Linear Algebra Examples

$T [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ y \\ 8 x + 4 y \end{array}]$

Step 1

The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre-image of the transformation).

$[\begin{array}{c} x \\ y \\ 8 x + 4 y \end{array}] = 0$

Step 2

Create a system of equations from the vector equation.

$x = 0$

$y = 0$

$8 x + 4 y = 0$

Step 3

Write the system as a matrix.

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 8 & 4 & 0 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.1.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 8 - 8 \cdot 1 & 4 - 8 \cdot 0 & 0 - 8 \cdot 0 \end{array}]$

Step 4.1.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 0 \end{array}]$

Step 4.2

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{3} = R_{3} - 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & 0 - 4 \cdot 0 \end{array}]$

Step 4.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 5

Use the result matrix to declare the final solution to the system of equations.

$x = 0$

$y = 0$

$0 = 0$

Step 6

Write a solution vector by solving in terms of the free variables in each row.

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

Step 7

Write as a solution set.

${[\begin{array}{c} 0 \\ 0 \end{array}]}$

Step 8

The kernel of $T$ is the subspace ${[\begin{array}{c} 0 \\ 0 \end{array}]}$ .

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