Suppose that \( A \) is an \( m \times n \) matrix, and let \( T \) be the associated matrix transformation \( T(\bbm x) = A\bbm x \). Then, \( T \) has these two properties:
To understand why \( T \) has these properties, we apply the properties of matrix-vector multiplication that we learned in Lecture 9. For vector addition, \[ T(\bbm u + \bbm v) = A(\bbm u + \bbm v) = A\bbm u + A\bbm v = T(\bbm u) + T(\bbm v). \]
For scalar multiplication, \[ T(c\ \bbm u) = A(c\ \bbm u) = c(A\bbm u) = c\ T(\bbm u). \]
We say that a transformation is "linear" if it has these two "respects" properties.
Definition. A transformation \( T : \mathbb R^n \to \mathbb R^m \) is a linear transformation if it has the following two properties:
We have seen that every matrix transformation is a linear transformation, but we don't yet know whether every linear transformation must be a matrix transformation for some matrix. We will consider this question in the next lecture.
Example 1. Let \( T : \mathbb R^2 \to \mathbb R^3 \) be defined by \( T(x_1,x_2) = (2x_2, x_1 - x_2, 3x_1) \). Prove that \( T \) is a linear transformation.
We must first show that \( T(\bbm u + \bbm v) = T(\bbm u) + T(\bbm v) \) for all \( \bbm u, \bbm v \in \mathbb R^n \). It will not be enough to choose specific numbers for the entries of \( \bbm u \) and \( \bbm v \); we must use generic vectors \( \bbm u = \vectwo {u_1} {u_2} \) and \( \bbm v = \vectwo {v_1} {v_2} \) so that our argument works no matter what vectors \( \bbm u \) and \( \bbm v \) are: \[ T(\bbm u + \bbm v) = T\left( \vectwo {u_1} {u_2} + \vectwo {v_1} {v_2} \right) = T\left( \vectwo {u_1+v_1} {u_2+v_2} \right) = \vecthree {2(u_2 + v_2)} {(u_1+v_1)-(u_2+v_2)} {3(u_1+v_1)} = \vecthree {2u_2 + 2v_2} {(u_1-u_2)+(v_1-v_2)} {3u_1 + 3v_1} = \vecthree {2u_2} {u_1-u_2} {3u_1} + \vecthree {2v_2} {v_1-v_2} {3v_1} = T(\bbm u) + T(\bbm v). \]
Next, we follow a similar process for showing that \( T(c\ \bbm u) = c\ T(\bbm u) \) for all \( \bbm u \in \mathbb R^n \) and all scalars \( c\): \[ T(c\ \bbm u) = T\left(c \vectwo {u_1} {u_2}\right) = T\left( \vectwo {c\ u_1} {c\ u_2}\right) = \vecthree {2(c u_1)} {cu_1-cu_2} {3(cu_1)} = \vecthree {c(2u_1)} {c(u_1-u_2)} {c(3u_1)} = c \vecthree {2u_2} {u_1-u_2} {3u_1} = c\ T(\bbm u).\ \Box \]
Example 2. Let \( T : \mathbb R^3 \to \mathbb R^2 \) be given by \( T\left( \vecthree {x_1}{x_2}{x_3} \right) = \vectwo {x_1+3x_2} {|x_1|-x_3} \). Prove that \( T \) is not a linear transformation.
To show that \( T \) is not linear, we need only show that \( T \) does not respect vector addition or that \( T\) does not respect scalar multiplication, since \( T \) must have both "respects" properties in order to be linear. To do this, we must either find an example of vectors \( \bbm u \) and \(\bbm v\) for which \( T(\bbm u + \bbm v) \ne T(\bbm u) + T(\bbm v) \) or a vector \( \bbm u \) and a scalar \( c\) for which \( T(c\ \bbm u) \ne c\ T(\bbm u) \)
The absolute value signs around \( x_1 \) in the formula for \( T \) might lead us to think about using negative values in the first entry to find the example we need. For example, if we let \( \bbm u = \vecthree {-1} 2 3 \) and \(\bbm u = \vecthree 456 \), we have: \[ T(\bbm u + \bbm v) = T\left( \vecthree 3 7 9 \right) = \vectwo {3+3(7)} {|3|-9} = \vectwo {24} {-6}, \quad \mbox{but} \quad T(\bbm u) + T(\bbm v) = T\left( \vecthree {-1} 2 3 \right) + T \left( \vecthree 456 \right) = \vectwo {(-1)+3(2)} {|-1|-3} + \vectwo {4+3(5)} {|4|-6} = \vectwo 5 {-2} + \vectwo {19} {-2} = \vectwo {24} {-4}. \]
Since, for this example, \( T(\bbm u + \bbm v) \ne T(\bbm u) + T(\bbm v) \), this shows that \( T \) does not respect vector addition and is therefore not linear.
