Linear Algebra Examples

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}]$

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 $[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] - λ I_{2})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 8 - λ & 6 + 0 \\ - 3 + 0 & 2 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 8 - λ & 6 \\ - 3 + 0 & 2 - λ \end{array}]$

Step 1.4.3.2

Add $- 3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 8 - λ & 6 \\ - 3 & 2 - λ \end{array}]$

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

$p (λ) = (8 - λ) (2 - λ) - (- 3 \cdot 6)$

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 $(8 - λ) (2 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

$p (λ) = 8 (2 - λ) - λ (2 - λ) - (- 3 \cdot 6)$

Step 1.5.2.1.1.2

Apply the distributive property.

$p (λ) = 8 \cdot 2 + 8 (- λ) - λ (2 - λ) - (- 3 \cdot 6)$

Step 1.5.2.1.1.3

Apply the distributive property.

$p (λ) = 8 \cdot 2 + 8 (- λ) - λ \cdot 2 - λ (- λ) - (- 3 \cdot 6)$

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 $8$ by $2$ .

$p (λ) = 16 + 8 (- λ) - λ \cdot 2 - λ (- λ) - (- 3 \cdot 6)$

Step 1.5.2.1.2.1.2

Multiply $- 1$ by $8$ .

$p (λ) = 16 - 8 λ - λ \cdot 2 - λ (- λ) - (- 3 \cdot 6)$

Step 1.5.2.1.2.1.3

Multiply $2$ by $- 1$ .

$p (λ) = 16 - 8 λ - 2 λ - λ (- λ) - (- 3 \cdot 6)$

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 16 - 8 λ - 2 λ - 1 \cdot - 1 λ \cdot λ - (- 3 \cdot 6)$

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 $λ$ .

$p (λ) = 16 - 8 λ - 2 λ - 1 \cdot - 1 (λ \cdot λ) - (- 3 \cdot 6)$

Step 1.5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 16 - 8 λ - 2 λ - 1 \cdot - 1 λ^{2} - (- 3 \cdot 6)$

Step 1.5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 16 - 8 λ - 2 λ + 1 λ^{2} - (- 3 \cdot 6)$

Step 1.5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 16 - 8 λ - 2 λ + λ^{2} - (- 3 \cdot 6)$

Step 1.5.2.1.2.2

Subtract $2 λ$ from $- 8 λ$ .

$p (λ) = 16 - 10 λ + λ^{2} - (- 3 \cdot 6)$

Step 1.5.2.1.3

Multiply $- (- 3 \cdot 6)$ .

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Step 1.5.2.1.3.1

Multiply $- 3$ by $6$ .

$p (λ) = 16 - 10 λ + λ^{2} - - 18$

Step 1.5.2.1.3.2

Multiply $- 1$ by $- 18$ .

$p (λ) = 16 - 10 λ + λ^{2} + 18$

Step 1.5.2.2

Add $16$ and $18$ .

$p (λ) = - 10 λ + λ^{2} + 34$

Step 1.5.2.3

Reorder $- 10 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 10 λ + 34$

Step 1.6

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$λ^{2} - 10 λ + 34 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Use the quadratic formula to find the solutions.

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

Step 1.7.2

Substitute the values $a = 1$ , $b = - 10$ , and $c = 34$ into the quadratic formula and solve for $λ$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot 34)}}{2 \cdot 1}$

Step 1.7.3

Simplify.

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Step 1.7.3.1

Simplify the numerator.

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Step 1.7.3.1.1

Raise $- 10$ to the power of $2$ .

$λ = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 34}}{2 \cdot 1}$

Step 1.7.3.1.2

Multiply $- 4 \cdot 1 \cdot 34$ .

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Step 1.7.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{10 \pm \sqrt{100 - 4 \cdot 34}}{2 \cdot 1}$

Step 1.7.3.1.2.2

Multiply $- 4$ by $34$ .

$λ = \frac{10 \pm \sqrt{100 - 136}}{2 \cdot 1}$

Step 1.7.3.1.3

Subtract $136$ from $100$ .

