mail  mail@stemandmusic.in

    

call  +91-9818088802

Change of Basis Vectors for a Vector

1. Any \(N\)-Dimensional Vector represented in terms of Standard Basis Vectors can always be represented in terms of Any Set of \(N\) Arbitrary Basis Vectors each of \(N\)-Dimensions and vice versa as demonstrated in the following
   
   Lets consider a \(N\)-Dimensional Vector \(A\) having Components \(A_1, A_2, \cdots, A_n\) when represented using Standard Basis Vectors as follows
   
   \(\Rightarrow A=A_1\begin{bmatrix}1\\0\\ \vdots \\ 0_n\end{bmatrix} + A_2\begin{bmatrix}0\\1\\ \vdots \\ 0_n\end{bmatrix} + \cdots + A_n\begin{bmatrix}0\\0\\ \vdots \\ 1_n\end{bmatrix}\)    ...(1)
   
   \(\Rightarrow A=\begin{bmatrix}1 & 0& \cdots & 0_n \\ 0 & 1 & \cdots & 0_n \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1_n\end{bmatrix} \begin{bmatrix}A_1 \\ A_2 \\ \vdots \\A_n\end{bmatrix}\)   ...(2)
   
   The Vector \(A\) as given above can also be represented in terms of \(N\) Arbitrary \(N\)-Dimensional Basis Vectors \(e_1, e_2, \cdots , e_n\), having Components \(A_1', A_2', \cdots, A_n'\) as follows
   
   \(A = A_1'e_1 + A_2'e_2 + \cdots + A_n'e_n\)     where \(e_1=\begin{bmatrix}e_{11}\\e_{12}\\ \vdots \\e_{1n}\end{bmatrix},\hspace{.5cm}e_2=\begin{bmatrix}e_{21}\\e_{22}\\ \vdots \\e_{2n}\end{bmatrix},\hspace{.5cm}\cdots\hspace{.5cm}e_n=\begin{bmatrix}e_{n1}\\e_{n2}\\ \vdots \\e_{nn}\end{bmatrix}\)   ...(3)
   
   \(\Rightarrow A=A_1'\begin{bmatrix}e_{11}\\e_{12}\\ \vdots \\e_{1n}\end{bmatrix} + A_2'\begin{bmatrix}e_{21}\\e_{22}\\ \vdots \\e_{2n}\end{bmatrix} + \cdots + A_n'\begin{bmatrix}e_{n1}\\e_{n2}\\ \vdots \\e_{nn}\end{bmatrix}\)   ...(4)
   
   \(\Rightarrow A = \begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1} \\ e_{12} & e_{22} & \cdots & e_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn}\end{bmatrix}\begin{bmatrix}A'_1\\A'_2\\ \vdots \\ A'_n\end{bmatrix}\)   ...(5)
   
   Since equations (1) to (5) above all represent Vector \(A\), using equations (2) and (5) we get
   
   \(\begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1} \\ e_{12} & e_{22} & \cdots & e_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn}\end{bmatrix}\begin{bmatrix}A'_1\\A'_2\\ \vdots \\ A'_n\end{bmatrix}=\begin{bmatrix}1 & 0& \cdots & 0_n \\ 0 & 1 & \cdots & 0_n \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1_n\end{bmatrix} \begin{bmatrix}A_1 \\ A_2 \\ \vdots \\A_n\end{bmatrix}\)
   
   \(\Rightarrow \begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1} \\ e_{12} & e_{22} & \cdots & e_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn}\end{bmatrix}\begin{bmatrix}A'_1\\A'_2\\ \vdots \\ A'_n\end{bmatrix}=\begin{bmatrix}A_1 \\ A_2 \\ \vdots \\A_n\end{bmatrix}\)   ...(6)
   
   Multiplying equation (6) with the Inverse of the Matrix of the Arbitrary Basis Vector Set on both sides we get
   
   \(\begin{bmatrix}A'_1\\A'_2\\ \vdots \\ A'_n\end{bmatrix}=\begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1} \\ e_{12} & e_{22} & \cdots & e_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn}\end{bmatrix}^{-1}\begin{bmatrix}A_1\\A_2\\ \vdots \\A_n\end{bmatrix}\)   ...(7)
   
   The equation (6) aboves gives us the formula for Finding the Components of Vector \(A\) in terms of Standard Basis Vectors when the Vector \(A\) is represented in Set of Arbitrary Basis Vectors and it's Correspnding Components. Hence, To Represent any Vector in Standard Basis Vectors, the Vector represented in Set of Arbitrary Basis Vectors needs to be Pre-Multiplied By the Matrix of the Arbitrary Basis Vector Set.
   
