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How can I prove that $\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$, if $A$ and $B$ are any two $n\times n$-matrices. Here, $\operatorname{adj} A$ means the adjugate of the matrix $A$.

I know how to prove it for non singular matrices, but I have no idea what to do in this case.

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7 Answers 7

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I thought of the following possibly simpler proof to the ones given above (inasmuch as it uses elementary linear algebra) which makes use of the Cauchy-Binet formula and importantly, works for non-invertible matrices too. In the following, $A^{ij}$ will indicate the matrix derived from some matrix $A$ after eliminating the $i$th row and the $j$th column, and $A^{i0}$ will indicate that only the $i$th row was eliminated (with all columns remaining). Similarly $A^{0j}$ will mean $A$ with the $j$th column eliminated.

For any $1\leq i,j \leq n$, $$\left(\operatorname{adj}\left(B\right)\operatorname{adj}\left(A\right)\right)_{ij}=\sum_{k=1}^{n}\left(\operatorname{adj}B\right)_{ik}\left(\operatorname{adj}A\right)_{kj}\\=\sum_k \left(-1\right)^{k+i}\det\left(B^{ki}\right)\left(-1\right)^{j+k}\det\left(A^{jk}\right)\\=\left(-1\right)^{i+j}\sum_k \det\left(A^{jk}\right)\det\left(B^{ki}\right)\\=\left(-1\right)^{i+j}\sum_k \det\left(A^{jk}B^{ki}\right)=\left(*\right)$$

But we notice that $\left(AB\right)^{ji}=A^{j0}B^{0i}$ and so by the Cauchy-Binet formula we have: $$\det\left(AB\right)^{ji}=\sum_k \det\left(A^{jk}B^{ki}\right)$$

which gives us: $$\left(*\right)=\left(-1\right)^{i+j}\det\left(AB\right)^{ji}=\left(\operatorname{adj}\left(AB\right)\right)_{ij}$$ and we are done.

I hope that this is correct and if so that it helps.

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The easiest technique for dealing with the adjugate matrix is to consider the field of rational functions in $2n^2$ indeterminates $K=F(X,Y)$, where $X$ and $Y$ denote the sets of indeterminates $X_{ij}$ and $Y_{ij}$, for $1\le i,j\le n$. Here $F$ is the base field, in your case probably $\mathbb{R}$ or $\mathbb{C}$.

Then the matrices $X=[X_{ij}]$ and $Y=[Y_{ij}]$ with coefficients in $K$ are invertible, because they have nonzero determinant. By general rule, $$\def\adj{\operatorname{adj}} (\det X) X^{-1}=\adj X $$ and similarly for $Y$ and $XY$. Thus $$ \adj(YX)=\det(XY)\cdot(YX)^{-1}=(\det X\cdot\det Y)X^{-1}Y^{-1} $$ while $$ (\det X)X^{-1}\cdot(\det Y)Y^{-1}=\adj X\cdot\adj Y. $$ Comparing the two expressions we get $$ \adj X\cdot\adj Y=\adj(YX). $$ Now these matrices have coefficients in $F[X,Y]$, the ring of polynomials in the $2n^2$ indeterminates above. Substituting the coefficients of $A$ for $X_{ij}$ and those of $B$ for $Y_{ij}$ gives your claim: $$ \adj A\cdot \adj B = \adj(BA). $$

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  • $\begingroup$ (+1) And considering $[X_{ij}],Y_{ij}]$ as matrices with coefficients in the polynomial ring $\Bbb Z[X,Y]$ ("with $2n^2$ indeterminates"), the final identity $\operatorname{adj} X\cdot\operatorname{adj} Y=\operatorname{adj} (YX)$ is absolute, i.e. proves $\operatorname{adj} A\cdot\operatorname{adj} B =\operatorname{adj} (BA)$ for matrices $A,B$ with coefficients in any commutative ring. (Of course you know that. I just wrote it down for future readers.) $\endgroup$ Dec 12, 2023 at 14:53
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    $\begingroup$ @AnneBauval Yes, one just needs to observe that the entries of the adjugate are polynomials in $X_{ij},Y_{ij}$ with integer coefficients. $\endgroup$
    – egreg
    Dec 12, 2023 at 15:59
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In addition to DonAntonio's answer, if you want just something to do with matrices, you can go through this

We know that $A~ Adj A = |A|~ I $ ( how ?)

If $A$ and $B$ are matrices of the same order, then :

$=> (AB)~ adj (AB) = |A|~|B|~I$

$=> (AB)~ adj (AB) = |A|~I~|B|~I$

$=> (AB)~ adj (AB) = A~(adj~ A)~|B|~I$

$=> B~(adj~AB) = (adj A)~|B|~I $

$=> B~(adj~AB) = (adj A)~|B|~I $

Can you work out from here?

