_{nn}) \neq 0$ (the invertible condition). It's also true that the invertible case comprises almost all cases, but it's still not the whole. So how can we proceed from this? – Vincent J. Ruan Jun 21 '14 at 01:55
  • @VincentJ.Ruan The matrices $A$ and $B$ are always invertible when $a_{11},\ldots,a_{nn},b_{11},\ldots,b_{nn}$ are $2n^2$ indeterminates. That is, if $K$ is the underlying field in the original question, then $A$ and $B$ are invertible over the field of fractions of the polynomial ring $K[a_{11},\ldots,a_{nn},b_{11},\ldots,b_{nn}]$. Hence $C(AB)-C(A)C(B)$ is a zero polynomial. So, when the $2n^2$ indeterminates are specialised to any values in $K$ (i.e. we perform an evaluation map), we get $C(AB)-C(A)C(B)=0$ for any $A,B$ over $K$. – user1551 Jun 28 '14 at 06:45
  • As an analogue, suppose $p\in\mathbb R[x]$ satisfies $p(x)(x-1)-(x^2-1)=0$. We want to find prove that $p(a)=a+1$ for every $a\in\mathbb R$. Since $x-1$ is not the zero polynomial, we may divide both sides by $x-1$ to obtain $p(x)=x+1$. Substitute $x=a$, we get $p(a)=a+1$. Note that in the previous argument, (polynomial) division precedes substitution of variable. Had we performed substitution first with $a=1$, we would get $p(1)(0)-(0)=0$ and we cannot divide by zero to obtain the value of $p(1)$. – user1551 Jun 28 '14 at 06:48