Exercise6.1
Show that if \(A \in \mathbb{C}^{n \times n}\) is non-singular, then
\begin{equation*} (A^{-1})^\star = (A^\star)^{-1}. \end{equation*}We recalled that, towards the end of Lecture 4, we observed that, if \(A\) is an \(n \times n\) matrix with \(n\) linearly independent rows, then there exists a matrix \(A^{-1}\) such that
\begin{equation*} A A^{-1} = A^{-1}A = I. \end{equation*}We say that \(A\) is non-singular or invertible. We then noted that multiplying by \(A^{-1}\) (or, indeed, by \(A\)) can be interpreted as a change of basis operation.
The transpose of the matrix \(A \in \mathbb{R}^{m \times n}\) is \(A^T \in \mathbb{R}\) where
\begin{equation*} (A)_{i,j} = (A^T)_{j,i}\text{.} \end{equation*}However, if \(A \in \mathbb{C}^{m \times n}\) we need to define the related concept of the Hermetian Transpose, \(A^\star\text{,}\) for which
\begin{equation*} (A)_{i,j} = \overline{(A^T)_{j,i}}\text{.} \end{equation*}Here \(\bar{z}\) denotes the complex conjugate of \(z\text{.}\) It is easy to show that
\begin{equation*} (AB)^\star = B^\star A^\star. \end{equation*}Show that if \(A \in \mathbb{C}^{n \times n}\) is non-singular, then
\begin{equation*} (A^{-1})^\star = (A^\star)^{-1}. \end{equation*}Find a definition of a vector space.
An inner product space (of vectors over a field \(\mathbb{F}\)) is a vector space, \(V\text{,}\) equipped with the function \((\cdot, \cdot) : V \times V \to \mathbb{F},\) such that, if \(x,y,z \in V\) and \(a \in \mathbb{F}\) then
\begin{equation*} \begin{split} (x,y) \amp =\overline{(x,y)}\\ (ax,y) \amp = a(x,y)\\ (x+y,z) \amp = (x,z)+(y,z)\\ (x,x) \amp \geq 0\\ (x,x) \amp = 0 \text{ iff } x=0. \end{split} \end{equation*}There are many possible inner products, but the most important, for vectors with entries in \(\mathbb{C}\) is the usual dot product:
\begin{equation*} (u,v) := u^\star v = \sum_{i=0}^n \bar{u}_i v_i. \end{equation*}For real-valued vectors it can be understood as the “angle” between the vectors in \(\mathbb{R}\text{.}\) That is, if \(\alpha\) is the angle between two vectors, \(u\) and \(v\text{,}\) then,
\begin{equation*} \cos(\alpha) = \frac{(u,v)}{\sqrt{(u,u)} \sqrt{(v,v)}}. \end{equation*}The most important/interesting situation when \((u,v)=0\text{,}\) in which case we say that \(u\) and \(v\) are orthogonal.
A matrix \(S \in \Cmm\) is “skew-hermitian” if \(S^\star = - S\text{.}\) Show that the eigenvalues of \(S\) are pure imaginary.