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    $\begingroup$ i thought about your proof ... after these steps, how should we proceed? cause i think this works just for invertible matrices ... $\endgroup$ Oct 21, 2016 at 11:09
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Another answer here:

Consider $(A+tI)^*, t\in R$. Every element of $(A+tI)^*$ is a polynomial in $t$, thus a continuous function of $t$. Thus we have $\lim_{t\to 0}(A+tI)^*=A^*,$ which implies that the Taylor series of $(A+tI)^*$ is $$ (A+tI)^*= A^*+ tC, t\in R, $$ for some matrix C. Similarly, there are some matrices $D,E$ such that $$ (B+tI)^*= B^*+ tD, t\in R, $$ $$ [(A+tI)(B+tI)]^*= (AB)^*+ tE, t\in R. $$

Now, note that there is $t_0>0$ such that both $A+tI$ and $B+tI$ are invertible if $t\in (t_0, \infty)$. Thus it is easy to show that
$$ [(A+tI)(B+tI)]^*= (B+tI)^* (A+tI)^*, t \in (t_0, \infty). $$ We have obtained $$(AB)^*+ tE= (B^*+ tD)(A^*+ tC)= B^*A^* +tF, t \in (t_0, \infty), $$ where $F= DA^* +B^*C + t DC$. Thus they should be equal on $t\in R$. Let $t=0$, we get what we want.

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  • $\begingroup$ But what about matrices with coefficients elsewhere that $\Bbb R$ (or an extension of $\Bbb R$)? For instance with coefficients in a field of prime characteristic? or in a non integral domain? $\endgroup$ Dec 12, 2023 at 14:43
  • $\begingroup$ + Your matrices $C,D,E,F$ are not constant, they depend on $t.$ So, your reasoning "Thus they should be equal on $t∈R$" sounds wrong. . $\endgroup$ Dec 12, 2023 at 15:08
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If we have an inner product space $\;V\;$ , then for all $\;v,u\in V\;$ :

$$\color{red}{\langle ABu,v\rangle}=\langle u,(AB)^*v\rangle$$

$$\langle u,B^*A^*v\rangle=\langle Bu,A^*v\rangle=\color{red}{\langle ABu,v\rangle}$$

Since the red terms in both lines are the same, then

$$\langle u,(AB)^*v\rangle=\langle u,B^*A^*v\rangle\implies \langle\,u\,,\,\left((AB)^*-(B^*A^*)\right)v\,\rangle=0\implies \ldots$$

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  • $\begingroup$ IT appears he's working specifically in finite dimensional cases so just with matrices; for these reasons this might be a bit confusing $\endgroup$
    – DanZimm
    May 17, 2014 at 17:09
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    $\begingroup$ The adjugate is not the transpose or hermitian transpose. $\endgroup$
    – egreg
    May 17, 2014 at 17:20
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    $\begingroup$ The OP never said whether he meant the adjugate or the adjoint, in spite of having been asked... $\endgroup$
    – DonAntonio
    May 17, 2014 at 18:25
  • $\begingroup$ The above proof works for adjoint of an operator in BL(X). $\endgroup$ Apr 22, 2018 at 11:26
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The case of A and B invertibale is easy using the eq. A*adj(A)=det(A)*I.

The case of A or B with rank <= n-2 is also easy becase in this case adj =0.

The last case: let A,B be matrices such that non of them with rank<= n-2 and not both invertibale. Fix i and j and define A',B' suche that A' equals to A except (maybe) the row i and B' equals to B except (maybe) the column j and satisfies: 1) A' and B' invertibale. or 2) A'or B' with rank <= n-2

In either case, we get that adj(A'B')=adj(B')adj(A'). We finish the prof by noticing that (denote $M_{i,j}(A)$ the minor $i,j$ of A and $R_i(A),C_j(A)$, the i'th row and j'th columnof A): $[adj(A'B')]_{j,i}=(-1)^{i+j}|M_{i,j}(A'B')|=(-1)^{i+j}|M_{i,j}(AB)|=[adj(A'B')]_{j,i}$ and $[adj(B')adj(A')]_{j,i}=R_{j}(adj(B'))C_i(adj(A'))=R_{j}(adj(B))C_i(adj(A))=[adj(B)adj(A)]_{j,i}$

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We have for any square matrix $A$ the following $$\DeclareMathOperator{\adj}{adj}A\adj A=\det A\cdot I$$ Therefore for two square matrices $A,B$ of the same order we get $$AB\cdot\adj AB=\det(AB)\cdot I=\det A\det B \cdot I=\det B\det A \cdot I=\det B\cdot A\adj A$$ But $$\det B\cdot A\adj A=A\cdot \det B\cdot\adj A=A\cdot (\det B\cdot I)\cdot\adj A=AB\cdot\adj B\adj A$$ Thus we obtain $$AB\cdot\adj AB=AB\cdot\adj B\adj A$$ If $AB$ is invertible then multiplying both sides by $(AB)^{-1}$ yields $$\adj AB=\adj B\adj A$$ If $(AB)$ is not invertible then we can at most say $$AB\cdot(\adj AB-\adj B\adj A)=0$$

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  • $\begingroup$ This does not adress the question. $\endgroup$ Dec 12, 2023 at 14:39

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