Answer to Question #149667 in Linear Algebra for Sourav Mondal

Question

Let T be a linear operator whose matrix with
respect to the basis {(1,1,1),(1,-1,0),(1,0,-1)} is the matrix [
4 0 0
0 1 0
0 0 1. ]
Obtain the matrix of T wrt the standard basis.

Answer 1

$\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix};$ $\begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};$

$\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix} \times T= \begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};$

$T=\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix} ^{-1} \begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};$

Consider the matrix $\begin{bmatrix} 1& 1& 1 \\ 1& -1& 0 \\ 1& 0& -1 \\ \end{bmatrix}$ is A. We use Linear Row Reduction to find the Inverse Matrix of A:

$A^{-1}=\begin{bmatrix} 1& \ \ \ 1&1 \ \ \ \ \ \ \ \ \ \ \ \ \ 1&0&0 \\ 1& -1& 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0&1&0 \\ 1&\ \ \ 0& -1 \ \ \ \ \ \ \ \ \ \ \ 0&0&1 \\ \end{bmatrix} ^{-1}$ = $\begin{Bmatrix} r_1=\frac{r_1+r_2+r_3}{3} \\ r_2=\frac{r_1+r_3-2\times r_2}{3} \\ r_3=\frac{r_1+r_2-2\times r_3}{3} \end{Bmatrix}$ =

$A^{-1}=\begin{bmatrix} 1&0&0\ \ \ \ \ \ \ \ \ \ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 0&1&0\ \ \ \ \ \ \ \ \ \ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ 0&0&1\ \ \ \ \ \ \ \ \ \ \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{bmatrix} ;$

$T=\begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{bmatrix} \begin{bmatrix} 4& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \\ \end{bmatrix};$

$T=\begin{bmatrix} \frac{4}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{4}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{4}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{bmatrix} ;$ .