Linear Algebra Examples

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

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.

x: $ℝ^{3} \to ℝ^{2}$

Step 2

First prove the transform preserves this property.

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

Step 3

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

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

Step 4

Add the two matrices.

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

Step 5

Apply the transformation to the vector.

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

Step 6

Simplify each element in the matrix.

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Step 6.1

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

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

Step 6.2

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

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

Step 7

Break the result into two matrices by grouping the variables.

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

Step 8

The addition property of the transformation holds true.

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

Step 9

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

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

Step 10

Factor the $p$ from each element.

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Step 10.1

Multiply $p$ by each element in the matrix.

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

Step 10.2

Apply the transformation to the vector.

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

Step 10.3

Simplify each element in the matrix.

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Step 10.3.1

Rearrange $2 (p a + p d)$ .

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

Step 10.3.2

Rearrange $- 4 (p b)$ .

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

Step 10.4

Factor each element of the matrix.

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Step 10.4.1

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

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

Step 10.4.2

Factor element $1, 0$ by multiplying $- 4 p b$ .

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

Step 11

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

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

Step 12

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

$x (0) = 0$

Step 13

Apply the transformation to the vector.

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

Step 14

Simplify each element in the matrix.

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Step 14.1

Rearrange $2 (0) + 0$ .

$x (0) = [\begin{array}{c} 0 \\ - 4 \cdot 0 \end{array}]$

Step 14.2

Rearrange $- 4 \cdot 0$ .

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

Step 15

The zero vector is preserved by the transformation.

$x (0) = 0$

Step 16

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

Linear Transformation