Linear Algebra Examples

$A = [\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}]$

Step 1

Find the eigenvalues.

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Step 1.1

Set up the formula to find the characteristic equation

$p (λ)$ .

$p (λ) = determinant (A - λ I_{2})$

Step 1.2

The identity matrix or unit matrix of size

$2$ is the

$2 \times 2$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 1.3

Substitute the known values into

$p (λ) = determinant (A - λ I_{2})$ .

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Step 1.3.1

Substitute

$[\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}]$ for

$A$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] - λ I_{2})$

Step 1.3.2

Substitute

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

$I_{2}$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Multiply

$- λ$ by each element of the matrix.

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 0 - λ & 7 + 0 \\ \frac{1}{7} + 0 & 0 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Subtract

$λ$ from

$0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 7 + 0 \\ \frac{1}{7} + 0 & 0 - λ \end{array}]$

Step 1.4.3.2

Add

$7$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 7 \\ \frac{1}{7} + 0 & 0 - λ \end{array}]$

Step 1.4.3.3

Add

$\frac{1}{7}$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 7 \\ \frac{1}{7} & 0 - λ \end{array}]$

Step 1.4.3.4

Subtract

$λ$ from

$0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 7 \\ \frac{1}{7} & - λ \end{array}]$

Step 1.5

Find the determinant.

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Step 1.5.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = - λ (- λ) - \frac{1}{7} \cdot 7$

Step 1.5.2

Simplify each term.

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Step 1.5.2.1

Rewrite using the commutative property of multiplication.

$p (λ) = - 1 \cdot - 1 λ \cdot λ - \frac{1}{7} \cdot 7$

Step 1.5.2.2

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 1.5.2.2.1

Move

$λ$ .

$p (λ) = - 1 \cdot - 1 (λ \cdot λ) - \frac{1}{7} \cdot 7$

Step 1.5.2.2.2

Multiply

$λ$ by

$λ$ .

$p (λ) = - 1 \cdot - 1 λ^{2} - \frac{1}{7} \cdot 7$

Step 1.5.2.3

Multiply

$- 1$ by

$- 1$ .

$p (λ) = 1 λ^{2} - \frac{1}{7} \cdot 7$

Step 1.5.2.4

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = λ^{2} - \frac{1}{7} \cdot 7$

Step 1.5.2.5

Cancel the common factor of

$7$ .

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Step 1.5.2.5.1

Move the leading negative in

$- \frac{1}{7}$ into the numerator.

$p (λ) = λ^{2} + \frac{- 1}{7} \cdot 7$

Step 1.5.2.5.2

Cancel the common factor.

$p (λ) = λ^{2} + \frac{- 1}{7} \cdot 7$

Step 1.5.2.5.3

Rewrite the expression.

$p (λ) = λ^{2} - 1$

Step 1.6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$λ^{2} - 1 = 0$

Step 1.7

Solve for

$λ$ .

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Step 1.7.1

Add

$1$ to both sides of the equation.

$λ^{2} = 1$

Step 1.7.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$λ = \pm \sqrt{1}$

Step 1.7.3

Any root of

$1$ is

$1$ .

$λ = \pm 1$

Step 1.7.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.7.4.1

First, use the positive value of the

$\pm$ to find the first solution.

$λ = 1$

Step 1.7.4.2

Next, use the negative value of the

$\pm$ to find the second solution.

$λ = - 1$

Step 1.7.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$λ = 1, - 1$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where

$N$ is the null space and

$I$ is the identity matrix.

$ε_{A} = N (A - λ I_{2})$

Step 3

Find the eigenvector using the eigenvalue

$λ = 1$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] - [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Subtract the corresponding elements.

$[\begin{array}{c} 0 - 1 & 7 - 0 \\ \frac{1}{7} - 0 & 0 - 1 \end{array}]$

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Subtract

$1$ from

$0$ .

$[\begin{array}{c} - 1 & 7 - 0 \\ \frac{1}{7} - 0 & 0 - 1 \end{array}]$

Step 3.2.2.2

Subtract

$0$ from

$7$ .

$[\begin{array}{c} - 1 & 7 \\ \frac{1}{7} - 0 & 0 - 1 \end{array}]$

Step 3.2.2.3

Subtract

$0$ from

$\frac{1}{7}$ .

