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3.6 Linear Operators (Transformation)

Recall that a function takes a member and maps it to another number. But an operator maps a function to another.
E.G: \(A = R[x]\). If \(f\in A,\, D(f) = g(x) \) where \(g = f'\,,\hspace {0.3cm} R[x] \longrightarrow R[x] \hspace {0.3cm} f\longmapsto g\).

Definition 3.6.1. A function \(T: \mathbb {R}^n \longrightarrow \mathbb {R}^m\) is said to be linear if:

(i)
for any \(v,\, w\in \mathbb {R}^n\, ,\hspace {0.3cm} T(v + w) = T(v) + T(w)\)
(ii)
for any \(v\in \mathbb {R}^n\, ,\hspace {0.2cm} \lambda \in \mathbb {R}\,,\, T(\lambda v) = \lambda T(v)\)

The map \(T\) is often called a linear transformation or linear operator.

A number of properties from the definition.

Proposition 3.6.2. If \(T: \mathbb {R}^n \longrightarrow \mathbb {R}^m\) is linear, then \(T(O_{\mathbb {R}^n}) \longrightarrow O_{\mathbb {R}^m}\)

Proof. Consider the following equations \(\, O_{\mathbb {R}^n} + O_{\mathbb {R}^n} = O_{\mathbb {R}^n}\). Apply \(T\) on both sides
\(T(O_{\mathbb {R}^n} + O_{\mathbb {R}^n}) = T(O_{\mathbb {R}^n})\) \begin {align*} \implies & T(O_{\mathbb {R}^n}) + T(O_{\mathbb {R}^n}) = T(O_{\mathbb {R}^n})\\\\ \implies & T(O_{\mathbb {R}^n}) + T(O_{\mathbb {R}^n}) - T(O_{\mathbb {R}^n}) = T(O_{\mathbb {R}^n}) - T(O_{\mathbb {R}^n})\\\\ \implies & T(O_{\mathbb {R}^n}) = O_{\mathbb {R}^m}\\\\ \end {align*} □

Example 3.6.3.

1.
The identity map \(Id : \mathbb {R}^n \longrightarrow \mathbb {R}^n\) is a linear operator.

Check: Let \(u,\, v\in \mathbb {R}^n\)

(i)
\(Id (u + v) = u + v = Id(u) + Id (v)\)
(ii)
\(Id(\lambda u) = \lambda u = \lambda \, Id(u)\).
2.
Define \(T: \mathbb {R}^3 \longrightarrow \mathbb {R}^3\,\) by \(T\begin {pmatrix} x\\ y\\ z\\ \end {pmatrix} = \begin {pmatrix} x\\ y\\ -z\\ \end {pmatrix}\)

Check: Let \(\, \begin {pmatrix} x_1\\ y_1\\ z_1\\ \end {pmatrix}\, \hspace {0.2cm} , \, \hspace {0.2cm} \begin {pmatrix} x_2\\ y_2\\ z_2\\ \end {pmatrix} \in \mathbb {R}^3\). Then

(i)
\(T \begin {pmatrix} \begin {pmatrix} x_1\\ y_1\\ z_1\\ \end {pmatrix} + \begin {pmatrix} x_2\\ y_2\\ z_2\\ \end {pmatrix} \end {pmatrix} = T \begin {pmatrix} x_1 + x_2\\ y_1 + y_2\\ z_1 + z_2\\ \end {pmatrix} = \begin {pmatrix} x_1 + x_2\\ y_1 + y_2\\ -(z_1 + z_2)\\ \end {pmatrix} = \begin {pmatrix} x_1\\ y_1\\ -z_1\\ \end {pmatrix} + \begin {pmatrix} x_2\\ y_2\\ -z_2\\ \end {pmatrix}\)

\( = T\begin {pmatrix} x_1\\ y_1\\ z_1\\ \end {pmatrix} + T\begin {pmatrix} x_2\\ y_2\\ z_2\\ \end {pmatrix}\)

(ii)
For a scalar \(\lambda \), \(\, T\begin {pmatrix} \lambda \begin {pmatrix} x\\ y\\ z\\ \end {pmatrix} \end {pmatrix} = T\begin {pmatrix} \lambda \, x\\ \lambda \, y\\ \lambda \, z\\ \end {pmatrix} = \begin {pmatrix} \lambda \, x\\ \lambda \, y\\ -\lambda \, z\\ \end {pmatrix} = \lambda \begin {pmatrix} x\\ y\\ -z\\ \end {pmatrix} = \lambda \, T\begin {pmatrix} x\\ y\\ z\\ \end {pmatrix}\)

3.
Consider an operator of a transformation from \(R[x]\) to \(R[x]\) defined by \[T : \begin {pmatrix} x_0\\ x_1\\ x_2\\ x_4\\ x_5\\ \end {pmatrix} \longmapsto \begin {pmatrix} 0\\ x_1\\ 2 x_2\\ 3x_3\\ 4 x_4\\ \end {pmatrix}\] Is this \(T\) a linear transformation on \(R[x]\)?
4.
The mapping \(\, T: \mathbb {R}^2 \longrightarrow \mathbb {R}^2\,\) defined by \(\, \displaystyle {T \binom {x}{y} = \binom {x}{x}}\,\) is linear.

