Linear transformation matrix wrto 2 basis $\beta = \{ (1,1,0), (1,0,1),(0,1,1)\}$ and $\beta' = \{ (2,1,1), (1,2,1),(-1,1,1)\}$

Question

Let $\beta = \{ (1,1,0), (1,0,1),(0,1,1)\}$ and $\beta' = \{ (2,1,1), (1,2,1),(-1,1,1)\}$ be two ordered basis of $R^3$. Then find a linear transformation $T:R^3 \rightarrow R^3$ which transforms from $\beta \rightarrow \beta'$. Use this matrix representation to find $T(x)$ , where $x=(2,3,1)$

My attempt:

Let $\bar e $ be standard basis

Let input basis be $x_1, x_2, x_3$ and output basis $\beta'= {y_1,y_2,y_3}$

$T(1,1,0) = (2,1,1)_{\bar e} = 1y_1 $

$T(1,0,1) = (1,2,1)_{\bar e} = 1y_2 $

$T(0,1,1) = (-1,1,1)_{\bar e} = 1y_3 $

So $[T]_{\beta}^{\beta'} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} $

can you pls tell me is this matrix of linear transformation correct?

Now $(2,3,1)=a(1,1,0)+b(1,0,1)+c(0,1,1) = (a+b, a+c,b+c) \implies a = 2, b=0, c=1$

$T(2,3,1) = T([2,0,1]_{\beta}) = 2 T(x_1)+0T(x_2)+1T(x_3)=(3,3,3)_{\bar e}$ i..e, wrto standard basis T(2,3,1) = (3,3,3)

Pls correct me if i am going wrong.

Answer 1

Following your method, we are looking for a matrix $T$ which transform vectors from the basis $\beta=\{x_1, x_2, x_3\}$ to the basis $\beta'=\{y_1, y_2, y_3\}$, the $T$ is such that

• $Tx_{1,\beta}= T(1,0,0)=(a_1,b_1,c_1)=x_{1,\beta'}$

• $Tx_{2,\beta}= T(0,1,0)=(a_2,b_2,c_2)=x_{2,\beta'}$

• $Tx_{3,\beta}= T(0,0,1)=(a_3,b_3,c_3)=x_{3,\beta'}$

thus we have

$$[T]_{\beta}^{\beta'} = \begin{bmatrix} a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \end{bmatrix}$$

where

$$a_1y_1+b_1y_2+c_1y_3=x_1$$ $$a_2y_1+b_2y_2+c_2y_3=x_2$$ $$a_3y_1+b_3y_2+c_3y_3=x_3$$

which can be written in matrix form as

$$N=[y_1\,y_2\,y_3]$$

$$M=[x_1\,x_2\,x_3]$$

$$NT=M\implies T=N^{-1}M$$

Notably we have

$$N=\begin{bmatrix} 2 & 1 & -1\\ 1 & 2 & 1\\ 1 & 1 & 1\end{bmatrix} \implies N^{-1} =\begin{bmatrix} \frac13 & -\frac23 & 1\\ 0 & 1 & -1\\ -\frac13 & -\frac13 & 1\end{bmatrix}$$

$$M=\begin{bmatrix} 1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix}$$

$$\implies T=N^{-1}M=\begin{bmatrix} \frac13 & -\frac23 & 1\\ 0 & 1 & -1\\ -\frac13 & -\frac13 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix}=\begin{bmatrix} -\frac13 & \frac43 & \frac13\\ 1 & -1 & 0\\ -\frac23 & \frac23 & \frac23\end{bmatrix}$$

Answer 2

Let solve the problem in matrix notation.

At first consider the matrix $M$ wich columns are the vectors of $\beta = \{ (1,1,0), (1,0,1),(0,1,1)\}$ that is

$$M=\begin{bmatrix} 1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix}$$

which, that's the key point, represent the change of basis from $\beta $ to the standard basis, thus

$$v_e=Mv_\beta \implies v_\beta=M^{-1}v_e$$

Now write down the corresponding matrix $N$ for the base $\beta^*$ and we have

$$N=\begin{bmatrix} 2 & 1 & -1\\ 1 & 2 & 1\\ 1 & 1 & 1\end{bmatrix}$$

$$v_e=Nv{_\beta*}\implies v_\beta*=N^{-1}v_e$$

Thus the transformation from $\beta \rightarrow \beta^*$ is espressed by

$$v_\beta*=N^{-1}v_e \quad v_e=Mv_\beta \implies v_\beta*=N^{-1}Mv_\beta $$

that is the matrix

$$[T]_{\beta}^{\beta'} = N^{-1}M=\begin{bmatrix} -\frac13 & \frac43 & \frac13\\ 1 & -1 & 0\\ -\frac23 & \frac23 & \frac23\end{bmatrix}$$