$[\begin{array}{c} - 1 & 7 \\ \frac{1}{7} & 0 - 1 \end{array}]$

Step 3.2.2.4

Subtract

$1$ from

$0$ .

$[\begin{array}{c} - 1 & 7 \\ \frac{1}{7} & - 1 \end{array}]$

Step 3.3

Find the null space when

$λ = 1$ .

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Step 3.3.1

Write as an augmented matrix for

$A x = 0$ .

$[\begin{array}{c} - 1 & 7 & 0 \\ \frac{1}{7} & - 1 & 0 \end{array}]$

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Multiply each element of

$R_{1}$ by

$- 1$ to make the entry at

$1, 1$ a

$1$ .

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Step 3.3.2.1.1

Multiply each element of

$R_{1}$ by

$- 1$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - - 1 & - 1 \cdot 7 & - 0 \\ \frac{1}{7} & - 1 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & - 7 & 0 \\ \frac{1}{7} & - 1 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation

$R_{2} = R_{2} - \frac{1}{7} R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 3.3.2.2.1

Perform the row operation

$R_{2} = R_{2} - \frac{1}{7} R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 7 & 0 \\ \frac{1}{7} - \frac{1}{7} \cdot 1 & - 1 - \frac{1}{7} \cdot - 7 & 0 - \frac{1}{7} \cdot 0 \end{array}]$

Step 3.3.2.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 7 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 3.3.3

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

$x - 7 y = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 7 y \\ y \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} 7 \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${y [\begin{array}{c} 7 \\ 1 \end{array}] | y \in R}$

Step 3.3.7

The solution is the set of vectors created from the free variables of the system.

${[\begin{array}{c} 7 \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue

$λ = - 1$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0 & 7 \\ \frac{1}{7} & 0 \end{array}] + [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 4.2

Simplify.

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Step 4.2.1

Add the corresponding elements.

$[\begin{array}{c} 0 + 1 & 7 + 0 \\ \frac{1}{7} + 0 & 0 + 1 \end{array}]$

Step 4.2.2

Simplify each element.

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Step 4.2.2.1

Add

$0$ and

$1$ .

$[\begin{array}{c} 1 & 7 + 0 \\ \frac{1}{7} + 0 & 0 + 1 \end{array}]$

Step 4.2.2.2

Add

$7$ and

$0$ .

$[\begin{array}{c} 1 & 7 \\ \frac{1}{7} + 0 & 0 + 1 \end{array}]$

Step 4.2.2.3

Add

$\frac{1}{7}$ and

$0$ .

$[\begin{array}{c} 1 & 7 \\ \frac{1}{7} & 0 + 1 \end{array}]$

Step 4.2.2.4

Add

$0$ and

$1$ .

$[\begin{array}{c} 1 & 7 \\ \frac{1}{7} & 1 \end{array}]$

Step 4.3

Find the null space when

$λ = - 1$ .

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Step 4.3.1

Write as an augmented matrix for

$A x = 0$ .

$[\begin{array}{c} 1 & 7 & 0 \\ \frac{1}{7} & 1 & 0 \end{array}]$

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

Perform the row operation

$R_{2} = R_{2} - \frac{1}{7} R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 4.3.2.1.1

Perform the row operation

$R_{2} = R_{2} - \frac{1}{7} R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & 7 & 0 \\ \frac{1}{7} - \frac{1}{7} \cdot 1 & 1 - \frac{1}{7} \cdot 7 & 0 - \frac{1}{7} \cdot 0 \end{array}]$

Step 4.3.2.1.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 7 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x + 7 y = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 7 y \\ y \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} - 7 \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${y [\begin{array}{c} - 7 \\ 1 \end{array}] | y \in R}$

Step 4.3.7

The solution is the set of vectors created from the free variables of the system.

${[\begin{array}{c} - 7 \\ 1 \end{array}]}$

Step 5

The eigenspace of

$A$ is the list of the vector space for each eigenvalue.

${[\begin{array}{c} 7 \\ 1 \end{array}], [\begin{array}{c} - 7 \\ 1 \end{array}]}$