5.3: Properties of Linear Transformations

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Outcomes

1. Use properties of linear transformations to solve problems.
2. 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, $$T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] = \left [ \begin{array}{r} -4 \\ -6 \\ 2 \\ -12 \end{array} \right ]$$.

Below is a video on finding linear combinations of a given linear transformation using properties.

Suppose two linear transformations act in the same way on $$\vec{x}$$ for all vectors. Then we say that these transformations are equal.

Definition $$\PageIndex{1}$$: Equal Transformations

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.

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: $\left [ \begin{array}{rr} 2 & 3 \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{r} 9 \\ 2 \end{array} \right ] = \left [ \begin{array}{r} 24 \\ 2 \end{array} \right ]\nonumber$

Below is a video on finding standard matrices for transformations and a composition of transformations.

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 $$A^{-1}$$.

Below is a video on finding the transformation matrix in $$R^2$$ given two transformations using an inverse matrix.

Below is a video on finding the transformation matrix in $$R^3$$ given three transformations using an inverse matrix.

This page titled 5.3: Properties of Linear Transformations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .