Linear Algebra Examples

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

Step 1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{2})$

Step 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 3

Substitute the known values into $p (λ) = determinant (A - λ I_{2})$ .

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Step 3.1

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

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

Step 3.2

Substitute $[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$ for $I_{2}$ .

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $- λ$ by each element of the 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 $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

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.

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