Calculate the change-of-basis matrix between two bases of \(\IR^n\text{.}\)
Subsection5.5.1Warm Up
Subsection5.5.2Class Activities
Remark5.5.1.
So far, when working with the Euclidean vector space \(\IR^n\text{,}\) we have primarily worked with the standard basis \(\mathcal{E}=\setList{\vec{e}_1,\dots, \vec{e}_n}\text{.}\) We can explore alternative perspectives more easily if we expand our toolkit to analyze different bases.
Activity5.5.2.
Let \(\cal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}=\setList{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}1\\-1\\1\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}}\text{.}\)
(a)
Is \(\cal{B}\) a basis of \(\IR^3\text{?}\)
Yes.
No.
(b)
Since \(\cal{B}\) is a basis, we know that if \(\vec{v}\in \IR^3\text{,}\) the following vector equation have will have a unique solution:
Given this, we define a map \(C_{\mathcal{B}}\colon\IR^3\to\IR^3\) via the rule that \(C_{\mathcal{B}}(\vec{v})\) is equal to the unique solution to the above vector equation. The map \(C_{\mathcal{B}}\) is a linear map.
Compute \(C_\mathcal{B}(\vec{e}_1),C_\mathcal{B}(\vec{e}_2), C_\mathcal{B}(\vec{e}_3)\) and, in doing so, write down the standard matrix \(M_\mathcal{B}\) of \(C_\mathcal{B}\text{.}\)
Observation5.5.3.
Note that one way to compute \(M_{\mathcal{B}}\) is calculate the RREF of the following matrix:
Given a basis \(\cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n}\) of \(\IR^n\text{,}\) the change of basis/coordinate transformation is the transformation \(C_\mathcal{B}\colon\IR^n\to\IR^n\) defined by the property that, for any vector \(\vec{v}\in\IR^n\text{,}\) the vector \(C_\mathcal{B}(\vec{v})\) is the unique solution to the vector equation:
More precisely, \(C_\mathcal{B}\) is the change-of-coordinate map from the standard basis \(\{\vec{e}_1,\dots, \vec{e}_n\}\)to the basis \(\mathcal{B}\text{;}\) the vector \(C_\mathcal{B}(\vec{v})\) is the \(\mathcal{B}\)-coordinates of \(\vec{v}\text{.}\) If you work with standard coordinates, and I work with \(\mathcal{B}\)-coordinates, then to build the vector that you call \(\vec{v}=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n\text{,}\) I would first compute \(C_\mathcal{B}(\vec{v})=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\) and then build \(\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n\text{.}\)
Let \(\vec{v}_1=\begin{bmatrix}1\\-2\\1\end{bmatrix},\ \vec{v}_2=\begin{bmatrix}-1\\0\\3\end{bmatrix},\ \vec{v}_3=\begin{bmatrix}0\\1\\-1\end{bmatrix}\text{,}\) and \(\mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}\)
(a)
Calculate \(M_{\mathcal{B}}\) using technology.
(b)
Use your result to calculate \(C_\mathcal{B}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right)\) and express the vector \(\begin{bmatrix}1\\1\\1\end{bmatrix}\) as a linear combination of \(\vec{v}_1,\vec{v}_2,\vec{v}_3\text{.}\)
Observation5.5.6.
Let \(T\colon\IR^n\to\IR^n\) be a linear transformation and let \(A\) denote its standard matrix. If \(\cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n}\) is some other basis, then we have:
In other words, the matrix \(M_{\mathcal{B}}AM_{\mathcal{B}}^{-1}\) is the matrix whose columns consist of \(\mathcal{B}\)-coordinate vectors of the image vectors \(T(\vec{v}_i)\text{.}\) The matrix \(M_{\mathcal{B}}AM_{\mathcal{B}}^{-1}\) is called the matrix of \(T\) with respect to \(\mathcal{B}\)-coordinates.
Activity5.5.7.
Let \(\mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}=\setList{\begin{bmatrix}1\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\0\\3\end{bmatrix},\begin{bmatrix}0\\1\\-1\end{bmatrix}}\) be basis from the previous Activity. Let \(T\) denote the linear transformation whose standard matrix is given by:
Calculate the matrix \(M_\mathcal{B}AM_{\mathcal{B}}^{-1}\text{.}\)
(b)
The matrix \(A\) describes how the standard basis of \(\IR^3\) is transformed by the linear transformation \(T\text{.}\) The matrix \(M_\mathcal{B}AM_{\mathcal{B}}^{-1}\) describes how \(\cal{B}\) is transformed (in \(\mathcal{B}\)-coordinates). Which of these two descriptions is easier for you to describe/understand/visualize and why?