\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Components of Vectors

These vectors span \(\mathbb{E}^{3}\). That is any vector \(\usf{x}\in \mathbb{E}^{3}\) can be written as a linear combination of these basis vector.

\[\begin{align} \usf{x}&=x_1 \hat{\usf{E}}_1+x_2 \hat{\usf{E}}_2+x_3 \hat{\usf{E}}_3\\ \usf{x}&=x_i \hat{\usf{E}}_i \end{align}\] \[\begin{align} [\usf{x}]_{\{\hat{\usf{E}}\}_{i=1,2,3}}&= \left[ \begin{array}{c} x_1\\ x_2\\ x_3 \end{array} \right] \end{align}\]

However if we are given a new basis, a basis different from \(\{\hat{\usf{E}}\}_{i=1,2,3}\), say \(\{\hat{\usf{F}}\}_{i=1,2,3}\). Would the components of \(\usf{x}\) in \(\{\hat{\usf{F}}\}_{i=1,2,3}\) be the same as in \(\{\hat{\usf{E}}\}_{i=1,2,3}\). That is, would

\begin{equation} \usf{x}\stackrel{?}{=}x_i \hat{\usf{F}}_i \end{equation}

The answer is no. Typically, in a different basis the components of \(\usf{x}\) will be different. This is same as measuring the size of an object using a scale that in centimeter unit and another scale which is units of inches. Thus, we will have

\(\begin{align} \usf{x}&=y_1 \hat{\usf{F}}_1+y_2 \hat{\usf{F}}_2+y_3 \hat{\usf{F}}_3\\ \usf{x}&=y_i \hat{\usf{F}}_i, \end{align}\) i.e.,

\[\begin{align} [\usf{x}]_{\{\hat{\usf{F}}\}_{i=1,2,3}}&= \left[ \begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right] \end{align}\]

and typically \(x_i \not= y_i\).

The question now arrises, if we know \(x_i\) how do we find \(y_i\)? The answer to this question involves the transformation matrix \([Q]\), such that,

Since \(\{\hat{\usf{E}}\}_{i=1,2,3}\) is a basis, and hence each of \(\hat{\usf{F}}_j\) can be written as a linear combination of \(\{\hat{\usf{E}}\}_{i=1,2,3}\).

\[\begin{align} \hat{\usf{F}}_1 &= Q_{11}\hat{\usf{E}}_1+ Q_{21}\hat{\usf{E}}_2+ Q_{31}\hat{\usf{E}}_3 \\ \hat{\usf{F}}_2 &= Q_{12}\hat{\usf{E}}_1+ Q_{22}\hat{\usf{E}}_2+ Q_{32}\hat{\usf{E}}_3 \\ \hat{\usf{F}}_3 &= Q_{13}\hat{\usf{E}}_1+ Q_{23}\hat{\usf{E}}_2+ Q_{33}\hat{\usf{E}}_3 \end{align}\]

The above set of equations can be compactly be written as:

\[\begin{align} \hat{\usf{F}}_j&=Q_{ij}\hat{\usf{E}}_i \end{align}\] \[\begin{align} \left[\hat{\usf{F}}_1\right]_{\{\hat{\usf{E}}\}_{i=1,2,3}}& &= \left[ \begin{array}{c} Q_{11}\\ Q_{21}\\ Q_{31} \end{array} \right] \end{align}\] \[\begin{align} \left[\hat{\usf{F}}_2\right]_{\{\hat{\usf{E}}\}_{i=1,2,3}}& &= \left[ \begin{array}{c} Q_{12}\\ Q_{22}\\ Q_{32} \end{array} \right] \end{align}\] \[\begin{align} \left[\hat{\usf{F}}_3\right]_{\{\hat{\usf{E}}\}_{i=1,2,3}}& &= \left[ \begin{array}{c} Q_{13}\\ Q_{23}\\ Q_{33} \end{array} \right] \end{align}\]

In summary we can write, \(\begin{align} Q= \left[ \begin{array}{lcr} \left[\hat{\usf{F}}_1\right]_{\{\hat{\usf{E}}\}_{i=1,2,3}} & \left[\hat{\usf{F}}_2\right]_{\{\hat{\usf{E}}\}_{i=1,2,3}} & \left[\hat{\usf{F}}_3\right]_{\{\hat{\usf{E}}\}_{i=1,2,3}} \end{array} \right] \end{align}\)

Now let’s try to find the components \(x_i\) from \(y_i\) of a general vector \(\usf{x}\).