   The equation (7) aboves gives us the formula for Finding the Components of Vector \(A\) in terms of Arbitrary Basis Vectors when the Vector \(A\) is given in terms of Standard Basis Vectors. Hence, To Represent any Vector in Any Set of Arbitrary Basis Vectors, the Vector represented in Standard Basis needs to be Pre-Multiplied By Inverse of the Matrix of the Arbitrary Basis Vector Set.
2. Any \(N\)-Dimensional Vector \(A\) represented using One Set of \(N\) Arbitrary Basis Vectors of \(N\)-Dimensions can be Represented using Any Other Set of \(N\) Arbitrary Basis Vectors \(N\)-Dimensions as demonstrated in the following
   
   Lets consider an \(N\)-Dimensional Vector \(A\) represented using One Set of \(N\) Arbitrary Basis Vectors of \(N\)-Dimensions \(e_1, e_2, ..., e_n\) through it's Corresponding Components \(A_1, A_2, \cdots, A_n\) and representated using Another Set of \(N\) Arbitrary Basis Vectors of \(N\)-Dimensions \(e_1', e_2', ..., e_n'\) through it's Corresponding Components \(A_1', A_2', \cdots, A_n'\) as follows
   
   \(A=A_1e_1 + A_2e_2 + ... + A_ne_n\)     where \(e_1=\begin{bmatrix}e_{11}\\e_{12}\\ \vdots \\e_{1n}\end{bmatrix},\hspace{.5cm}e_2=\begin{bmatrix}e_{21}\\e_{22}\\ \vdots \\e_{2n}\end{bmatrix},\hspace{.5cm}...\hspace{.5cm},\hspace{.5cm}e_n=\begin{bmatrix}e_{n1}\\e_{n2}\\ \vdots \\e_{nn}\end{bmatrix}\)   ...(8)
   
   Also, \(A=A'_1e_1' + A'_2e_2' + ... + A'_ne_n'\)     where \(e_1'=\begin{bmatrix}e'_{11}\\e'_{12}\\ \vdots \\e'_{1n}\end{bmatrix},\hspace{.5cm}e_2'=\begin{bmatrix}e'_{21}\\e'_{22}\\ \vdots \\e'_{2n}\end{bmatrix},\hspace{.5cm}...\hspace{.5cm},\hspace{.5cm}e_n'=\begin{bmatrix}e'_{n1}\\e'_{n2}\\ \vdots \\e'_{nn}\end{bmatrix}\)   ...(9)
   
   Since both equations (8) and (9) above represent Vector \(A\), we get
   
   \(A'_1e_1' + A'_2e_2' + ... + A'_ne_n'=A_1e_1 + A_2e_2 + ... + A_ne_n\)   ...(8)
   
   \(\Rightarrow A'_1\begin{bmatrix}e'_{11}\\e'_{12}\\ \vdots \\e'_{1n}\end{bmatrix} + A'_2\begin{bmatrix}e'_{21}\\e'_{22}\\ \vdots \\e'_{2n}\end{bmatrix} + ... + A'_n\begin{bmatrix}e'_{n1}\\e'_{n2}\\ \vdots \\e'_{nn}\end{bmatrix}=A_1\begin{bmatrix}e_{11}\\e_{12}\\ \vdots \\e_{1n}\end{bmatrix} + A_2\begin{bmatrix}e_{21}\\e_{22}\\ \vdots \\e_{2n}\end{bmatrix} + ... + A_n\begin{bmatrix}e_{n1}\\e_{n2}\\ \vdots \\e_{nn}\end{bmatrix}\)   ...(9)
   
   \(\Rightarrow \begin{bmatrix}e'_{11} & e'_{21} & \cdots & e'_{n1}\\ e'_{12} & e'_{22} & \cdots & e'_{n2}\\\vdots & \vdots & \ddots & \vdots \\ e'_{1n} & e'_{2n} & \cdots & e'_{nn} \end{bmatrix}\begin{bmatrix}A'_1\\A'_2\\ \vdots \\A'_n\end{bmatrix}=\begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1}\\ e_{12} & e_{22} & \cdots & e_{n2}\\\vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn} \end{bmatrix}\begin{bmatrix}A_1\\A_2\\ \vdots \\A_n\end{bmatrix}\)   ...(10)
   
   \(\Rightarrow \begin{bmatrix}A'_1\\A'_2\\ \vdots \\A'_n\end{bmatrix}=\begin{bmatrix}e'_{11} & e'_{21} & \cdots & e'_{n1}\\ e'_{12} & e'_{22} & \cdots & e'_{n2}\\\vdots & \vdots & \ddots & \vdots \\ e'_{1n} & e'_{2n} & \cdots & e'_{nn} \end{bmatrix}^{-1}\begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1}\\ e_{12} & e_{22} & \cdots & e_{n2}\\\vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn} \end{bmatrix}\begin{bmatrix}A_1\\A_2\\ \vdots \\A_n\end{bmatrix}\)   ...(11)
   
