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.
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 \(x_1,x_2,x_3\text{?}\)
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:
Since the solution vector \(C_{\mathcal B}(\vec v)=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}\) describes the β\(\mathcal B\)-coordinatesβ of \(\vec v\text{,}\) we will write
The vector \(C_\mathcal{B}(\vec{v})\) describes the β\(\mathcal{B}\)-coordinatesβ of \(\vec{v}\text{.}\) If you work with standard coordinates, and I work with \(\mathcal{B}\)-coordinates, then you might write
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 \(\mathcal{B}=\setList{\vec{b}_1,\vec{b}_2,\vec{b}_3}\)
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, 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.
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:
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).
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 \(P^{-1}\text{.}\)
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:
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.
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.