$λ = \frac{10 \pm \sqrt{- 36}}{2 \cdot 1}$

Step 1.7.3.1.4

Rewrite $- 36$ as $- 1 (36)$ .

$λ = \frac{10 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Step 1.7.3.1.5

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$λ = \frac{10 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

Step 1.7.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$λ = \frac{10 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Step 1.7.3.1.7

Rewrite $36$ as $6^{2}$ .

$λ = \frac{10 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

Step 1.7.3.1.8

Pull terms out from under the radical, assuming positive real numbers.

$λ = \frac{10 \pm i \cdot 6}{2 \cdot 1}$

Step 1.7.3.1.9

Move $6$ to the left of $i$ .

$λ = \frac{10 \pm 6 i}{2 \cdot 1}$

Step 1.7.3.2

Multiply $2$ by $1$ .

$λ = \frac{10 \pm 6 i}{2}$

Step 1.7.3.3

Simplify $\frac{10 \pm 6 i}{2}$ .

$λ = 5 \pm 3 i$

Step 1.7.4

The final answer is the combination of both solutions.

$λ = 5 + 3 i, 5 - 3 i$

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 $λ = 5 + 3 i$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] - (5 + 3 i) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the distributive property.

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + (- 1 \cdot 5 - (3 i)) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 3.2.1.2

Multiply $- 1$ by $5$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + (- 5 - (3 i)) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 3.2.1.3

Multiply $3$ by $- 1$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + (- 5 - 3 i) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 3.2.1.4

Multiply $- 5 - 3 i$ by each element of the matrix.

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} (- 5 - 3 i) \cdot 1 & (- 5 - 3 i) \cdot 0 \\ (- 5 - 3 i) \cdot 0 & (- 5 - 3 i) \cdot 1 \end{array}]$

Step 3.2.1.5

Simplify each element in the matrix.

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Step 3.2.1.5.1

Multiply $- 5 - 3 i$ by $1$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 - 3 i & (- 5 - 3 i) \cdot 0 \\ (- 5 - 3 i) \cdot 0 & (- 5 - 3 i) \cdot 1 \end{array}]$

Step 3.2.1.5.2

Multiply $- 5 - 3 i$ by $0$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 - 3 i & 0 \\ (- 5 - 3 i) \cdot 0 & (- 5 - 3 i) \cdot 1 \end{array}]$

Step 3.2.1.5.3

Multiply $- 5 - 3 i$ by $0$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 - 3 i & 0 \\ 0 & (- 5 - 3 i) \cdot 1 \end{array}]$

Step 3.2.1.5.4

Multiply $- 5 - 3 i$ by $1$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 - 3 i & 0 \\ 0 & - 5 - 3 i \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 8 - 5 - 3 i & 6 + 0 \\ - 3 + 0 & 2 - 5 - 3 i \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $5$ from $8$ .

$[\begin{array}{c} 3 - 3 i & 6 + 0 \\ - 3 + 0 & 2 - 5 - 3 i \end{array}]$

Step 3.2.3.2

Add $6$ and $0$ .

$[\begin{array}{c} 3 - 3 i & 6 \\ - 3 + 0 & 2 - 5 - 3 i \end{array}]$

Step 3.2.3.3

Add $- 3$ and $0$ .

$[\begin{array}{c} 3 - 3 i & 6 \\ - 3 & 2 - 5 - 3 i \end{array}]$

Step 3.2.3.4

Subtract $5$ from $2$ .

$[\begin{array}{c} 3 - 3 i & 6 \\ - 3 & - 3 - 3 i \end{array}]$

Step 3.3

Find the null space when $λ = 5 + 3 i$ .

