[[Hermitian Adjoint]]: Needs to know what conjugate transpose is doing to the matrix. Theorem 2.1.9 in book \<Matrix Analysis\> --- ### **Intro** > This is the theroem statement: > $ > A\in \mathbb{M}_n \quad A^{-1} \sim A^H \iff > B\in \mathbb{M}_n: A = B^{-1}B^H > $ > The inverse of matrix $A$ is similar to $A^H$ if and only if $A$ can be written as the product of $B^{-1}B^H$. The assumptions are: * Matrix $A$ is invertible. * Matrix $B$ is invertible. **Proof $\leftarrow$** $ \begin{aligned} A &= B^{-1}B^H \\ A^H &= (B^{-1}B^H)^H = (BB^{-H}) \\ A^{-1} &= B^{-H}B \\ (B^H)\underbrace{B^{-H}B}_{A^{-1}}(B^{-H}) &= \underbrace{BB^{-H}}_{A^H} \end{aligned} $ If $A = B^{-1}B$ then $A^{-1}$ is similar to the matrix $A^H$ by the transformation represented by $B^H$ $\blacksquare$ **Proof $\rightarrow$** Note: This direction is way harder to proof. We wish to show that: > $ > A^{-1}\sim A^{H}\implies A = B^{-1}B^H > $ Our task is to look for the matrix $B$. It's procedurals. Firstly we would like to find this representation of: $ H = A^HHA \quad \text{ Where: }\quad H^H = H \wedge \exists \; H^{-1} \tag{1} $ Let's start by considering that: $ \begin{aligned} & A \sim A^{H} \\ \implies & SAS^{-1} = A^H \wedge S = A^HSA \\ \text{Let: } & S_\theta = e^{i\theta}S \\ \implies & S_\theta + S_\theta^H = A^HS_\theta A + A^H S_\theta^H A \\ & S_\theta + S_\theta^H = A^H(S_\theta + S_\theta^H)A \end{aligned}\tag{2} $ Consider the choice of $\theta$ to make $S_\theta + S_\theta^H$ invertible, then: $ \begin{aligned} & \exists \; x \neq \mathbf{0}: H_\theta x = \mathbf{0} \\ \implies & (S_\theta + S_\theta^H)x =\mathbf{0} \\ & x = -S^{-1}_\theta S_\theta^{H}x \\ & x = e^{-2i\theta} S^{-1}S^Hx \\ & S^{-1}S^{H}x = -e^{2i\theta}x \end{aligned}\tag{3} $ Therefore, to make it invertible we will have to choose $\theta\in (0, \pi]$ such that it's not an eigenvalue for the matrix $S^{-1}S^H$ Choose that $\theta$, then we consider $S$ using that $\theta^+$ which is then used to construct the matrix $H$: > $ > H = A^HHA\tag{4} > $ We are halfway there, let's talk about the motivation for the second part of the proof. **Motivations and Tricks to Reach there** Let's consider: $ \begin{aligned} B &= (I - A^H)H \\ B^H &= H(I - A) \end{aligned}\tag{5} $ Then: $ \begin{aligned} A &= B^{-1}B^H \\ BA &= B^H \end{aligned}\tag{6} $ To this regard, it would mean that: $ \begin{aligned} BA &= (I - A^H)HA \\ &= (HA - A^HHA) \\ &= (HA - H) \\ &= H(A - I) \\ B^H &= H(I - A) \end{aligned}\tag{7} $ Notice that, we are one constant way by expression (7), however, $B$ is invertible when the eigen value of $A$ contains 1, in that case, we are farther away from where we want to get, hence, let's consider: **Some Extra Ingredients** > $ > \begin{aligned} > B &= \beta (\alpha I - A^H)H > \end{aligned}\tag{8} > $ > Where we need $|\alpha| = 1$ and $\alpha \in \mathbb{C}$, and we have the free choice for $\beta \in \mathbb{C}$, which will reveal itself after a while. Let's consier again on the values before which is going to be: $ \begin{aligned} BA &= \beta(\alpha I - A^H)HA \\ &= \beta(\alpha HA - A^HHA) \\ &= \beta(\alpha HA - H) \\ &= H(\beta\alpha A - \beta I) \\ B^H &= H(\bar{\beta} \bar{\alpha}I - \bar{\beta}A) \\ \implies H(\bar{\beta} \bar{\alpha} - \bar{\beta}A) &= H(\beta\alpha A - \beta I) \\ \implies \beta &= -\bar{\beta}\bar{\alpha} \end{aligned}\tag{9} $ If we were to assume the choice of $\alpha = e^{i\varphi}$ then it means that: $ \beta = \exp \left( \frac{i(\pi - \varphi)}{2} \right) $ And that concludes the proof both ways. $\blacksquare$