5.3: Properties of Linear Transformations

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, $T \left [ \begin{array}{r} -7 \\ 3 \\ -9 \end{array} \right ] = \left [ \begin{array}{r} -4 \\ -6 \\ 2 \\ -12 \end{array} \right ]$.

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:

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

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 $A^{-1}$.