Learning Outcomes
-
Calculate the change-of-basis matrix for the standard basis to a non-standard basis of \(\mathbb R^n\text{.}\)
Section 4.3 Change of Basis (MX3)
Subsection 4.3.1 Warm Up
Activity 4.3.1.
Let \(T\colon\IR^4\to\IR^4\) be the linear bijection given by the standard matrix:
\begin{equation*}
A=\left[\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 1 & 0 & -1 & -4 \\ 1 & 1 & 0 & -4 \\ 1 & -1 & -1 & 2 \end{array}\right].
\end{equation*}
(a)
If \(\vec{b}=\begin{bmatrix}1\\0\\-1\\2\end{bmatrix}\text{,}\) what is the meaning of the vector \(T^{-1}(\vec{b})\text{?}\)
(b)
Explain and demonstrate how to find the third column of \(A^{-1}\text{.}\)
Subsection 4.3.2 Class Activities
Remark 4.3.2.
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.
An interactive that visualizes the change of basis from the standard basis to a custom basis in the plane.
Activity 4.3.3.
Consider the non-standard basis
\begin{equation*}
\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}=\setList{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}1\\-1\\1\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}}
\end{equation*}
for \(\mathbb R^3\text{.}\)
(a)
Since \(\mathcal{B}\) is a basis, how many ways can we write some arbitrary \(\vec v\in \IR^3\) in terms of \(\mathcal B\) vectors?
\begin{equation*}
\vec v= x_1\vec{b}_1+x_2\vec{b}_2+x_3\vec{b}_3=
x_1\begin{bmatrix}1\\0\\1\end{bmatrix}+
x_2\begin{bmatrix}1\\-1\\1\end{bmatrix}+
x_3\begin{bmatrix}0\\1\\1\end{bmatrix}
\end{equation*}
-
Zero
-
At most one
-
Exactly one
-
At least one
-
Infinitely-many
(b)
If \(\vec x=\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]\) and \(B = \left[\begin{array}{ccc}\vec b_1& \vec b_2&\vec b_3\end{array}\right]=\left[\begin{array}{ccc}1&1&0\\0&-1&1\\1&1&1\end{array}\right]\text{,}\) which of these matrix equations can be used to find \(x_1,x_2,x_3\text{?}\)
-
\(\displaystyle \vec v=B\vec x\)
-
\(\displaystyle B\vec v=\vec x\)
-
\(\displaystyle \vec v=B^{-1}\vec x\)
-
\(\displaystyle B^{-1}\vec v=\vec x\)
-
A or D
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B or C
(c)
Let \(\vec v=\begin{bmatrix}1\\2\\3\end{bmatrix}\) and use this equation to find
\begin{equation*}
x_1=\unknown,x_2=\unknown,x_3=\unknown\text{.}
\end{equation*}
Definition 4.3.4.
Given a basis \(\mathcal{B}=\setList{\vec{b}_1,\dots, \vec{b}_n}\) of \(\IR^n\) and corresponding matrix \(B=\begin{bmatrix}\vec b_1&\cdots&\vec b_n\end{bmatrix}\text{,}\) 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 \(C_\mathcal{B}(\vec{v})\) describes the unique way to write \(\vec v\) in terms of the basis, that is, the unique solution to the vector equation:
\begin{equation*}
\vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n\text{.}
\end{equation*}
Since the solution vector \(C_{\mathcal B}(\vec v)=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\) describes the “\(\mathcal B\)-coordinates” of \(\vec v\text{,}\) we will write
\begin{equation*}
\vec v=x_1\vec{b}_1+\dots+x_n\vec{b}_n=
B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}_{\mathcal B}
\end{equation*}
Remark 4.3.5.
As was just shown, the standard matrix \(M_{\mathcal B}\) for this transformation is exactly the inverse matrix \(B^{-1}\text{.}\)
The vector \(C_\mathcal{B}(\vec{v})\) describes the “\(\mathcal{B}\)-coordinates” of \(\vec{v}\text{.}\) If you work with standard coordinates, and I work with \(\mathcal{B}\)-coordinates, then you might write
\begin{equation*}
\vec{v}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}
\end{equation*}
and I might instead write
\begin{equation*}
\vec{v}=x_1\vec{b}_1+\cdots+x_n\vec{b}_n=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}_{\mathcal B}=B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}
\end{equation*}
To convert from your standard coordinates to my \(\mathcal B\)-coordinates, we need simply compute:
\begin{equation*}
\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
C_{\mathcal B}\left(\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}\right)=
M_{\mathcal B}\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=
B^{-1}\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}\text{.}
\end{equation*}
And similarly we can convert backwards:
\begin{equation*}
\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=
C_{\mathcal B}^{-1}\left(\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\right)=
M_{\mathcal B}^{-1}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}=
B\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\text{.}
\end{equation*}
Activity 4.3.6.
Let \(\vec{b}_1=\begin{bmatrix}-1\\1\\2\end{bmatrix},\ \vec{b}_2=\begin{bmatrix}0\\-1\\-5\end{bmatrix},\ \vec{b}_3=\begin{bmatrix}-4\\2\\-1\end{bmatrix}\text{,}\) and \(\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}\)
(a)
Calculate \(M_{\mathcal{B}}=B^{-1}\) using technology.