Alternatively, we could have tried to show that \( T \) does not respect scalar multiplication. For example, let \( \bbm u = \vecthree {-1} 2 3 \) and \( c = -2 \): \[ T(c\ \bbm u) = T\left(-2 \vecthree {-1} 2 3\right) = T\left( \vecthree 2 {-4} {-6}\right) = \vectwo {2+3(-4)} {|2|-(-6)} = \vectwo {-10} 8, \quad \mbox{but} \quad c\ T(\bbm u) = -2 T\left(\vecthree {-1} 2 3\right) = -2 \vectwo {-1+3(2)} {|-1|-3} = -2 \vectwo {5} {-2} = \vectwo {-10} 4. \]
Since, for this example, \( T(c\ \bbm u) \ne c\ T(\bbm u) \), this shows that \( T \) does not respect scalar multiplication and is therefore not linear. \( \Box \)
Keep in mind the difference between what is required to show that a given transformation is linear or not linear:
In addition to the "respects" properties that define linear transformations, they have some additional useful properties.
Theorem (Properties of Linear Transformations). Suppose that \( T : \mathbb R^n \to \mathbb R^m \) is a linear transformation. Then,
Proof. For (1), let \( \bbm u \) be any vector in \( \mathbb R^n \). Then, \[ T(\bbm 0) = T(\bbm u + (-\bbm u)) = T(\bbm u) + T(-\bbm u) = T(\bbm u) - T(\bbm u) = \bbm 0. \]
For (2), let vectors \( \bbm u, \bbm v \in \mathbb R^n \) and scalars \( c\) and \( d\) be given. Then, \[ T(c\ \bbm u + d\ \bbm v) = T(c\ \bbm u) + T(d\ \bbm v) = c\ T(\bbm u) + d\ T(\bbm v).\ \Box \]
In fact, property (2) of the previous theorem can be extended, using mathematical induction, to show that \( T \) respects all linear combinations: \[ T(c_1 \bbm v_1 + \cdots + c_p \bbm v_p) = c_1 T(\bbm v_1) + \cdots + c_p T(\bbm v_p). \]
To illustrate the meaning of the properties of linear transformations, suppose that we have a company that produces two products, X and Y. Suppose further that, for each product, the company budgets money for materials, labor, and overhead:
| Cost per unit | Product X | Product Y |
|---|---|---|
| Materials | $0.45 | $0.40 |
| Labor | $0.25 | $0.35 |
| Overhead | $0.15 | $0.15 |
This naturally leads to a matrix \( A = \begin{bmatrix} 0.45 & 0.4 \\ 0.25 & 0.35 \\ 0.15 & 0.15 \end{bmatrix} \). If \( \bbm v = \vectwo x y \) is a production vector that represents how much of each unit we plan to product, the total cost in each category is given by \( T(\bbm v) = A\bbm v \).
Suppose that \( \bbm j = \vectwo {150} {90} \) represents the production in January and \( \bbm f = \vectwo {120} {200} \) represents production in February. How could we compute the combined costs for these two months? We could compute the costs separately and add: \( T(\bbm j) + T(\bbm f) \). However, since \( T \) is linear, we can also add these production vectors and compute the cost using that vector: \( T(\bbm j + \bbm f) \).
Suppose that our planned production for March is given by \( \bbm m = \vectwo {130} {105} \). What would be the cost of doubling production? We could double \( \bbm m \) and then compute the cost: \( T(2 \bbm m) \). However, since \( T \) is linear, we could also compute the cost using \( \bbm m \) and then double the result: \( 2 T(\bbm m) \).
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