# $\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$

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.

## 7 Answers

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: $$\adj A\cdot \adj B = \adj(BA).$$

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.

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?

• i thought about your proof ... after these steps, how should we proceed? cause i think this works just for invertible matrices ... – Arman Malekzadeh Oct 21 '16 at 11:09

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.

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

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

• IT appears he's working specifically in finite dimensional cases so just with matrices; for these reasons this might be a bit confusing – DanZimm May 17 '14 at 17:09
• The adjugate is not the transpose or hermitian transpose. – egreg May 17 '14 at 17:20
• The OP never said whether he meant the adjugate or the adjoint, in spite of having been asked... – DonAntonio May 17 '14 at 18:25
• The above proof works for adjoint of an operator in BL(X). – Jayanth Kumar Apr 22 '18 at 11:26

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 $$AB\cdot(\adj AB-\adj B\adj A)=0$$

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 $$[adj(B')adj(A')]_{j,i}=R_{j}(adj(B'))C_i(adj(A'))=R_{j}(adj(B))C_i(adj(A))=[adj(B)adj(A)]_{j,i}$$