(b)
Use this matrix to write
\begin{equation*}
\begin{bmatrix}-1\\0\\3\end{bmatrix}=\begin{bmatrix}\unknown\\\unknown\\\unknown\end{bmatrix}_{\mathcal B}=\unknown\vec b_1+\unknown\vec b_2+\unknown\vec b_3
\end{equation*}
as a linear combination of \(\vec{b}_1,\vec{b}_2,\vec{b}_3\text{.}\)
Activity 4.3.7.
While defining linear transformations in terms of their standard matrix \(A\) is convenient when working with standard coordinates, it would be helpful to be able to apply transformations directly to non-standard bases/coordinates as well.
Let \(\mathcal B=\setList{\vec b_1,\cdots\vec b_n}\) be a basis, and consider the matrix \(B=\begin{bmatrix}\vec b_1&\cdots&\vec b_n\end{bmatrix}\) \(M_{\mathcal B}=B^{-1}\text{.}\)
(a)
Given \(\vec v\) representing \(\mathcal B\)-coordinates, which of these expressions would correctly compute the transformation of \(\vec v\) by a standard matrix \(A\) with output in standard coordinates?
-
\(\displaystyle AB^{-1}\vec v=AM_{\mathcal B}\vec v\)
-
\(\displaystyle AB\vec v=AM_{\mathcal B}^{-1}\vec v\)
-
\(\displaystyle B^{-1}A\vec v=M_{\mathcal B}A\vec v\)
-
\(\displaystyle BA\vec v=M_{\mathcal B}^{-1}A\vec v\)
(b)
Therefore, which matrix would directly calculate the transformation of vectors by a linear map with standard matrix \(A\text{,}\) but where inputs and outputs are all given in \(\mathcal B\)-coordinates?
-
\(\displaystyle B^{-1}AB=M_{\mathcal B}AM_{\mathcal B}^{-1}\)
-
\(\displaystyle BAB^{-1}=M_{\mathcal B}^{-1}AM_{\mathcal B}\)
-
\(\displaystyle AB^{-1}B=AM_{\mathcal B}M_{\mathcal B}^{-1}\)
-
\(\displaystyle ABB^{-1}=AM_{\mathcal B}^{-1}M_{\mathcal B}\)
Observation 4.3.8.
Let \(T\colon\IR^n\to\IR^n\) be a linear transformation and let \(A\) denote its standard matrix. If \(\mathcal{B}=\setList{\vec{b}_1,\dots, \vec{v}_n}\) is some other basis and \(B=\begin{bmatrix}\vec b_1&\cdots&\vec b_n\end{bmatrix}\text{,}\) then \(M_{\mathcal B}AM_{\mathcal B}^{-1}=B^{-1}AB\) is the \(\mathcal B\)-coordinate matrix for \(T\text{,}\) which applies the transformation \(T\) where inputs and outputs are all given in \(\mathcal B\)-coordinates.
Activity 4.3.9.
Let \(\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}=\setList{\begin{bmatrix}1\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\0\\3\end{bmatrix},\begin{bmatrix}0\\1\\-1\end{bmatrix}}\) 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 \(T\) transforms the standard basis of \(\IR^3\text{.}\) The matrix \(M_\mathcal{B}AM_{\mathcal{B}}^{-1}\) describes how \(T\) transforms the basis \(\mathcal{B}\) (in \(\mathcal{B}\)-coordinates).
Which of these two descriptions of \(T\) is most helpful to you in describing/understanding/visualizing the transformation \(T\) and why?
Definition 4.3.10.
Suppose that \(A\) and \(B\) are two \(n\times n\) matrix. We say that \(A\) is similar to \(B\) if there exists an invertible matrix \(P\) that satisfies:
\begin{equation*}
PAP^{-1}=B.
\end{equation*}
The results of this section demonstrate that similar matrices can be viewed as describing the same linear transformation with respect to different bases. Specifically, if \(A\) describes a transformation with respect to the standard basis of \(\IR^n\text{,}\) then the matrix \(B\) describes the same linear transformation with respect to the basis consisting of the columns of \(P^{-1}\text{.}\)
Subsection 4.3.3 Individual Practice
Activity 4.3.11.
Suppose that \(T\colon\IR^3\to\IR^3\) is a linear transformation and you knew that \(\mathcal{B}=\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}\) was a basis of \(\IR^3\) that satisfied:
\begin{equation*}
T(\vec{v}_1)=3\vec{v}_1,\ T(\vec{v}_2)=-5\vec{v}_2,\ T(\vec{v}_3)=7\vec{v}_3.
\end{equation*}
If \(A\) is the standard matrix of \(T\text{,}\) do you have enough information to determine the matrix \(M_{\mathcal{B}}AM_{\mathcal{B}}^{-1}\text{?}\) If yes, write it down; if not, describe what additional information is needed.
Subsection 4.3.4 Videos
Video coming soon to this YouTube playlist.
Subsection 4.3.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/MX3/
Subsection 4.3.6 Mathematical Writing Explorations
Activity 4.3.12.
Suppose that \(A\) is similar to \(B\text{.}\) Prove that \(B\) is also similar to \(A\text{.}\) Thus, we may simply that \(A\) and \(B\) are similar matrices.
Subsection 4.3.7 Sample Problem and Solution
Sample problem Example B.1.20.
