Question 4.3.7: The equation x1 + x2 + x3 = 0 defines a plane V in R^3. In t......
The equation x_1 + x_2 + x_3 = 0 defines a plane V in ℝ^3. In this plane, consider the two bases
𝔄 = (\vec{a}_1, \vec{a}_2)=\left(\begin{bmatrix}0\\1\\−1\end{bmatrix},\begin{bmatrix}1\\0\\−1\end{bmatrix}\right) and 𝔅 =(\vec{b}_1, \vec{b}_2)=\left(\begin{bmatrix}1\\2\\−3\end{bmatrix},\begin{bmatrix}4\\-1\\−3\end{bmatrix}\right).
a. Find the change of basis matrix S from 𝔅 to 𝔄.
b. Verify that the matrix S in part (a) satisfies the equation
[\vec{b}_1 \ \ \vec{b}_2]=[\vec{a}_1 \ \ \vec{a}_2]S.
a. By inspection, we find that \vec{b}_1 = 2\vec{a}_1 + \vec{a}_2 and \vec{b}_2 = −\vec{a}_1 +4\vec{a}_2. so that
\vec{b}_1 \ \ \ \vec{b}_2
S_{ \mathfrak{B}\rightarrow \mathfrak{A} }=\left [ \begin{matrix} &2&-1&\\&1&4& \end{matrix} \right ] \begin{matrix} \ \ \ \vec{a}_1 \\ \ \ \ \vec{a}_2\end{matrix}.
b. We can verify that
\begin{bmatrix}\vec{b}_1&\vec{b}_2\end{bmatrix}=\begin{bmatrix}1&4\\2&-1\\-3&-3\end{bmatrix}=\begin{bmatrix}0&1\\1&0\\-1&-1\end{bmatrix}\begin{bmatrix}2&-1\\1&4\end{bmatrix}=\begin{bmatrix}\vec{a}_1&\vec{a}_2\end{bmatrix}S.
This equation reflects the fact that the two columns of S are the coordinate vectors of \vec{b}_1 and \vec{b}_2 with respect to the basis 𝔄=(\vec{a}_1, \vec{a}_2). We can illustrate this equation with a commutative diagram, where \vec{x} represents a vector in V:
Let us remind ourselves where the equation \vec{x} =\begin{bmatrix}\vec{b}_1&\vec{b}_2\end{bmatrix} [\vec{x}]_𝔅 comes from: If c_1, c_2 are the coordinates of \vec{x} with respect to 𝔅, then
\vec{x} = c_1\vec{b}_1 + c_2\vec{b}_2 =\begin{bmatrix}\vec{b}_1&\vec{b}_2\end{bmatrix} \begin{bmatrix}c_1\\c_2\end{bmatrix}=\begin{bmatrix}\vec{b}_1&\vec{b}_2\end{bmatrix}[\vec{x}]_𝔅.