Learning Outcomes
- Calculate the change-of-basis matrix between two bases of \(\IR^n\text{.}\)
Section 5.5 Change of Basis (GT5)
Subsection 5.5.1 Warm Up
Subsection 5.5.2 Class Activities
Remark 5.5.1.
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.
Activity 5.5.2.
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:
\begin{equation*}
x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\vec{v}
\end{equation*}
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 \(C_{\mathcal{B}}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right).\)
(c)
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 \(C_\mathcal{B}\text{.}\)
Observation 5.5.3.
Note that one way to compute \(M_{\mathcal{B}}\) is calculate the RREF of the following matrix:
\begin{equation*}
\left[\begin{array}{ccc|ccc}1&1&0&1&0&0\\
0&-1&1&0&1&0\\
1&1&1&0&0&1\end{array}\right]\sim\left[\begin{array}{ccc|ccc}1&0&0&2&1&-1\\
0&1&0&-1&-1&1\\
0&0&1&-1&0&1\end{array}\right]
\end{equation*}
Thus, the matrix \(M_{\mathcal{B}}\) is the inverse of the matrix \([\vec{v}_1\ \vec{v}_2\ \vec{v}_3]\text{.}\) That is:
\begin{equation*}
M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \vec{v}_2\ \vec{v}_3].
\end{equation*}
Definition 5.5.4.
Given a basis \(\cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n}\) of \(\IR^n\text{,}\) the change of basis/coordinate transformation from the standard basis to \(\mathcal{B}\) 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:
\begin{equation*}
x_1\vec{v}_1+\dots+x_n\vec{v}_n=\vec{v}.
\end{equation*}
Its standard matrix is called the change-of-basis matrix from the standard basis to \(\mathcal{B}\) and is denoted by \(M_{\mathcal{B}}\text{.}\) It satisfies the following:
\begin{equation*}
M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \cdots\ \vec{v}_n].
\end{equation*}
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 \(\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n\text{.}\)
In particular, notation as above, we would have:
\begin{equation*}
a_1\vec{e}_1+\cdots a_n\vec{e}_n=\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n.
\end{equation*}
Activity 5.5.5.
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 \(\vec{v}_1,\vec{v}_2,\vec{v}_3\text{.}\)
Observation 5.5.6.
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:
\begin{align*}
M_\mathcal{B}AM_{\mathcal{B}}^{-1} \amp= M_\mathcal{B}A[\vec{v_1}\cdots\vec{v}_n] \\
\amp= M_\mathcal{B}[T(\vec{v}_1)\cdots T(\vec{v}_n)]\\
\amp= [C_\mathcal{B}(T(\vec{v}_1))\cdots C_\mathcal{B}(T(\vec{v}_n))]
\end{align*}
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.
Activity 5.5.7.
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:
\begin{equation*}
A=\begin{bmatrix}9&4&4\\6&9&2\\-18&-16&-9\end{bmatrix}.
\end{equation*}
(a)
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?
Subsection 5.5.3 Sample Problem and Solution
Example 5.5.8.
Let \(\mathcal{B}=\setList{\begin{bmatrix}-2\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\-2\\-1\end{bmatrix},\begin{bmatrix}1\\3\\2\end{bmatrix}}\text{,}\) and \(\vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}\text{.}\)
(a)
Explain and demonstrate how to verify that \(\mathcal{B}\) is a basis of \(\IR^3\) and how to calculate \(M_\mathcal{B}\text{,}\) the change-of-basis matrix from the standard basis of \(\IR^3\) to \(\mathcal{B}\text{.}\)
(b)
Explain and demonstrate how to use \(M_\mathcal{B}\) to express \(\vec{v}\) in terms of \(\mathcal{B}\)-basis vectors.
Solution.
(a)
We can accomplish both tasks by calculating the RREF of the following matrix:
\begin{equation*}
\RREF\left[\begin{array}{ccc|ccc}-2&-1&1&1&0&0\\
-2&-2&3&0&1&0\\1&-1&2&0&0&1\end{array}\right]
=
\left[\begin{array}{ccc|ccc}1&0&0&1&-1&1\\
0&1&0&-7&5&-4\\
0&0&1&-4& 3&-2\end{array}\right].
\end{equation*}
The fact that the matrix to the left of the vertical bar is the identity matrix tells that \(\mathcal{B}\) is a basis. The matrix on the right hand side of the bar is equal to the change-of-basis matrix:
\begin{equation*}
M_\mathcal{B}=\left[\begin{array}{ccc}1&-1&1\\
-7&5&-4\\
-4& 3&-2\end{array}\right].
\end{equation*}
(b)
By definition of the change of basis matrix, if \(\vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}\text{,}\) then the coordinates of \(\vec{v}\) with respect to \(\mathcal{B}\) are given by:
\begin{equation*}
M_\mathcal{B}\vec{v}=M_\mathcal{B}=\left[\begin{array}{ccc}1&-1&1\\
-7&5&-4\\
-4& 3&-2\end{array}\right]\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}2\\-9\\-4\end{bmatrix}.
\end{equation*}
It follows that:
\begin{equation*}
\begin{bmatrix}1\\2\\3\end{bmatrix}=2\begin{bmatrix}-2\\-2\\1\end{bmatrix} -9\begin{bmatrix}-1\\-2\\-1\end{bmatrix} -4\begin{bmatrix}1\\3\\2\end{bmatrix}.
\end{equation*}
