Linear Algebra Examples

$x - [\begin{array}{c} 0 & 1 & 5 & - 6 \\ 5 & 8 & 7 & - 1 \end{array}] = [\begin{array}{c} 1 & 2 & 3 & 6 \\ - 8 & - 4 & - 2 & - 3 \end{array}]$

Step 1

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

x: $ℝ^{2} \to ℝ^{2}$

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 $x$ .

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

Step 4

Add the two matrices.

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

Step 5

Apply the transformation to the vector.

$x (x + y) = [\begin{array}{c} 1 \\ - 8 \end{array}]$

Step 6

Break the result into two matrices by grouping the variables.

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

Step 7

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

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