6.3: Properties of Linear Transformations
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Outcomes
- Use properties of linear transformations to solve problems.
- Find the composite of transformations and the inverse of a transformation.
Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. Then there are some important properties of \(T\) which will be examined in this section. Consider the following theorem.
Theorem \(\PageIndex{1}\): Properties of Linear Transformations
Properties of Linear Transformationsproperties Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation and let \(\vec{x} \in \mathbb{R}^n\).
- \(T\) preserves the zero vector. \[T(0\vec{x}) = 0 T(\vec{x}). \mbox{ Hence }T(\vec{0}) = \vec{0}\nonumber\]
- \(T\) preserves the negative of a vector: \[T( (-1)\vec{x})=(-1)T(\vec{x}). \mbox{ Hence }T(-\vec{x}) = -T(\vec{x}).\nonumber\]
- \(T\) preserves linear combinations: \[\mbox{Let }\vec{x}_1, ..., \vec{x}_k \in \mathbb{R}^n \mbox{ and }a_1, ..., a_k \in \mathbb{R}.\nonumber\] \[\mbox{Then if }\vec{y} = a_1\vec{x}_1 + a_2\vec{x}_2 + ...+a_k \vec{x}_k, \mbox{it follows that }\nonumber\] \[T(\vec{y}) = T(a_1\vec{x}_1 + a_2\vec{x}_2 + ...+a_k \vec{x}_k) = a_1T(\vec{x}_1) + a_2T(\vec{x}_2) + ...+a_k T(\vec{x}_k).\nonumber\]
These properties are useful in determining the action of a transformation on a given vector. Consider the following example.
Example \(\PageIndex{1}\): Linear Combination
Let \(T:\mathbb{R}^3 \mapsto \mathbb{R}^4\) be a linear transformation such that \[T \left [ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right ] = \left [ \begin{array}{r} 4 \\ 4 \\ 0 \\ -2 \end{array} \right ], T \left [ \begin{array}{r} 4 \\ 0 \\ 5 \end{array} \right ] = \left [ \begin{array}{r} 4 \\ 5 \\ -1 \\ 5 \end{array} \right ]\nonumber\] Find \(T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ]\).
Solution
Using the third property in Theorem 9.6.1 , we can find \(T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ]\) by writing \(\left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ]\) as a linear combination of \(\left [ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right ]\) and \(\left [ \begin{array}{r} 4 \\ 0 \\ 5 \end{array} \right ]\).
Therefore we want to find \(a,b \in \mathbb{R}\) such that \[\left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] = a \left [ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right ] + b \left [ \begin{array}{r} 4 \\ 0 \\ 5 \end{array} \right ]\nonumber \]
The necessary augmented matrix and resulting reduced row-echelon form are given by: \[\left [ \begin{array}{rr|r} 1 & 4 & -7 \\ 3 & 0 & 3 \\ 1 & 5 & -9 \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array} \right ]\nonumber \]
Hence \(a = 1, b = -2\) and \[\left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] = 1 \left [ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right ] + (-2) \left [ \begin{array}{r} 4 \\ 0 \\ 5 \end{array} \right ]\nonumber \]
Now, using the third property above, we have \[\begin{aligned} T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] &=T \left( 1 \left [ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right ] + (-2) \left [ \begin{array}{r} 4 \\ 0 \\ 5 \end{array} \right ] \right) \\ &= 1T \left [ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right ] -2T \left [ \begin{array}{r} 4 \\ 0 \\ 5 \end{array} \right ] \\ &= \left [ \begin{array}{r} 4 \\ 4 \\ 0 \\ -2 \end{array} \right ] -2 \left [ \begin{array}{r} 4 \\ 5 \\ -1 \\ 5 \end{array} \right ] \\ &= \left [ \begin{array}{r} -4 \\ -6 \\ 2 \\ -12 \end{array} \right ]\end{aligned}\]
Therefore, \(T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] = \left [ \begin{array}{r} -4 \\ -6 \\ 2 \\ -12 \end{array} \right ]\).
Suppose two linear transformations act in the same way on \(\vec{x}\) for all vectors. Then we say that these transformations are equal.
Suppose two linear transformations act on the same vector \(\vec{x}\), first the transformation \(T\) and then a second transformation given by \(S\). We can find the composite transformation that results from applying both transformations.
Definition \(\PageIndex{2}\): Composition of Linear Transformations
Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Then the composite of \(S\) and \(T\) is \[S \circ T: \mathbb{R}^k \mapsto \mathbb{R}^m\nonumber\] The action of \(S \circ T\) is given by \[(S \circ T) (\vec{x}) = S(T(\vec{x})) \; \mbox{for all} \; \vec{x} \in \mathbb{R}^k\nonumber\]
Notice that the resulting vector will be in \(\mathbb{R}^m\). Be careful to observe the order of transformations. We write \(S \circ T\) but apply the transformation \(T\) first, followed by \(S\).
