91影视

Consider a linear transformation T from V to V, where V is an n-dimensional linear space .Let B be a basis of V.

Consider the linear transformation$L_B鈭�T鈭�{L_B}^{鈭�1}$from$R^n$to$R^n$with standard matrix B

Which implies$B\stackrel{鈫�}{x}=L_B\left[T\left({L_B}^{鈭�1}\left(\stackrel{鈫�}{x}\right)\right)\right]鈭赌\stackrel{鈫�}{x}\text{鈥夆赌�}in\text{鈥夆赌�}{R}^n$

This matrix B is called the B matrix of transformation T.

Consider two bases U and B of an n-dimensional linear space V.

Consider the linear transformation$L_U鈭�{L_B}^{鈭�1}$ from$R^n$ to $R^n$with standard matrix S,meaning that$S\stackrel{鈫�}{x}=L_U\left[{L_B}^{鈭�1}\left(\stackrel{鈫�}{x}\right)\right]\text{鈥夆赌�}鈭赌\text{鈥夆赌夆赌�}\stackrel{鈫�}{x}\text{鈥夆赌�}in\text{鈥夆赌�}{R}^n$

This invertible matrix S is called the change of basis matrix from B to U,sometimes denoted by$S_{B鈫�U}$

Consider the bases as B and U as follows

$\begin{array}{l}B=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥�}\\ 2\text{鈥夆赌夆赌�}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌�}2\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌�}\end{array}\right]\\ U=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥�}\\ 2\text{鈥夆赌夆赌�}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌�}2\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌�}\end{array}\right]\end{array}$

Now by inspection we find out the change of matrix from B to U

$\begin{array}{l}{S}_{B鈫�U}=\left[\begin{array}{l}a\\ 鈭�a+b+c\\ b\\ b+d\end{array}\right]\\ S_{B鈫�U}=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\\ 鈭�1\text{鈥夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌�}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}1\end{array}\right]\end{array}$

Hence the solution.

Consider the matrix as follows

$\begin{array}{l}B=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥�}\\ 2\text{鈥夆赌夆赌�}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌�}2\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌�}\end{array}\right]\\ A=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥�}\\ 2\text{鈥夆赌夆赌�}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌�}2\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌�}\end{array}\right]\end{array}$

Here both the A and B matrix is same.

$S=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\\ 鈭�1\text{鈥夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌�}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}1\end{array}\right]$

$\begin{array}{l}SB=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\\ 鈭�1\text{鈥夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌�}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}1\end{array}\right]\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥�}\\ 2\text{鈥夆赌夆赌�}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌�}2\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌�}\end{array}\right]\\ SB=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 1\text{鈥夆赌夆赌夆赌夆赌�}1\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\\ 0\text{鈥夆赌夆赌夆��}3\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}3\end{array}\right]\end{array}$

Similarly,AS becomes as follows

$\begin{array}{l}AS=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥�}\\ 2\text{鈥夆赌夆赌�}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌�}2\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}2\text{鈥夆赌夆赌�}\end{array}\right]\left[\begin{array}{l}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆赌夆赌�}0\\ 鈭�1\text{鈥夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}1\text{鈥夆赌夆赌夆赌夆赌�}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆赌夆赌夆赌夆��}1\text{鈥夆赌夆赌夆赌夆赌夆赌�}0\text{鈥夆赌夆赌夆��}1\end{array}\right]\\ AS=\left[\begin{array}{l}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 1\text{鈥夆赌夆赌夆赌夆赌�}1\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\\ 0\text{鈥夆赌夆赌夆��}1\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}1\\ 0\text{鈥夆赌夆赌夆��}3\text{鈥夆赌夆赌夆��}0\text{鈥夆赌夆赌夆��}3\end{array}\right]\end{array}$

Hence the proof