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\title{\bf \large Linear transformation and the change of basis}
\maketitle

\abstract{This short note supplements the set of ZCA 110 lecture notes on linear algebra. It expounds in understandable language the idea of linear transformation involving different bases.}


\section{Going from one basis to another}

Consider two bases, $W = (W_1,W_2, \cdots , W_n)$ and $Z = (Z_1,Z_2, \cdots, Z_n)$ in the vector space V$_n(R)$. Note that here $Z_i,W_i$ are colume vectors of $n-$components. The connection between the two bases can be worked out via the following consideration: \\

Consider a general $n$-vector in V$_n$(R), call it $Q$. Q can be represented in the W-basis or in the Z-basis. \\

In the W-basis: $Q = W \cdot Q_W$,where $Q_W$ is the coordinate vector of $Q$ relative to the W-basis. \\
In the Z-basis: $Q = Z \cdot Q_Z$,where $Q_Z$ is the coordinate vector of $Q$ relative to the Z-basis. 
The vector $Q$ is the same vector irrespective of its basis representation, hence 
	\begin{eqnarray}
Q &=& W \cdot Q_W = Z \cdot Q_Z \nonumber \\ 
\Rightarrow Q_Z &=& (Z^{-1}\cdot W)Q_W \equiv P Q_W. \label{PQW}
\end{eqnarray}
Eq. (\ref{PQW}) relates the coordinate vector of an general vector $Q$ in the Z-basis to that in the W-basis. In other words, if the basis $Z$ and $W$ is given, we can form a matrix 
\begin{eqnarray}\label{P} 
P = Z^{-1}\cdot X \end{eqnarray}
which allows the  coodinate vector in one basis to be determined if the coordinate vector in the other is given. Eq. (\ref{PQW}) can also equivalently be stated as \[Q_W = P^{-1}Q_Z. \label{PQZ}\]

\section{Linear Transformation}

A transformation can be carried out in any basis. Consider a vector $X$ is tranformed into vector Y. Such a transformatino can be represented in both the W-basis and the Z-basis. In each basis representationm the transformation takes on different forms. Say $A$ is the transformation matrix in the X-basis representation, whereas $B$ is the correspoinding transformation in the Z-basis representation. The linear transformations in both bases are given by: 



\begin{center}
\begin{math}\begin{array}{clrr} 
\mbox{W-basis}                       & \mbox{Z-basis} \\
                              &          \\
W = (W_1,W_2,\cdots, W_n)     & Z = (Z_1,Z_2,\cdots, Z_n) \\       
X \stackrel{A}{\rightarrow}Y  & X \stackrel{B}{\rightarrow}Y \\
															&          \\
\mbox{In component form}      & \mbox{In component form} \\
X_W \stackrel{A}{\rightarrow}Y_W  & X_Z \stackrel{B}{\rightarrow}Y_Z \\
&          \\
\mbox{In matrix form}      & \mbox{In matrix form} \\
Y_W = A X_W                & Y_Z = B X_Z
\end{array}\end{math}
\end{center}

\vspace{8pt}
Now, we shall prove that: If 
\begin{eqnarray}
X_W &\stackrel{A}{\rightarrow}& Y_W \mbox{ in the W-basis} \nonumber \\ 
X_Z &\stackrel{B}{\rightarrow}& Y_Z \mbox{ in the Z-basis}, \nonumber 
\end{eqnarray}
then the two transformation $A,B$ are similar, i.e.,
\[ B = Q^{-1} A Q, \] 
where $Q=P^{-1} = (Z^{-1}W)^{-1} = W^{-1}Z$.

The proof is as followed: 
We begin with 
	\begin{equation}\label{YZB}
	Y_Z = B X_Z. 
	\end{equation}
	In Eq. (\ref{YZB}), the LHS, i.e. $Y_Z$ is related to $Y_W$ via $Y_Z = P Y_W$, whereas $X_Z$ in the RHS is related to $X_W$ via $X_Z = P X_W$. Hence, Eq. (\ref{YZB}) can be written as 
	\begin{eqnarray}\label{PYW}
	          && P Y_W = B (P X_W) \nonumber \\
\Rightarrow &&   Y_W = (P^{-1} B P) X_W.
	\end{eqnarray}
Eq. (\ref{PYW}) is just the tranformation of $X_W$ into $Y_W$ by $A$ (in the W-basis), i.e. 
\begin{eqnarray}\label{xxx}
	          &&  Y_W = (P^{-1} B P) X_W = A X_W.
	\end{eqnarray}
Hence, we can identify \[ A = P^{-1} B P,\] or \[ B = P A P^{-1} = Q^{-1} A Q,\] where $P$ is given by Eq. (\ref{P}).


			     
	










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\[
Z_i = \bordermatrix{\cr \cr &\alpha_1 \cr & \alpha_2  \cr & \vdots  \cr & \alpha_n}
\]