Theorem \(\PageIndex{2}\): Composition of Transformations
Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations such that \(T\) is induced by the matrix \(A\) and \(S\) is induced by the matrix \(B\). Then \(S \circ T\) is a linear transformation which is induced by the matrix \(BA\).
Consider the following example.
Example \(\PageIndex{2}\): Composition of Transformations
Let \(T\) be a linear transformation induced by the matrix \[A = \left [ \begin{array}{rr} 1 & 2 \\ 2 & 0 \end{array} \right ]\nonumber \] and \(S\) a linear transformation induced by the matrix \[B = \left [ \begin{array}{rr} 2 & 3 \\ 0 & 1 \end{array} \right ]\nonumber \] Find the matrix of the composite transformation \(S \circ T\). Then, find \((S \circ T)(\vec{x})\) for \(\vec{x} = \left [ \begin{array}{r} 1 \\ 4 \end{array} \right ]\).
Solution
By Theorem \(\PageIndex{2}\) , the matrix of \(S \circ T\) is given by \(BA\). \[BA = \left [ \begin{array}{rr} 2 & 3 \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{rr} 1 & 2 \\ 2 & 0 \end{array} \right ] = \left [ \begin{array}{rr} 8 & 4 \\ 2 & 0 \end{array} \right ]\nonumber \]
To find \((S \circ T)(\vec{x})\), multiply \(\vec{x}\) by \(BA\) as follows \[\left [ \begin{array}{rr} 8 & 4 \\ 2 & 0 \end{array} \right ] \left [ \begin{array}{rr} 1 \\ 4 \end{array} \right ] = \left [ \begin{array}{r} 24 \\ 2 \end{array} \right ]\nonumber \]
To check, first determine \(T(\vec{x})\): \[\left [ \begin{array}{rr} 1 & 2 \\ 2 & 0 \end{array} \right ] \left [ \begin{array}{r} 1 \\ 4 \end{array} \right ] = \left [ \begin{array}{r} 9 \\ 2 \end{array} \right ]\nonumber \]
Then, compute \(S(T(\vec{x}))\) as follows: \[\left [ \begin{array}{rr} 2 & 3 \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{r} 9 \\ 2 \end{array} \right ] = \left [ \begin{array}{r} 24 \\ 2 \end{array} \right ]\nonumber \]
Consider a composite transformation \(S \circ T\), and suppose that this transformation acted such that \((S \circ T) (\vec{x}) = \vec{x}\). That is, the transformation \(S\) took the vector \(T(\vec{x})\) and returned it to \(\vec{x}\). In this case, \(S\) and \(T\) are inverses of each other. Consider the following definition.
Definition \(\PageIndex{3}\): Inverse of a Transformation
Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^n\) and \(S:\mathbb{R}^n \mapsto \mathbb{R}^n\) be linear transformations. Suppose that for each \(\vec{x} \in \mathbb{R}^n\), \[(S \circ T)(\vec{x}) = \vec{x}\nonumber \] and \[(T \circ S)(\vec{x}) = \vec{x}\nonumber \] Then, \(S\) is called an inverse of \(T\) and \(T\) is called an inverse of \(S\). Geometrically, they reverse the action of each other.
The following theorem is crucial, as it claims that the above inverse transformations are unique.
Theorem \(\PageIndex{3}\): Inverse of a Transformation
Let \(T:\mathbb{R}^n \mapsto \mathbb{R}^n\) be a linear transformation induced by the matrix \(A\). Then \(T\) has an inverse transformation if and only if the matrix \(A\) is invertible. In this case, the inverse transformation is unique and denoted \(T^{-1}: \mathbb{R}^n \mapsto \mathbb{R}^n\). \(T^{-1}\) is induced by the matrix \(A^{-1}\).
Consider the following example.
Example \(\PageIndex{3}\): Inverse of a Transformation
Let \(T: \mathbb{R}^2 \mapsto \mathbb{R}^2\) be a linear transformation induced by the matrix \[A = \left [ \begin{array}{rr} 2 & 3 \\ 3 & 4 \end{array} \right ]\nonumber\] Show that \(T^{-1}\) exists and find the matrix \(B\) which it is induced by.
Solution
Since the matrix \(A\) is invertible, it follows that the transformation \(T\) is invertible. Therefore, \(T^{-1}\) exists.
You can verify that \(A^{-1}\) is given by: \[A^{-1} = \left [ \begin{array}{rr} -4 & 3 \\ 3 & -2 \end{array} \right ]\nonumber \] Therefore the linear transformation \(T^{-1}\) is induced by the matrix \(A^{-1}\).