Linear Algebra Examples

$S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} a - b - c \\ a - b + c \\ a + b + 5 c \end{array}]$

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

S: $ℝ^{3} \to ℝ^{3}$

First prove the transform preserves this property.

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

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

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

Add the two matrices.

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

Apply the transformation to the vector.

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

Simplify each element in the matrix.

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

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

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

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

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

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

Break the result into two matrices by grouping the variables.

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

The addition property of the transformation holds true.

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

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

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

Factor the $p$ from each element.

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Multiply $p$ by each element in the matrix.

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

Apply the transformation to the vector.

$S (p x) = [\begin{array}{c} (p a) - (p b) - (p c) \\ (p a) - (p b) + p c \\ p a + p b + 5 (p c) \end{array}]$

Simplify each element in the matrix.

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Rearrange $(p a) - (p b) - (p c)$ .

$S (p x) = [\begin{array}{c} p a - p b - p c \\ (p a) - (p b) + p c \\ p a + p b + 5 (p c) \end{array}]$

Rearrange $(p a) - (p b) + p c$ .

$S (p x) = [\begin{array}{c} p a - p b - p c \\ p a - p b + p c \\ p a + p b + 5 (p c) \end{array}]$

Rearrange $p a + p b + 5 (p c)$ .

$S (p x) = [\begin{array}{c} p a - p b - p c \\ p a - p b + p c \\ p a + p b + 5 p c \end{array}]$

Factor each element of the matrix.

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Factor element $0, 0$ by multiplying $p a - p b - p c$ .

$S (p x) = [\begin{array}{c} p (a - b - c) \\ p a - p b + p c \\ p a + p b + 5 p c \end{array}]$

Factor element $1, 0$ by multiplying $p a - p b + p c$ .

$S (p x) = [\begin{array}{c} p (a - b - c) \\ p (a - b + c) \\ p a + p b + 5 p c \end{array}]$

Factor element $2, 0$ by multiplying $p a + p b + 5 p c$ .

$S (p x) = [\begin{array}{c} p (a - b - c) \\ p (a - b + c) \\ p (a + b + 5 c) \end{array}]$

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

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

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

$S (0) = 0$

Apply the transformation to the vector.

$S (0) = [\begin{array}{c} (0) - (0) - (0) \\ (0) - (0) + 0 \\ 0 + 0 + 5 (0) \end{array}]$

Simplify each element in the matrix.

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Rearrange $(0) - (0) - (0)$ .

$S (0) = [\begin{array}{c} 0 \\ (0) - (0) + 0 \\ 0 + 0 + 5 (0) \end{array}]$

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

$S (0) = [\begin{array}{c} 0 \\ 0 \\ 0 + 0 + 5 (0) \end{array}]$

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

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

The zero vector is preserved by the transformation.

$S (0) = 0$

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

Linear Transformation

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