   \(\Rightarrow \begin{bmatrix}e_{11} & e_{21} & \cdots & e_{n1}\\ e_{12} & e_{22} & \cdots & e_{n2}\\\vdots & \vdots & \ddots & \vdots \\ e_{1n} & e_{2n} & \cdots & e_{nn} \end{bmatrix}^{-1}\begin{bmatrix}e'_{11} & e'_{21} & \cdots & e'_{n1}\\ e'_{12} & e'_{22} & \cdots & e'_{n2}\\\vdots & \vdots & \ddots & \vdots \\ e'_{1n} & e'_{2n} & \cdots & e'_{nn} \end{bmatrix}\begin{bmatrix}A'_1\\A'_2\\ \vdots \\A'_n\end{bmatrix}=\begin{bmatrix}A_1\\A_2\\ \vdots \\A_n\end{bmatrix}\)   ...(12)
   
   The equation (11) aboves gives the formula for Finding the Components of Vector \(A\) corresponding to Basis Vectors \(e_1', e_2', ..., e_n'\) when Components of Vector \(A\) corresponding to Basis Vectors \(e_1, e_2, ..., e_n\) are given.
   
   The equation (12) aboves gives the formula for Finding the Components of Vector \(A\) corresponding to Basis Vectors \(e_1, e_2, ..., e_n\) when Components of Vector \(A\) corresponding to Basis Vectors \(e_1', e_2', ..., e_n'\) are given.
3. In general, any Vector represented in terms of One Set of Basis Vectors (Source Basis Vector Set) can be represented in terms of Any Other Set of Basis Vectors (Target Basis Vector Set) under following conditions
   1. The Dimension of Basis Vector Matrix formed by the Source Basis Vector Set (also called the Source Basis Vector Matrix) must be same as the Dimension of Basis Vector Matrix formed by the Target Basis Vector Set (also called the Target Basis Vector Matrix).
   2. The given Vector must belong to the Column Space of both the Source and Target the Basis Vector Matrices. This can only happen if the Source Basis Vector Matrix represents either the Same Vector Space as the Target Basis Vector Matrix or a Vector Subspace of the Target Basis Vector Matrix.
   Provided the conditions given above are met, given a Vector \(X\) in terms of Source Basis Vector Matrix \(S\), its Components \(X'\) in terms of Target Basis Matrix \(T\) can be calculated by Pre-Multiplying \(X\) with a Matrix \(L\) (which is called the Vector Component Transformation Matrix) as
   
   \(X'=LX\)   ...(13)
   
   the Vector Component Transformation Matrix \(L\) can be calculated as
   
   \(L = {T_D}^T S\)   ...(14) (where \(T_D\) is Dual Matrix of \(T\))
   
   If the Target Basis Vector Matrix \(T\) is a Complex Matrix, then the Vector Component Transformation Matrix \(L\) can also be calculated as
   
   \(L = {T_D}^{\dagger}S\)   ...(15)
   
   If the Target Basis Vector Matrix \(T\) is a Square Matrix, then the formuale given in equation (14) and (15) get reduced to
   
   \(L = T^{-1}S\)   ...(16)
   
   If Both Basis Vector Matrices \(S\) and \(T\) represent the Same Vector Space/Subspace then
   1. The Vector Component Transformation Matrix \(L\) is a Square Matrix.
   2. The Vector \(X'\) given in terms of Target Basis Vector Matrix \(T\) can be converted back to \(X\) in terms of Source Basis Vector Matrix \(S\) by Pre-Mutiplying \(X'\) with Inverse of Vector Component Transformation Matrix \(L\) (i.e. \(L^{-1}\)). That is
      
      \(X =L^{-1}X'\)   ...(17)
      
      The Matrix \(L^{-1}\) can also be calculated as
      
      \(L^{-1} = {S_D}^T T\)   ...(18) (where \(S_D\) is Dual Matrix of \(S\))
      
      If the Source Basis Vector Matrix \(S\) is a Complex Matrix, then \(L^{-1}\) can also be calculated as
      
      \(L^{-1} = {S_D}^{\dagger}T\)   ...(19)
      
      If the Source Basis Vector Matrix \(S\) is a Square Matrix, then the formuale given in equation (18) and (19) get reduced to
      
      \(L^{-1} = S^{-1}T\)   ...(20)