TeachingLinear Algebra – Eigenvalues and eigenvectors – Eigenvectors and linear transformations
Let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be a linear transformation with standard matrix \(A\). Then \(A\) is an \(m\times n\) matrix given by \(\Bigg(T(\mathbf{e}_1)\;\ldots\;T(\mathbf{e}_n)\Bigg)\) with \(\{\mathbf{e}_1,\ldots,\mathbf{e}_n\}\) the standard basis of \(\mathbb{R}^n\). Now suppose that \(\mathcal{B}=\{\mathbf{b}_1,\ldots,\mathbf{b}_n\}\) is any basis of \(\mathbb{R}^n\) and that \(\mathcal{C}=\{\mathbf{c}_1,\ldots,\mathbf{c}_m\}\) is any basis of \(\mathbb{R}^m\). Then: what is the connection between \([\mathbf{x}]_{\mathcal{B}}\) and \(\left[T(\mathbf{x})\right]_{\mathcal{C}}\) for an arbitrary vector \(\mathbf{x}\in\mathbb{R}^n\)?
If \(\mathbf{x}=r_1\mathbf{b}_1+\cdots+r_n\mathbf{b}_n\), then \([\mathbf{x}]_{\mathcal{B}}=\begin{pmatrix}r_1\\\vdots\\r_n\end{pmatrix}\) and \(T(\mathbf{x})=T(r_1\mathbf{b}_1+\cdots+r_n\mathbf{b}_n)=r_1T(\mathbf{b}_1)+\cdots+r_nT(\mathbf{b}_n)\). This implies that
\[\left[T(\mathbf{x})\right]_{\mathcal{C}}=r_1\left[T(\mathbf{b}_1)\right]_{\mathcal{C}}+\cdots+r_n\left[T(\mathbf{b}_n)\right]_{\mathcal{C}} \quad\Longrightarrow\quad\left[T(\mathbf{x})\right]_{\mathcal{C}}=M[\mathbf{x}]_{\mathcal{B}}\quad\text{with}\quad M=\Bigg(\left[T(\mathbf{b}_1)\right]_{\mathcal{C}}\;\ldots\;\left[T(\mathbf{b}_n)\right]_{\mathcal{C}}\Bigg).\]This matrix \(M\) is the matrix representation of the linear transformation \(T\) relative to the bases \(\mathcal{B}\) and \(\mathcal{C}\).
In the special case that \(m=n\) we have a linear transformation \(T:\mathbb{R}^n\to\mathbb{R}^n\) and we can choose the basis \(\mathcal{C}\) the same as \(\mathcal{B}\). In that case the matrix \(M=\left[T\right]_{\mathcal{B}}\) is called the matrix representation of the linear transformation \(T\) relative to the basis \(\mathcal{B}\) and we have: \(\left[T(\mathbf{x})\right]_{\mathcal{B}}=\left[T\right]_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\).
Theorem: Suppose that \(A=PDP^{-1}\), where \(D\) is a diagonal \(n\times n\) matrix. If \(\mathcal{B}\) is the basis of \(\mathbb{R}^n\) consisting of the columns of \(P\), then \(D\) is the matrix representation of the linear transformation \(\mathbf{x}\mapsto A\mathbf{x}\) relative to the basis \(\mathcal{B}\).
Proof: Let \(P=\Bigg(\mathbf{b}_1\;\ldots\;\mathbf{b}_n\Bigg)\) and \(\mathcal{B}=\{\mathbf{b}_1,\ldots,\mathbf{b}_n\}\). Then we have: \(P[\mathbf{x}]_{\mathcal{B}}=\mathbf{x}\) and \([\mathbf{x}]_{\mathcal{B}}=P^{-1}\mathbf{x}\). If \(T(\mathbf{x})=A\mathbf{x}\) for \(\mathbf{x}\in\mathbb{R}^n\), then:
\[\left[T\right]_{\mathcal{B}}=\Bigg(\left[T(\mathbf{b}_1)\right]_{\mathcal{B}}\;\ldots\;\left[T(\mathbf{b}_n)\right]_{\mathcal{B}}\Bigg) =\Bigg(\left[A\mathbf{b}_1\right]_{\mathcal{B}}\;\ldots\;\left[A\mathbf{b}_n\right]_{\mathcal{B}}\Bigg) =\Bigg(P^{-1}A\mathbf{b}_1\;\ldots\;P^{-1}A\mathbf{b}_n\Bigg)=P^{-1}A\Bigg(\mathbf{b}_1\;\ldots\;\mathbf{b}_n\Bigg)=P^{-1}AP.\]Since \(A=PDP^{-1}\), this implies that \(\left[T\right]_{\mathcal{B}}=P^{-1}AP=D\).
Note that the proof does not use the fact that \(D\) is a diagonal matrix. So, this matrix can be replaced by any matrix \(B\) that is similar to \(A\). In fact, if \(A=PBP^{-1}\) and \(\mathcal{B}\) is the basis of \(\mathbb{R}^n\) formed by the columns of \(P\), then \(\left[T\right]_{\mathcal{B}}=B\).
Last modified on April 5, 2021
