TeachingLinear Algebra – Systems of linear equations – Matrix equations
Definition: If \(A\) is an \(m\times n\) matrix with columns \(\mathbf{a}_1,\mathbf{a}_2,\ldots,\mathbf{a}_n\) and \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}\in\mathbb{R}^n\), then the matrix-vector product \(A\mathbf{x}\) is defined as
\[A\mathbf{x}=\Bigg(\mathbf{a}_1\;\mathbf{a}_2\;\ldots\;\mathbf{a}_n\Bigg)\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix} =x_1\mathbf{a}_1+x_2\mathbf{a}_2+\cdots+x_n\mathbf{a}_n.\]Example: If \(A=\begin{pmatrix}1&1\\0&1\\-1&0\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}1\\2\\1\end{pmatrix}\) and \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}\in\mathbb{R}^2\), then we have:
\[A\mathbf{x}=\begin{pmatrix}1&1\\0&1\\-1&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=x_1\begin{pmatrix}1\\0\\-1\end{pmatrix} +x_2\begin{pmatrix}1\\1\\0\end{pmatrix}.\]Hence we have:
\[x_1\mathbf{a}_1+x_2\mathbf{a}_2=\mathbf{b}\quad\Longleftrightarrow\quad x_1\begin{pmatrix}1\\0\\-1\end{pmatrix}+x_2\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\2\\1\end{pmatrix} \quad\Longleftrightarrow\quad A\mathbf{x}=\mathbf{b}.\]Theorem: If \(A=\Bigg(\mathbf{a}_1\;\mathbf{a}_2\;\ldots\;\mathbf{a}_n\Bigg)\) is an \(m\times n\) matrix and \(\mathbf{b}\in\mathbb{R}^n\), then we have:
\[A\mathbf{x}=\mathbf{b}\quad\Longleftrightarrow\quad x_1\mathbf{a}_1+x_2\mathbf{a}_2+\cdots+x_n\mathbf{a}_n=\mathbf{b} \quad\Longleftrightarrow\quad\Bigg(\mathbf{a}_1\;\mathbf{a}_2\;\ldots\;\mathbf{a}_n\;\Bigg|\;\mathbf{b}\Bigg).\]Hence we have:\(A\mathbf{x}=\mathbf{b}\) is consistent \(\;\Longleftrightarrow\;\) \(\mathbf{b}\in\{\mathbf{a}_1,\mathbf{a}_2,\ldots,\mathbf{a}_n\}\).
Example: Let \(A=\begin{pmatrix}1&1\\0&1\\-1&0\end{pmatrix}\). For which \(\mathbf{b}\in\mathbb{R}^n\) is \(A\mathbf{x}\) consistent?
\[A\mathbf{x}=\mathbf{b}\quad\Longleftrightarrow\quad\left(\left.\begin{matrix}1&1\\0&1\\-1&0\end{matrix}\;\right|\;\begin{matrix}b_1\\b_2\\b_3\end{matrix}\right) \sim\left(\left.\begin{matrix}1&1\\0&1\\0&1\end{matrix}\;\right|\;\begin{matrix}b_1\\b_2\\b_1+b_3\end{matrix}\right) \sim\left(\left.\begin{matrix}1&1\\0&1\\0&0\end{matrix}\;\right|\;\begin{matrix}b_1\\b_2\\b_1-b_2+b_3\end{matrix}\right).\]Hence: \(A\mathbf{x}=\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\) is only consistent if \(b_1-b_2+b_3=0\) \(\;\Longrightarrow\;\) \(\left\{\begin{array}{l}b_1=b_2-b_3\\b_2\;\text{is free}\\b_3\;\text{is free.}\end{array}\right.\)
This impiies:
\[\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=\begin{pmatrix}b_2-b_3\\b_2\\b_3\end{pmatrix}=b_1\begin{pmatrix}1\\1\\0\end{pmatrix} +b_3\begin{pmatrix}-1\\0\\1\end{pmatrix}.\]The set \(\{\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\,:\,b_1-b_2+b_3=0\}=\text{Span}\left\{\begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}-1\\0\\1\end{pmatrix}\right\}\) is a plane in \(\mathbb{R}^3\) through the origin.
Theorem: Let \(A\) be an \(m\times n\) matrix. Then the following statements are logically equivalent:
- \(A\mathbf{x}=\mathbf{b}\) is consistent for every \(\mathbf{b}\in\mathbb{R}^m\).
- Each \(\mathbf{b}\in\mathbb{R}^m\) is a linear combination of the columns of \(A\).
- The columns of \(A\) span \(\mathbb{R}^m\).
- \(A\) has a pivot position in every row.
Computation of the matrix-vector product
Example: Let \(A=\begin{pmatrix}1&0&2\\-1&1&0\\0&2&-1\end{pmatrix}\) and \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\), then we have:
\[A\mathbf{x}=\begin{pmatrix}1&0&2\\-1&1&0\\0&2&-1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =x_1\begin{pmatrix}1\\-1\\0\end{pmatrix}+x_2\begin{pmatrix}0\\1\\2\end{pmatrix}+x_3\begin{pmatrix}2\\0\\-1\end{pmatrix} =\begin{pmatrix}x_1+2x_3\\-x_1+x_2\\2x_2-x_3\end{pmatrix}.\]Note that this is equal to
\[A\mathbf{x}=\begin{pmatrix}1&0&2\\-1&1&0\\0&2&-1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =\begin{pmatrix}1\cdot x_1+0\cdot x_2+2\cdot x_3\\-1\cdot x_1+1\cdot x_2+0\cdot x_3\\0\cdot x_1+2\cdot x_2-1\cdot x_3\end{pmatrix}=\begin{pmatrix}x_1+2x_3\\-x_1+x_2\\2x_2-x_3\end{pmatrix}.\]Example: Let \(I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\) and \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\), then we have:
\[I\mathbf{x}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\mathbf{x}.\]The matrix \(I\) is called an identity matrix.
Theorem: If \(A\) is an \(m\times n\) matrix, \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in \(\mathbb{R}^n\), and \(c\) is a scalar, then:
- \(A(\mathbf{u}+\mathbf{v})=A\mathbf{u}+A\mathbf{v}\);
- \(A(c\mathbf{u})=c(A\mathbf{u})\).
Last modified on March 22, 2021
