Linear Algebra Examples

$p [\begin{array}{c} 1 \\ 5 \\ 8 \end{array}] = [\begin{array}{c} \frac{41}{14} \\ \frac{26}{14} \\ \frac{101}{14} \end{array}]$

Step 1

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.

p: $ℝ^{3} \to ℝ^{3}$

Step 2

First prove the transform preserves this property.

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

Step 3

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

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

Step 4

Add the two matrices.

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

Step 5

Apply the transformation to the vector.

$p (x + y) = [\begin{array}{c} \frac{41}{14} \\ \frac{26}{14} \\ \frac{101}{14} \end{array}]$

Step 6

Rearrange $\frac{26}{14}$ .

$p (x + y) = [\begin{array}{c} \frac{41}{14} \\ \frac{13}{7} \\ \frac{101}{14} \end{array}]$

Step 7

Break the result into two matrices by grouping the variables.

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

Step 8

Since the transformation addition property does not hold, this is not a linear transformation.

$p (x + y) \neq p (x) + p (y)$