Proof

(i)
Let \(\, \displaystyle {a = \binom {x_1}{y_1}\, , \, b = \binom {x_2}{y_2}\, , \hspace {0.3cm} a\, , \, b \in \mathbb {R}^2}\). Then \begin {align*} T(a + b) & = T\begin {pmatrix} \begin {pmatrix} x_1\\ y_1\\ \end {pmatrix} + \begin {pmatrix} x_2\\ y_2\\ \end {pmatrix} \end {pmatrix} = T \begin {pmatrix} x_1 + x_2\\ y_1 + y_2\\ \end {pmatrix} = \begin {pmatrix} x_1 + x_2\\ x_1 + x_2\\ \end {pmatrix}\\\\ & = \begin {pmatrix} x_1\\ x_1\\ \end {pmatrix} + \begin {pmatrix} x_2\\ x_2\\ \end {pmatrix} = T\binom {x_1}{y_1} + T\binom {x_2}{y_2}\\\\ & = T(a) + T(b) \end {align*}
(ii)
for \(\, \lambda \in \mathbb {R}\), we have \(T(\lambda \, a) = T\begin {pmatrix} \lambda \, x_1\\ \lambda \, y_1\\ \end {pmatrix} = \begin {pmatrix} \lambda \, x_1\\ \lambda \, x_1\\ \end {pmatrix} = \lambda \begin {pmatrix} x_1\\ x_1\\ \end {pmatrix} = \lambda \, T\begin {pmatrix} x_1\\ y_1\\ \end {pmatrix} = \lambda \, T(a)\)

\(\therefore \, T\) as defined is linear.

5.
\(T : \mathbb {R}^2 \longrightarrow \mathbb {R}^2\,\) defined by \(\, \displaystyle {\binom {x}{y} \longmapsto \binom {x - 1}{y - 1}}\,\) is not linear.

Let \(\, \displaystyle {a = \binom {x_1}{y_1}\, \hspace {0.3cm} , \, \hspace {0.3cm} b = \binom {x_2}{y_2}\, , \hspace {0.3cm} a\, ,\, b \in \mathbb {R}^2}\). Then \begin {align*} T(a + b) & = T\begin {pmatrix} x_1 + x_2\\ y_1 + y_2\\ \end {pmatrix} = \begin {pmatrix} x_1 + x_2 - 1\\ y_1 + y_2 - 1\\ \end {pmatrix} = \begin {pmatrix} x_1\\ y_1\\ \end {pmatrix} + \begin {pmatrix} x_2 - 1\\ y_2 - 1\\ \end {pmatrix}\\\\ & \neq \begin {pmatrix} x_1 - 1\\ y_1 - 1\\ \end {pmatrix} + \begin {pmatrix} x_2 - 1\\ y_2 - 1\\ \end {pmatrix} = T\begin {pmatrix} x_1\\ y_1\\ \end {pmatrix} + T\begin {pmatrix} x_2\\ y_2\\ \end {pmatrix} = T(a) + T(b) \end {align*}

\(\therefore \, T \) is not linear.

6.
\(T : \mathbb {R}^2 \longrightarrow \mathbb {R}^2\,\) defined by \(\displaystyle {\binom {x}{y} \longmapsto \binom {x}{x^2}}\,\) is not linear.

 

Proof. Let \(\, a\, ,\, b\in \mathbb {R}^2\,\) such that \(\, \displaystyle {a = \binom {x_1}{y_1}\, \hspace {0.3cm},\hspace {0.3cm} b = \binom {x_2}{y_2}}\). Then □

\begin {align*} T(a + b) & = T\begin {pmatrix} x_1 + x_2\\ y_1 + y_2\\ \end {pmatrix} = \begin {pmatrix} x_1 + x_2\\ x_1^2 + x_2^2 + 2x_1x_2\\ \end {pmatrix}\\\\ & \neq \begin {pmatrix} x_1 + x_2\\ x_1^2 + x_2^2\\ \end {pmatrix} = \begin {pmatrix} x_1\\ x_1^2\\ \end {pmatrix} + \begin {pmatrix} x_2\\ x_2^2\\ \end {pmatrix} = T\binom {x_1}{y_1} + T{x_2}{y_2} = T(a) + T(b) \end {align*}

\(\therefore \, T\) as defined is not linear.


Example 3.6.4. Let \(A\) be an \(m\times n\) matrix. Then \(A\) defines a linear map \(\, L_A : \mathbb {R}^n \longrightarrow \mathbb {R}^m\) \((``\) left multiplication by \(A")\) denoted by \(\, L_A(v) = A_v,\, \) for \(\, v\in \mathbb {R}^n\).

Theorem 3.6.5 (Recall the fundamental Homomorphism). \(\phi : G \longrightarrow H, \, G\) and \(H\) groups, \(\, \, G\big /ker\, \phi \, \cong \, H\).

Proof. Since \(ker\, \phi \) is an ideal of \(R,\hspace {0.2cm} R/ker\, \phi \,\) is a ring. Recall that a ring is a group under addition. \(Im\, \phi \) is also a subring thus it is a group under addition.
Now, define a map \(\, h = R/ker\, \phi \longrightarrow Im\, \phi \).
\(k + R \, \longrightarrow \, \phi (R)\).
We check that indeed \(h\) as defined respects the map operations. let \(k_1 + R\) and \(k_2 + R\) be in \(\, R/ker\, \phi \) \begin {align*} h(k_1 + R)(k_2 + R) & = h(k_1k_2 + k_1R + Rk_2 + R)\\ & = h(k_1k_2 + R)\\ & = \phi (R) \end {align*}

Clearly, \(h\) is well behaved. We involve the fundamental homonorphism theorem, we have that \(h\) is a isomorphism and therefore \(\, R/\ker \, \phi \cong Im\, \phi \).
□

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