I need to show that $\text{Adj}(AB) = \text{Adj}(B)\text{Adj}(A)$, where $A$ is an $N$x$M$ matrix, and $B$ is an $M$x$K$ matrix, and $\text{Adj}(A)$ denotes the adjoint of $A$.
I have a feeling that I should expand the matrix multiplication into the sum, and then it I should be able to some multiplication since the the indeces are just simple numbers and I can move them around. But I'm not quite sure how to go about it.
Okay, I think I got it now! Assuming $C = AB$, I have that
$$ C_{ij} = \sum\limits_{k} A_{ik}B_{kj} $$
Then
$$ \text{Adj}(C_{ij}) = \text{Adj} \left( \sum\limits_{k} A_{ik}B_{kj} \right) = \sum\limits_{k} \text{Adj}(A_{ik})\text{Adj}(B_{kj}) = \sum\limits_{k} A^*_{ki}B^*_{jk} = \sum\limits_{k} B^*_{jk}A^*_{ki} = \text{Adj}(B)\text{Adj}(A) $$
$\endgroup$ 52 Answers
$\begingroup$For general Hilbert spaces
$\newcommand{\adjun}{^{*}}$I denote $\operatorname{Adj}(A)$ as $A\adjun$ and $(x,y)_{H_i}$ is the scalarproduct of the Hilbertspace $H_i$. So $A: H_2 \to H_3$ and $B: H_1 \to H_2$. \begin{align*} (ABx,y)_{H_3} = (Bx,A\adjun y)_{H_2} = (x,B\adjun A\adjun y)_{H_1} \end{align*} So by definition $(AB)\adjun =B\adjun A\adjun$
Concrete for matrices in finite vectorspaces
Let $A \in \mathbb{C}^{n\times m}$ and $B \in \mathbb{C}^{m \times k}$ then $AB = \mathbb{C}^{n\times k}$ and \begin{align*} (AB)_{i,j} = \sum_{l = 1}^m A_{i,l}\cdot B_{l,j} \end{align*} Additionally we know by your definition that $((AB)\adjun)_{i,j} = \overline{AB_{j,i}}$ where for $z= x + \mathrm{i}y \in \mathbb{C}$ $\overline{z}= x - \mathrm{i} y$ (Complex conjugate). We know that the Complex conjugate is a homomorphism for the field $\mathbb{C}$. This means amongst others
- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
- $\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}$
Endowed with this knowledge we have \begin{align*} ((AB)\adjun)_{i,j} = \overline{AB_{j,i}} = \overline{\sum_{l = 1}^m A_{j,l} \cdot B_{l,i}} = \sum_{l = 1}^m \overline{A_{j,l}} \cdot\overline{B_{l,i}} \end{align*} On the other hand we have $(A\adjun)_{i,j} = \overline{A_{j,i}}$ and the same counts for $B$ \begin{align*} (B\adjun A\adjun)_{i,j} = \sum_{l=1}^m (B\adjun)_{i,l}\cdot(A\adjun)_{l,j} = \sum_{l=1}^m \overline{B_{l,i}}\cdot\overline{A_{j,l}} =\sum_{l = 1}^m \overline{A_{j,l}} \cdot\overline{B_{l,i}} \end{align*} and this does the trick :)
$\endgroup$ 5 $\begingroup$Let us remember that, where $A_{n \times n}$ is a matrix,$$ \DeclareMathOperator{\adjoint}{adj} A \cdot \adjoint{(A)} = \adjoint{(A)} \cdot A = |A| \cdot I_{n} $$Hence$$ \adjoint{(AB)} \cdot AB = |AB| \cdot I_{n} = |A|\cdot |B| \cdot I_{n} \\ \implies \adjoint{(AB)} \cdot AB \cdot \adjoint(B) = |A|\cdot |B| \cdot \adjoint(B) \\ = \adjoint{(AB)} \cdot A \cdot |B| \cdot I_{n} = |A|\cdot |B| \cdot \adjoint(B) \\ \implies |B| \cdot \adjoint{(AB)} \cdot A \cdot \adjoint(A) = |A|\cdot |B| \cdot \adjoint(B) \cdot \adjoint(A) \\ = |B| \cdot \adjoint{(AB)} \cdot |A| \cdot I_{n} = |A|\cdot |B| \cdot \adjoint(B) \cdot \adjoint(A) \\ = |A|\cdot |B| \cdot \adjoint{(AB)} = |A|\cdot |B| \cdot \adjoint(B) \cdot \adjoint(A) $$If neither $A$ nor $B$ is singular i.e. $0\notin \{|A|,|B|\}$, then we may simply divide on both sides to obtain$$ \boxed{ \adjoint{(AB)} = \adjoint(B) \cdot \adjoint(A) } $$If either of $A$ or $B$ is singular, we make them non-singular by applying a limit on them. We may do this since $\adjoint$ is a continuous function on the entries of the pre-image. The details of this been left as an exercise for the reader. :)
$\endgroup$
