Subsection 2.1.3 Dimensionality of a Vector Space
A vector space is said to be an \(n\)- dimensional if it contains \(n\) linearly independent vectors. A vector space is called an infinite - dimensional if there exists an arbitrary large number of linearly independent vectors in the space. If an arbitrary vector \(\phi\) in \(V_{n}\) can be represented as a linear combination of vectors \(\left\{\psi_{i}\right\}\) in \(V_{n}\) and scalars \(\left\{\alpha_{i}\right\}\text{,}\) \i.e.
\begin{equation*}
\phi=\sum\limits_{i=1}^{n}\alpha_{i}\psi_{i} =\alpha_{1}\psi_{1}+\alpha_{2}\psi_{2}+\alpha_{3}\psi_{3}+\cdots+\alpha_{n}\psi_{n}
\end{equation*}
then \(\left\{\psi_{i}\right\}\) is said to be a span of the vector space \(V_{n}\text{.}\) A linearly independent set of vectors \(\left\{\psi_{i}\right\}\) that spans a vector space \(V_{n}\) is called a basis for \(V_{n}\text{.}\) For example, the unit vectors \(i{,}j{,}\) and \(k\) of a position vector \('r'\) are the basis for the three - dimensional vector space. The three mutually perpendicular vectors forms an orthogonal basis for a three - dimensional vector space. In other words, if the scalar product of two vectors is zero, the vectors are said to be an orthogonal to each other. The orthogonal bases of unit magnitude such as \(\hat{i} {,} \hat{j}{,}\) and \(\hat{k}\) forms a normal orthogonal basis, called an orthonormal basis. A scalar has one component (magnitude only) and hence zero basis vector per component, A vector has 3 components (magnitude and one direction) in 3D and hence has 1 basis vector per component. A tensor of rank 2 (dyad) has \(3^{2} = 9\) components (magnitude and two directions) in 3D hence has 2 basis vectors per component. A tensor of rank 3 (triad) has \(3^{3} = 27\) components (magnitude and three directions) in 3D hence has 3 basis vectors per component.
Subsubsection 2.1.3.1 Inner Product
The inner or scalar product of two vectors u and v is denoted by \(\lt u,v \gt\) in vector space \(V_{n}\text{,}\) which hold the following properties - \(\lt u,v+\xi \gt=\lt u,v \gt + \lt u,\xi \gt; \) \(\lt u+v,\xi \gt = \lt u,\xi \gt + \lt v,\xi \gt \) and \(\lt u,u \gt \geq 0\) (unless \(u=0\)). The length (or Norm), \(\parallel u \parallel\text{,}\) of a vector \(u\) is defined as \(\parallel u \parallel = \sqrt{ \lt u,u \gt }\text{.}\) The inner product of two vectors equals zero, \(\lt u,v \gt =0\) for \(u \neq 0\) and \(v \neq 0\text{,}\) then the vectors are said to form an orthogonal set. If the norm within an orthogonal set is unity, i.e. \(\parallel u \parallel = 1\text{,}\) then the set is called an orthonormal set.
Subsubsection 2.1.3.2 Gram Schmidt’s Orthogonalization
An orthogonal basis is the best basis for a vector space because the coefficients of which can easily be expressed a vector as a linear combination of basis vectors. However, we are not always given an orthogonal basis. Gram-Schmidt orthogonalization is a process used to transform a set of linearly independent vectors into a set of orthonormal vectors, which may not be orthogonal to each other. This process is named after Jørgen Pedersen Gram and Erhard Schmidt, who independently developed it in the late \(19^{th}\) century. Let \(u_{1},u_{2},u_{3},\cdots, u_{n}\) be a set of linearly independent vectors which are not necessarily orthogonal to each other. Now, It is required to obtain a set of orthogonal vectors \(v_{1},v_{2},v_{3},\cdots, v_{n}\) from the original set of vectors \(u_{1},u_{2},u_{3},\cdots, u_{n}\) by following the steps below.
In step (1) take \(v_{1} = u_{1}\text{,}\) in step (2) let \(v_{2} = u_{2}+\lambda v_{1}\) where \(\lambda\) is a constant to be determined from the condition that \(v_{2}\) to be orthogonal to \(v_{1}\text{.}\) That is, \((v_{1},v_{2}=0)\) or, \(\lt v_{1}, u_{2} +\lambda v_{1}\gt =0=\lt v_{1}, u_{2}\gt +\lambda \lt v_{1}, v_{1} \gt \text{,}\) and in step (3) let \(v_{3} = u_{3}+\lambda_{1} v_{1}+\lambda_{2} v_{2}\) where \(\lambda_{1}\) and \(\lambda_{2}\) are constants to be determined from the conditions that \(v_{3}\) is orthogonal to \(v_{1}\) and \(v_{2}\text{.}\) This gives - \(\lt v_{1}, v_{3} \gt = 0 = \lt v_{1}, u_{3} \gt +\lambda_{1} \lt v_{1}, v_{1} \gt +
\lambda_{2} \lt v_{1}, v_{2}\gt \text{,}\) since \(\lt v_{1}, v_{2}\gt = 0.\) Therefore, we have -
\begin{equation}
\Rightarrow \lambda_{1}=-\frac{\lt v_{1}, u_{3}\gt}{\lt v_{1}, v_{1}\gt}\tag{2.1.4}
\end{equation}
\begin{equation*}
\text{Also,} \quad \lt v_{2}, v_{3}\gt = 0 = \lt v_{2}, u_{3}\gt +\lambda_{1} \lt v_{2}, v_{1}\gt
+\lambda_{2} \lt v_{2}, v_{2}\gt
\end{equation*}
\begin{equation}
\Rightarrow \lambda_{2}=-\frac{\lt v_{2}, u_{3}\gt}{\lt v_{2}, v_{2}\gt } \quad
[\because \lt v_{2}, v_{1}\gt = \lt v_{1}, v_{2}\gt = 0]\tag{2.1.5}
\end{equation}
Now, we have three mutually orthogonal vectors \(v_{1},v_{2}\) and \(v_{3}\text{.}\) The same procedure can be continued to obtain other othogonal vectors. Finally, all the vectors can be normalized to obtain an orthonormal set \(\left\{x_{i}\right\}\text{,}\) where \(x_{i}= \frac{v_{i}}{\parallel v_{i} \parallel}\text{.}\)
