Question

Let B1 ={(1,1),(1,2)} and B2 ={(1,0),(2,1)}. Find the matrix of the change of basis from B1 to B2

Answer 1

Answer on Question #42864 – Math - Linear Algebra

Let $B1 = \{(1,1),(1,2)\}$ and $B2 = \{(1,0),(2,1)\}$. Find the matrix of the change of basis from B1 to B2

Solution

To find the matrix $P_{B_2 \leftarrow B_1}$ of the change of basis from $B_1 = \{e_1; e_2\} = \left\{\binom{1}{1}, \binom{1}{2}\right\}$ to $B_2 = \{u_1; u_2\} = \left\{\binom{1}{0}, \binom{2}{1}\right\}$, we first express the basis vectors

$$u_1 = \alpha_1 e_1 + \alpha_2 e_2$$ $$u_2 = \beta_1 e_1 + \beta_2 e_2 \quad \text{or}$$ $$\binom{1}{0} = \alpha_1 \binom{1}{1} + \alpha_2 \binom{1}{2},$$ $$\binom{2}{1} = \beta_1 \binom{1}{1} + \beta_2 \binom{1}{2},$$

It means

$$\begin{array}{cc} \binom{1}{1} & \binom{1}{2} \binom{\alpha_1}{\alpha_2} = \binom{1}{0}, \\ \binom{1}{1} & \binom{1}{2} \binom{\beta_1}{\beta_2} = \binom{2}{1}. \end{array}$$

Multiply equalities by the corresponding inverse matrix $\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}^{-1}$ and get

$$\begin{array}{l} \binom{\alpha_1}{\alpha_2} = \binom{1}{1} \binom{1}{2} ^{-1} \binom{1}{0}, \\ \binom{\beta_1}{\beta_2} = \binom{1}{1} \binom{1}{2} ^{-1} \binom{2}{1} \end{array}$$

Find matrix $\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}^{-1}$ by row reduction:

$$\begin{array}{cccccc} \binom{1}{1} & \binom{1}{2} & 0 & 0 & \sim & R_1 \leftarrow R_1 \\ & & & & & R_2 \leftarrow R_2 - R_1 \end{array} \sim \begin{array}{cccccc} \binom{1}{0} & \binom{1}{1} & 1 & 0 & \sim & R_1 \leftarrow R_1 - R_2 \\ & & & & & R_2 \leftarrow R_2 \end{array} \sim \begin{array}{cccccc} \binom{1}{0} & 0 & 2 & -1 \\ 0 & 1 & -1 & 1 \end{array}.$$

So, $\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$.

Finally

$$\binom{\alpha_1}{\alpha_2} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \begin{pmatrix} \beta_1 \\ \beta_2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}^{-1} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}.$$

Thus, $\begin{pmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{pmatrix}^T = \begin{pmatrix} \alpha_1 & \beta_1 \\ \alpha_2 & \beta_2 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ -1 & -1 \end{pmatrix}$ is the matrix of change of basis from B1 to B2

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