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Section6Orthogonal Vectors (NM)

Subsection6.1Some bases are better than others

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*}

Subsection6.3Vector spaces and inner product spaces


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.