Linear Algebra Examples

$T ([\begin{array}{c} a \\ b \\ d \end{array}]) = [\begin{array}{c} 2 a - b + 5 d \\ - 4 a + 2 b - 10 d \end{array}]$

Step 1

The transformation defines a map from $ℝ^{3}$ to $ℝ^{2}$ . To prove the transformation is linear, the transformation must preserve scalar multiplication, addition, and the zero vector.

T: $ℝ^{3} \to ℝ^{2}$

Step 2

First prove the transform preserves this property.

$T (x + y) = T (x) + T (y)$

Step 3

Set up two matrices to test the addition property is preserved for $T$ .

$T ([\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}] + [\begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array}])$

Step 4

Add the two matrices.

$T [\begin{array}{c} x_{1} + y_{1} \\ x_{2} + y_{2} \\ x_{3} + y_{3} \end{array}]$

Step 5

Apply the transformation to the vector.

$T (x + y) = [\begin{array}{c} 2 (x_{1} + y_{1}) - (x_{2} + y_{2}) + 5 (x_{3} + y_{3}) \\ - 4 (x_{1} + y_{1}) + 2 (x_{2} + y_{2}) - 10 (x_{3} + y_{3}) \end{array}]$

Step 6

Simplify each element in the matrix.

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Step 6.1

Rearrange $2 (x_{1} + y_{1}) - (x_{2} + y_{2}) + 5 (x_{3} + y_{3})$ .

$T (x + y) = [\begin{array}{c} 2 x_{1} - x_{2} + 5 x_{3} + 2 y_{1} - y_{2} + 5 y_{3} \\ - 4 (x_{1} + y_{1}) + 2 (x_{2} + y_{2}) - 10 (x_{3} + y_{3}) \end{array}]$

Step 6.2

Rearrange $- 4 (x_{1} + y_{1}) + 2 (x_{2} + y_{2}) - 10 (x_{3} + y_{3})$ .

$T (x + y) = [\begin{array}{c} 2 x_{1} - x_{2} + 5 x_{3} + 2 y_{1} - y_{2} + 5 y_{3} \\ - 4 x_{1} + 2 x_{2} - 10 x_{3} - 4 y_{1} + 2 y_{2} - 10 y_{3} \end{array}]$

Step 7

Break the result into two matrices by grouping the variables.

$T (x + y) = [\begin{array}{c} 2 x_{1} - x_{2} + 5 x_{3} \\ - 4 x_{1} + 2 x_{2} - 10 x_{3} \end{array}] + [\begin{array}{c} 2 y_{1} - y_{2} + 5 y_{3} \\ - 4 y_{1} + 2 y_{2} - 10 y_{3} \end{array}]$

Step 8

The addition property of the transformation holds true.

$T (x + y) = T (x) + T (y)$

Step 9

For a transformation to be linear, it must maintain scalar multiplication.

$T (p x) = T (p [\begin{array}{c} a \\ b \\ d \end{array}])$

Step 10

Factor the $p$ from each element.

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Step 10.1

Multiply $p$ by each element in the matrix.

$T (p x) = T ([\begin{array}{c} p a \\ p b \\ p d \end{array}])$

Step 10.2

Apply the transformation to the vector.

$T (p x) = [\begin{array}{c} 2 ((p a) - (p b) + 5 (p d)) \\ - 4 ((p a) + 2 (p b) - 10 (p d)) \end{array}]$

Step 10.3

Simplify each element in the matrix.

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Step 10.3.1

Rearrange $2 ((p a) - (p b) + 5 (p d))$ .

$T (p x) = [\begin{array}{c} 2 a p - 2 b p + 10 d p \\ - 4 ((p a) + 2 (p b) - 10 (p d)) \end{array}]$

Step 10.3.2

Rearrange $- 4 ((p a) + 2 (p b) - 10 (p d))$ .

$T (p x) = [\begin{array}{c} 2 a p - 2 b p + 10 d p \\ - 4 a p - 8 b p + 40 d p \end{array}]$

Step 10.4

Factor each element of the matrix.

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Step 10.4.1

Factor element $0, 0$ by multiplying $2 a p - 2 b p + 10 d p$ .

$T (p x) = [\begin{array}{c} p (2 a - 2 b + 10 d) \\ - 4 a p - 8 b p + 40 d p \end{array}]$

Step 10.4.2

Factor element $1, 0$ by multiplying $- 4 a p - 8 b p + 40 d p$ .

$T (p x) = [\begin{array}{c} p (2 a - 2 b + 10 d) \\ p (- 4 a - 8 b + 40 d) \end{array}]$

Step 11

The second property of linear transformations is preserved in this transformation.

$T (p [\begin{array}{c} a \\ b \\ d \end{array}]) = p T (x)$

Step 12

For the transformation to be linear, the zero vector must be preserved.

$T (0) = 0$

Step 13

Apply the transformation to the vector.

$T (0) = [\begin{array}{c} 2 (0) - (0) + 5 (0) \\ - 4 \cdot 0 + 2 (0) - 10 \cdot 0 \end{array}]$

Step 14

Simplify each element in the matrix.

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Step 14.1

Rearrange $2 (0) - (0) + 5 (0)$ .

$T (0) = [\begin{array}{c} 0 \\ - 4 \cdot 0 + 2 (0) - 10 \cdot 0 \end{array}]$

Step 14.2

Rearrange $- 4 \cdot 0 + 2 (0) - 10 \cdot 0$ .

$T (0) = [\begin{array}{c} 0 \\ 0 \end{array}]$

Step 15

The zero vector is preserved by the transformation.

$T (0) = 0$

Step 16

Since all three properties of linear transformations are not met, this is not a linear transformation.

Linear Transformation