Linear Algebra Examples

[\begin{matrix} 0 & 1 - 1 & \sqrt{2} \end{matrix}]

$[\begin{array}{c} 0 & 1 \\ - 1 & \sqrt{2} \end{array}]$

Step 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 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 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 3.1

Substitute

[\begin{matrix} 0 & 1 - 1 & \sqrt{2} \end{matrix}]

$[\begin{array}{c} 0 & 1 \\ - 1 & \sqrt{2} \end{array}]$ for

A

$A$ .

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

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

Step 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} 0 & 1 - 1 & \sqrt{2} \end{matrix}] - λ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}])

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

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

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply

- λ

$- λ$ by each element of the matrix.

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

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

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

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

Step 4.1.2.2

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

0

$0$ by

- 1

$- 1$ .

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

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

Step 4.1.2.2.2

Multiply

0

$0$ by

λ

$λ$ .

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

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

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

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

Step 4.1.2.3

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

0

$0$ by

- 1

$- 1$ .

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

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

Step 4.1.2.3.2

Multiply

0

$0$ by

λ

$λ$ .

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

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

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

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

Step 4.1.2.4

Multiply

$- 1$ by

$1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Subtract

$λ$ from

$0$ .

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

Step 4.3.2

Add

$1$ and

$0$ .

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

Step 4.3.3

Add

$- 1$ and

$0$ .

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

Step 5

Find the determinant.

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Step 5.1

The determinant of a

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

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

$p (λ) = - λ (\sqrt{2} - λ) - (- 1 \cdot 1)$

Step 5.2

Simplify the determinant.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Apply the distributive property.

$p (λ) = - λ \sqrt{2} - λ (- λ) - (- 1 \cdot 1)$

Step 5.2.1.2

Rewrite using the commutative property of multiplication.

$p (λ) = - λ \sqrt{2} - 1 \cdot - 1 λ \cdot λ - (- 1 \cdot 1)$

Step 5.2.1.3

Simplify each term.

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Step 5.2.1.3.1

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.2.1.3.1.1

Move

$λ$ .

$p (λ) = - λ \sqrt{2} - 1 \cdot - 1 (λ \cdot λ) - (- 1 \cdot 1)$

Step 5.2.1.3.1.2

Multiply

$λ$ by

$λ$ .

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

Step 5.2.1.3.2

Multiply

$- 1$ by

$- 1$ .

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

Step 5.2.1.3.3

Multiply

$λ^{2}$ by

$1$ .

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

Step 5.2.1.4

Multiply

$- (- 1 \cdot 1)$ .

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Step 5.2.1.4.1

Multiply

$- 1$ by

$1$ .

$p (λ) = - λ \sqrt{2} + λ^{2} - - 1$

Step 5.2.1.4.2

Multiply

$- 1$ by

$- 1$ .

$p (λ) = - λ \sqrt{2} + λ^{2} + 1$

Step 5.2.2

Reorder

$- λ \sqrt{2}$ and

$λ^{2}$ .

$p (λ) = λ^{2} - λ \sqrt{2} + 1$

Step 6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$λ^{2} - λ \sqrt{2} + 1 = 0$

Step 7

Solve for

$λ$ .

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Step 7.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7.2

Substitute the values

$a = 1$ ,

$b = - \sqrt{2}$ , and

$c = 1$ into the quadratic formula and solve for

$λ$ .

$\frac{\sqrt{2} \pm \sqrt{{(- \sqrt{2})}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 7.3

Simplify.

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Step 7.3.1

Simplify the numerator.

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Step 7.3.1.1

Apply the product rule to

$- \sqrt{2}$ .

$λ = \frac{\sqrt{2} \pm \sqrt{{(- 1)}^{2} {\sqrt{2}}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.2

Raise

$- 1$ to the power of

$2$ .

$λ = \frac{\sqrt{2} \pm \sqrt{1 {\sqrt{2}}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.3

Multiply

${\sqrt{2}}^{2}$ by

$1$ .

$λ = \frac{\sqrt{2} \pm \sqrt{{\sqrt{2}}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.4

Rewrite

${\sqrt{2}}^{2}$ as

$2$ .

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Step 7.3.1.4.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{2}$ as

$2^{\frac{1}{2}}$ .

$λ = \frac{\sqrt{2} \pm \sqrt{{(2^{\frac{1}{2}})}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.4.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$λ = \frac{\sqrt{2} \pm \sqrt{2^{\frac{1}{2} \cdot 2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.4.3

Combine

$\frac{1}{2}$ and

$2$ .

$λ = \frac{\sqrt{2} \pm \sqrt{2^{\frac{2}{2}} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.4.4

Cancel the common factor of

$2$ .

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Step 7.3.1.4.4.1

Cancel the common factor.

$λ = \frac{\sqrt{2} \pm \sqrt{2^{\frac{2}{2}} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.4.4.2

Rewrite the expression.

$λ = \frac{\sqrt{2} \pm \sqrt{2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.4.5

Evaluate the exponent.

$λ = \frac{\sqrt{2} \pm \sqrt{2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.5

Multiply

$- 4 \cdot 1 \cdot 1$ .

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Step 7.3.1.5.1

Multiply

$- 4$ by

$1$ .

$λ = \frac{\sqrt{2} \pm \sqrt{2 - 4 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.5.2

Multiply

$- 4$ by

$1$ .

$λ = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2 \cdot 1}$

Step 7.3.1.6

Subtract

$4$ from

$2$ .

$λ = \frac{\sqrt{2} \pm \sqrt{- 2}}{2 \cdot 1}$

Step 7.3.1.7

Rewrite

$- 2$ as

$- 1 (2)$ .

$λ = \frac{\sqrt{2} \pm \sqrt{- 1 \cdot 2}}{2 \cdot 1}$

Step 7.3.1.8

Rewrite

$\sqrt{- 1 (2)}$ as

$\sqrt{- 1} \cdot \sqrt{2}$ .

$λ = \frac{\sqrt{2} \pm \sqrt{- 1} \cdot \sqrt{2}}{2 \cdot 1}$

Step 7.3.1.9

Rewrite

$\sqrt{- 1}$ as

$i$ .

$λ = \frac{\sqrt{2} \pm i \sqrt{2}}{2 \cdot 1}$

Step 7.3.2

Multiply

$2$ by

$1$ .

$λ = \frac{\sqrt{2} \pm i \sqrt{2}}{2}$

Step 7.4

The final answer is the combination of both solutions.

$λ = \frac{\sqrt{2} + i \sqrt{2}}{2}, \frac{\sqrt{2} - i \sqrt{2}}{2}$