Derivation 1

\[\begin{align} \usf{x}&=x_i \hat{\usf{E}}_i\\ \usf{x}&=y_i \hat{\usf{F}}_i\\ \hat{\usf{F}}_i&= Q_{ji}\hat{\usf{E}}_j\\ \usf{x}&=y_i Q_{ji}\hat{\usf{E}}_j\\ x_j&=y_iQ_{ji}\\ x_j&=Q_{ji}y_i\\ x_i&=Q_{ij}y_j\\ \end{align}\] \[\begin{align} \left[ \begin{array}{c} x_1\\ x_2\\ x_3 \end{array} \right] = \left[ \begin{array}{lcr} Q_{11} & Q_{12} & Q_{13} \\ Q_{21} & Q_{22} & Q_{23} \\ Q_{31} & Q_{32} & Q_{33} \\ \end{array} \right] \left[ \begin{array}{c} y_1\\ y_2\\ y_3 \end{array} \right] \end{align}\]

Derivation 2

\[\begin{align} \usf{x}&=x_i \hat{\usf{E}}_i\\ \usf{x}&=y_i \hat{\usf{F}}_i\\ \hat{\usf{F}}_i&= Q_{ji}\hat{\usf{E}}_j\\ \usf{x}&=y_i Q_{ji}\hat{\usf{E}}_j\\ \end{align}\]

Recall that \(\usf{x}=x_1 \hat{\usf{E}}_1+x_2 \hat{\usf{E}}_2+x_3 \hat{\usf{E}}_3\), which can be compactly be written as \(\usf{x}=x_p \hat{\usf{E}}_p\). Thus, we have

\[\begin{align} x_p \hat{\usf{E}}_p&= Q_{ji} y_i\hat{\usf{E}}_j\\ x_p \hat{\usf{E}}_p&= Q_{ij} y_j\hat{\usf{E}}_i\\ (x_p \hat{\usf{E}}_p)\cdot \hat{\usf{E}}_q&= (Q_{ij} y_j\hat{\usf{E}}_i)\cdot \hat{\usf{E}}_q\\ (x_p \hat{\usf{E}}_p\cdot \hat{\usf{E}}_q)&= (Q_{ij} y_j)\hat{\usf{E}}_i\cdot \hat{\usf{E}}_q\\ (x_p \delta_{pq}&= (Q_{ij} y_j)\delta_{iq}\\ x_q &= Q_{qj} y_j\\ x_i &= Q_{ij} y_j\\ \end{align}\]

Examples from AL notes.

Now let’s assume that

\[\begin{align} \hat{\usf{E}}_i=P_{ji}\hat{\usf{F}}_j \end{align}\]

Following the same steps as before we arrive at the equation, \(\begin{align} y_i &= P_{ij} x_j\\ \end{align}\)

Combining the results, \(y_i = P_{ij} x_j\) and \(x_i = Q_{ij} x_j\) we have that,

\[\begin{align} y_i &= P_{ij} Q_{jk}y_k\\ \delta_{im}y_m &= P_{ij} Q_{jk}y_k\\ \delta_{im}y_m -P_{ij} Q_{jk}y_k&=0\\ (\delta_{ik} -P_{ij} Q_{jk})y_{k}&=0\\ \delta_{ik} -P_{ij} Q_{jk}&=0\\ \end{align}\]

The last equation can be written as \(\begin{align} [P][Q]&=[I], \end{align}\)

which implies that \([P]\) is the left inverse of \([Q]\).

\[\begin{align} x_i &= Q_{ij} P_{jk}x_k\\ \delta_{ik}x_k &= Q_{ij} P_{jk}x_k\\ \delta_{ik}x_k -P_{ij} Q_{jk}x_k&=0\\ (\delta_{ik} -Q_{ij} P_{jk})x_{k}&=0\\ \delta_{ik} -Q_{ij} P_{jk}&=0\\ \end{align}\]

The last equation can be written as \(\begin{align} [Q][P]&=[I], \end{align}\)

which implies that \([P]\) is also the right inverse of \([Q]\).

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