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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$S\stackrel{鈫�}{x}=L_U\left[{L_B}^{鈭�1}\left(\stackrel{鈫�}{x}\right)\right]\text{鈥夆赌�}鈭赌\text{鈥夆赌夆��}\stackrel{鈫�}{x}\text{鈥夆赌�}in\text{鈥夆赌�}{R}^n$

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

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

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

$\begin{array}{l}{S}_{B鈫�U}=\left[\begin{array}{l}a\\ b+c\\ c\end{array}\right]\\ S_{B鈫�U}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\end{array}$

Hence the solution.

Consider the matrix as follows

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

Here both the A and B matrix is same.

$S=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)$

$\begin{array}{l}SB=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\left[\begin{array}{l}1\text{鈥夆赌夆�夆赌夆��}0\text{鈥夆赌夆�夆赌�}0\\ 0\text{鈥夆赌夆�夆赌夆��}1\text{鈥夆赌夆�夆赌夆��}2\\ 0\text{鈥夆赌夆�夆赌夆��}0\text{鈥夆赌夆�夆赌�}3\end{array}\right]\\ SB=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 5\\ 0& 0& 3\end{array}\right)\end{array}$

Similarly, AS becomes as follows

$\begin{array}{l}AS=\left[\begin{array}{l}1\text{鈥夆赌夆�夆赌夆��}0\text{鈥夆赌夆�夆赌�}0\\ 0\text{鈥夆赌夆�夆赌夆��}1\text{鈥夆赌夆�夆赌夆��}2\\ 0\text{鈥夆赌夆�夆赌夆��}0\text{鈥夆赌夆�夆赌�}3\end{array}\right]\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\\ AS=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 5\\ 0& 0& 3\end{array}\right)\end{array}$

Hence the proof

Consider the bases as B and U as follows

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

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

$\begin{array}{l}{S^{鈭�1}}_{U鈫�B}=\left[\begin{array}{l}a\\ b鈭�c\\ c\end{array}\right]\\ {S^{鈭�1}}_{U鈫�B}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 鈭�1\\ 0& 0& 1\end{array}\right)\end{array}$

Hence the solution.