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Matrix Theory From Eigenvalues to Data Insights

Yilan Tan

Section 6.1 The matrix of a Linear map

Subsection 6.1.1 Linear Transformation

The definition of Linear Transformation.

Let \(V\) and \(W\) be vector spaces. The function \(T: V \rightarrow W\) is a linear transformation of \(V\) into \(W\) when the two properties below are true for all \(\mathbf{u}\) and \(\mathbf{v}\) in \(V\) and for any scalar \(c\text{,}\)
\begin{equation*} T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\quad\text{ and } \quad T(c \mathbf{u})=c T(\mathbf{u}). \end{equation*}
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Example 6.1.1.

Show that the map \(T:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}\) defined by
\begin{equation*} T\left(\begin{array}{r} x_1 \\ x_2 \\ x_3\\ x_4 \end{array}\right)=\left(\begin{array}{r} x_1+x_2 \\ x_2+3x_3+x_4 \\ -x_1+x_3-x_4\\ 2x_2+2x_3-2x_4 \end{array}\right) \end{equation*}
is a linear transformation.
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Thus it is a necessary condition for \(T\) to be a linear transformation.
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Example 6.1.3.

Is the map \(T:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}\) linear? where \(T\) is defined by
\begin{equation*} T\left(\begin{array}{r} x_1 \\ x_2 \\ x_3\\ x_4 \end{array}\right)=\left(\begin{array}{r} x_1+x_2+2\\ x_2+3x_3+x_4 \\ -x_1+x_3-x_4\\ 2x_2+2x_3-2x_4 \end{array}\right) \end{equation*}
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Example 6.1.4.

  1. Show that the map \(T:P_{2}(x)\rightarrow P_{2}(x)\) defined by
    \begin{equation*} T\left(f(x)\right)=(x-2)f'(x)+f(x) \end{equation*}
    is a linear transformation.
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  2. Compute all polynomials \(f(x)=x^{2}+bx+c\) such that \(T(f(x))=3f(x)\)
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To study the linear transformation, we only need to know the image of a basis vector of \(V\text{,}\) which explained that the linear transformation is the easiest type of function between two vector spaces.
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Proof.

\(\forall v\in V\text{,}\) since \(\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_k\}\) is a basis of \(V\text{,}\)
\begin{equation*} v = x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\ldots+x_{n}\mathbf{v}_{n} \end{equation*}
Then we have
\begin{align*} T(v)=\amp T(x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\ldots+x_{n}\mathbf{v}_{n})\\ =\amp T(x_{1}\mathbf{v}_{1})+T(x_{2}\mathbf{v}_{2})+\ldots+T(x_{n}\mathbf{v}_{n})\\ =\amp x_{1}T(\mathbf{v}_{1})+x_{2}T(\mathbf{v}_{2})+\ldots+x_{n}T(\mathbf{v}_{n}) \end{align*}
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In next subsection, we study the special case: \(V=\mathbb{R}^{n}\) and \(W = \mathbb{R}^{m}\text{.}\)
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Subsection 6.1.2 Linear Transformation \(T:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}\)

Standard matrix.

Let \(T\) be a linear transformation from \(\mathbb{R}^{n}\) to \(\mathbb{R}^{m}\text{,}\) and \(\{\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n\}\) be the standard basis of \(\mathbb{R}^{n}\text{.}\) Then \(\forall v\in V,\)
\begin{equation*} T(v)=Av, \end{equation*}
where \(A=\Big[T(\mathbf{e}_1)\, T(\mathbf{e}_2)\,\ldots\, T(\mathbf{e}_n)\Big]\text{.}\)
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The matrix \(A\) is called the standard matrix of \(T\text{.}\)
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Example: Let \(T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}\) be a linear transformation such that
\begin{equation*} T\left(\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right)=\left(\begin{array}{r} 1 \\ 3\\ 0 \end{array}\right),T\left(\begin{array}{r} 0 \\ 1 \\ 0 \end{array}\right)=\left(\begin{array}{r} 3 \\ 1 \\ 0 \end{array}\right),T\left(\begin{array}{r} 0 \\ 0 \\ 1 \end{array}\right)=\left(\begin{array}{r} 0 \\ 0 \\ -2 \end{array}\right). \end{equation*}
Find the standard matrix \(A\text{.}\)
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Exercise: Let \(T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}\) be a linear transformation such that
\begin{equation*} T\left(\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right)=\left(\begin{array}{r} 1 \\ 3\\ 0 \end{array}\right),T\left(\begin{array}{r} 1 \\ 1 \\ 0 \end{array}\right)=\left(\begin{array}{r} 4 \\ 4 \\ 0 \end{array}\right),T\left(\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{r} 4 \\ 4 \\ -2 \end{array}\right). \end{equation*}
Find the standard matrix \(A\text{.}\)
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Exercise: Find the standard matrix of the linear transformation in Example 6.1.1.
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Exercise: Find the standard matrix of the linear transformation in Example 6.1.4. In next chapter, you will find matrix is much easy to manipulate than the linear transformation.
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