Linear Algebra Examples

[\begin{matrix} 1 & 1 0 & 1 \end{matrix}]

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

Step 1

Find the eigenvalues.

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Step 1.1

Set up the formula to find the characteristic equation

p (λ)

$p (λ)$ .

p (λ) = determinant (A - λ I_{2})

$p (λ) = determinant (A - λ I_{2})$

Step 1.2

The identity matrix or unit matrix of size

2

$2$ is the

2 \times 2

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

[\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

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

Step 1.3

Substitute the known values into

p (λ) = determinant (A - λ I_{2})

$p (λ) = determinant (A - λ I_{2})$ .

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Step 1.3.1

Substitute

[\begin{matrix} 1 & 1 0 & 1 \end{matrix}]

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

A

$A$ .

p (λ) = determinant ([\begin{matrix} 1 & 1 0 & 1 \end{matrix}] - λ I_{2})

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

Step 1.3.2

Substitute

[\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

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

I_{2}

$I_{2}$ .

p (λ) = determinant ([\begin{matrix} 1 & 1 0 & 1 \end{matrix}] - λ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}])

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

p (λ) = determinant ([\begin{matrix} 1 & 1 0 & 1 \end{matrix}] - λ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}])

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

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

$- λ$ by each element of the matrix.

p (λ) = determinant ([\begin{matrix} 1 & 1 0 & 1 \end{matrix}] + [\begin{matrix} - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ([\begin{matrix} 1 & 1 0 & 1 \end{matrix}] + [\begin{matrix} - λ & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

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

Step 1.4.1.2.2

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ([\begin{matrix} 1 & 1 0 & 1 \end{matrix}] + [\begin{matrix} - λ & 0 λ - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

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

Step 1.4.1.2.2.2

Multiply

$0$ by

$λ$ .

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

Step 1.4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply

$0$ by

$- 1$ .

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

Step 1.4.1.2.3.2

Multiply

$0$ by

$λ$ .

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

Step 1.4.1.2.4

Multiply

$- 1$ by

$1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add

$1$ and

$0$ .

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

Step 1.4.3.2

Add

$0$ and

$0$ .

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

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

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

$p (λ) = (1 - λ) (1 - λ) + 0 \cdot 1$

Step 1.5.2

Simplify the determinant.

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Step 1.5.2.1

Simplify each term.

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Step 1.5.2.1.1

Expand

$(1 - λ) (1 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

$p (λ) = 1 (1 - λ) - λ (1 - λ) + 0 \cdot 1$

Step 1.5.2.1.1.2

Apply the distributive property.

$p (λ) = 1 \cdot 1 + 1 (- λ) - λ (1 - λ) + 0 \cdot 1$

Step 1.5.2.1.1.3

Apply the distributive property.

$p (λ) = 1 \cdot 1 + 1 (- λ) - λ \cdot 1 - λ (- λ) + 0 \cdot 1$

Step 1.5.2.1.2

Simplify and combine like terms.

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Step 1.5.2.1.2.1

Simplify each term.

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Step 1.5.2.1.2.1.1

Multiply

$1$ by

$1$ .

$p (λ) = 1 + 1 (- λ) - λ \cdot 1 - λ (- λ) + 0 \cdot 1$

Step 1.5.2.1.2.1.2

Multiply

$- λ$ by

$1$ .

$p (λ) = 1 - λ - λ \cdot 1 - λ (- λ) + 0 \cdot 1$

Step 1.5.2.1.2.1.3

Multiply

$- 1$ by

$1$ .

$p (λ) = 1 - λ - λ - λ (- λ) + 0 \cdot 1$

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 1 - λ - λ - 1 \cdot - 1 λ \cdot λ + 0 \cdot 1$

Step 1.5.2.1.2.1.5

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 1.5.2.1.2.1.5.1

Move

$λ$ .

$p (λ) = 1 - λ - λ - 1 \cdot - 1 (λ \cdot λ) + 0 \cdot 1$

Step 1.5.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = 1 - λ - λ - 1 \cdot - 1 λ^{2} + 0 \cdot 1$

Step 1.5.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

$p (λ) = 1 - λ - λ + 1 λ^{2} + 0 \cdot 1$

Step 1.5.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = 1 - λ - λ + λ^{2} + 0 \cdot 1$

Step 1.5.2.1.2.2

Subtract

$λ$ from

$- λ$ .

$p (λ) = 1 - 2 λ + λ^{2} + 0 \cdot 1$

Step 1.5.2.1.3

Multiply

$0$ by

$1$ .

$p (λ) = 1 - 2 λ + λ^{2} + 0$

Step 1.5.2.2

Add

$1 - 2 λ + λ^{2}$ and

$0$ .

$p (λ) = 1 - 2 λ + λ^{2}$

Step 1.5.2.3

Move

$1$ .

$p (λ) = - 2 λ + λ^{2} + 1$

Step 1.5.2.4

Reorder

$- 2 λ$ and

$λ^{2}$ .

$p (λ) = λ^{2} - 2 λ + 1$

Step 1.6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$λ^{2} - 2 λ + 1 = 0$

Step 1.7

Solve for

$λ$ .

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Step 1.7.1

Factor using the perfect square rule.

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Step 1.7.1.1

Rewrite

$1$ as

$1^{2}$ .

$λ^{2} - 2 λ + 1^{2} = 0$

Step 1.7.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 λ = 2 \cdot λ \cdot 1$

Step 1.7.1.3

Rewrite the polynomial.

$λ^{2} - 2 \cdot λ \cdot 1 + 1^{2} = 0$

Step 1.7.1.4

Factor using the perfect square trinomial rule

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where

$a = λ$ and

$b = 1$ .

${(λ - 1)}^{2} = 0$

Step 1.7.2

Set the

$λ - 1$ equal to

$0$ .

$λ - 1 = 0$

Step 1.7.3

Add

$1$ to both sides of the equation.

$λ = 1$

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.

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

Step 3.2

Simplify.

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Step 3.2.1

Subtract the corresponding elements.

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

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Subtract

$1$ from

$1$ .

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

Step 3.2.2.2

Subtract

$0$ from

$1$ .

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

Step 3.2.2.3

Subtract

$0$ from

$0$ .

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

Step 3.2.2.4

Subtract

$1$ from

$1$ .

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

Step 3.3

Find the null space when

$λ = 1$ .

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Step 3.3.1

Write as an augmented matrix for

$A x = 0$ .

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

Step 3.3.2

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

$y = 0$

$0 = 0$

Step 3.3.3

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

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

Step 3.3.4

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = x [\begin{array}{c} 1 \\ 0 \end{array}]$

Step 3.3.5

Write as a solution set.

${x [\begin{array}{c} 1 \\ 0 \end{array}] | x \in R}$

Step 3.3.6

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

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

Step 4

The eigenspace of

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

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