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Step 3.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 3 - 3 i & 6 & 0 \\ - 3 & - 3 - 3 i & 0 \end{array}]$

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 $\frac{1}{3 - 3 i}$ 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 $\frac{1}{3 - 3 i}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3 - 3 i}{3 - 3 i} & \frac{6}{3 - 3 i} & \frac{0}{3 - 3 i} \\ - 3 & - 3 - 3 i & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 + i & 0 \\ - 3 & - 3 - 3 i & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 + i & 0 \\ - 3 + 3 \cdot 1 & - 3 - 3 i + 3 (1 + i) & 0 + 3 \cdot 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.3

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

$x + (1 + i) y = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = 5 - 3 i$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] - (5 - 3 i) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 4.2

Simplify.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Apply the distributive property.

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + (- 1 \cdot 5 - (- 3 i)) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 4.2.1.2

Multiply $- 1$ by $5$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + (- 5 - (- 3 i)) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 4.2.1.3

Multiply $- 3$ by $- 1$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + (- 5 + 3 i) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 4.2.1.4

Multiply $- 5 + 3 i$ by each element of the matrix.

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} (- 5 + 3 i) \cdot 1 & (- 5 + 3 i) \cdot 0 \\ (- 5 + 3 i) \cdot 0 & (- 5 + 3 i) \cdot 1 \end{array}]$

Step 4.2.1.5

Simplify each element in the matrix.

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Step 4.2.1.5.1

Multiply $- 5 + 3 i$ by $1$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 + 3 i & (- 5 + 3 i) \cdot 0 \\ (- 5 + 3 i) \cdot 0 & (- 5 + 3 i) \cdot 1 \end{array}]$

Step 4.2.1.5.2

Multiply $- 5 + 3 i$ by $0$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 + 3 i & 0 \\ (- 5 + 3 i) \cdot 0 & (- 5 + 3 i) \cdot 1 \end{array}]$

Step 4.2.1.5.3

Multiply $- 5 + 3 i$ by $0$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 + 3 i & 0 \\ 0 & (- 5 + 3 i) \cdot 1 \end{array}]$

Step 4.2.1.5.4

Multiply $- 5 + 3 i$ by $1$ .

$[\begin{array}{c} 8 & 6 \\ - 3 & 2 \end{array}] + [\begin{array}{c} - 5 + 3 i & 0 \\ 0 & - 5 + 3 i \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 8 - 5 + 3 i & 6 + 0 \\ - 3 + 0 & 2 - 5 + 3 i \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract $5$ from $8$ .

$[\begin{array}{c} 3 + 3 i & 6 + 0 \\ - 3 + 0 & 2 - 5 + 3 i \end{array}]$

Step 4.2.3.2

Add $6$ and $0$ .

$[\begin{array}{c} 3 + 3 i & 6 \\ - 3 + 0 & 2 - 5 + 3 i \end{array}]$

Step 4.2.3.3

Add $- 3$ and $0$ .

$[\begin{array}{c} 3 + 3 i & 6 \\ - 3 & 2 - 5 + 3 i \end{array}]$

Step 4.2.3.4

Subtract $5$ from $2$ .

$[\begin{array}{c} 3 + 3 i & 6 \\ - 3 & - 3 + 3 i \end{array}]$

Step 4.3

Find the null space when $λ = 5 - 3 i$ .

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Step 4.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 3 + 3 i & 6 & 0 \\ - 3 & - 3 + 3 i & 0 \end{array}]$

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

Multiply each element of $R_{1}$ by $\frac{1}{3 + 3 i}$ to make the entry at $1, 1$ a $1$ .

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Step 4.3.2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{3 + 3 i}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3 + 3 i}{3 + 3 i} & \frac{6}{3 + 3 i} & \frac{0}{3 + 3 i} \\ - 3 & - 3 + 3 i & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 - i & 0 \\ - 3 & - 3 + 3 i & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 - i & 0 \\ - 3 + 3 \cdot 1 & - 3 + 3 i + 3 (1 - i) & 0 + 3 \cdot 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

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

Step 4.3.3

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

$x + (1 - i) y = 0$

$0 = 0$

Step 4.3.4

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

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

Step 4.3.5

Write the solution as a linear combination of vectors.

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

Step 4.3.6

Write as a solution set.

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

Step 4.3.7

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

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

Step 5

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

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