Opening the book…
- 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.
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.
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 \(\PageIndex{1}\), 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, \(T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] = \left [ \begin{array}{r} -4 \\ -6 \\ 2 \\ -12 \end{array} \right ]\).
Watch the accompanying video for a visual demonstration of this concept.
- Video Length: 2 minutes 17 seconds.
- Context: This video demonstrates how to use the properties of a linear transformation:
Suppose \(T\) is a linear transformation with \(T(\vec u) = \langle 5, 2 \rangle\) and \(T(\vec v) = \langle -5, 2 \rangle\). Find the following:
- \(T(-\vec u)\)
- \(T(5\vec u - 6\vec v )\)
Suppose two linear transformations act in the same way on \(\vec{x}\) for all vectors. Then we say that these transformations are equal.
Let \(S\) and \(T\) be linear transformations from \(\mathbb{R}^n\) to \(\mathbb{R}^m\). Then \(S = T\) if and only if for every \(\vec{x} \in \mathbb{R}^n\), \[S \left( \vec{x} \right) = T \left( \vec{x} \right)\nonumber \]
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.
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 \(S\).
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.
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: \[\left [ \begin{array}{rr} 2 & 3 \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{r} 9 \\ 2 \end{array} \right ] = \left [ \begin{array}{r} 24 \\ 2 \end{array} \right ]\nonumber \]
Watch the accompanying video for a visual demonstration of the this concept.
- Video Length: 7 minutes 41 seconds.
- Context: This video demonstrates how to find the standard matrices for the composition of transformations:
Let \(T_1 \left [ \begin{array}{r} x_1 \\ x_2 \end{array} \right ] = \left [ \begin{array}{r} 10x_1 - 3x_2 \\ 27x_1 - 8x_2 \end{array} \right ]\) and \(T_2 \left [ \begin{array}{r} x_1 \\ x_2 \end{array} \right ] = \left [ \begin{array}{r} 1x_1 + 0x_2 \\ -3x_1 + 1x_2 \end{array} \right ]\). Write standard matrices for \( T_1, T_2, \text{ and for } T_1 \circ T_2 \).
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.
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.
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.
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 \(A^{-1}\).
This page titled 5.3: Properties of Linear Transformations was last modified on Thu, 10 Apr 2025 05:35:56 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.
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