Linear Algebra Examples

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}]$

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 $[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] - λ I_{2})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.3 + 0 \\ 0.2 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $0.3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.3 \\ 0.2 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3.2

Add $0.2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.3 \\ 0.2 & 0.7 - λ \end{array}]$

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

$p (λ) = (0.8 - λ) (0.7 - λ) - 0.2 \cdot 0.3$

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 $(0.8 - λ) (0.7 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

$p (λ) = 0.8 (0.7 - λ) - λ (0.7 - λ) - 0.2 \cdot 0.3$

Step 1.5.2.1.1.2

Apply the distributive property.

$p (λ) = 0.8 \cdot 0.7 + 0.8 (- λ) - λ (0.7 - λ) - 0.2 \cdot 0.3$

Step 1.5.2.1.1.3

Apply the distributive property.

$p (λ) = 0.8 \cdot 0.7 + 0.8 (- λ) - λ \cdot 0.7 - λ (- λ) - 0.2 \cdot 0.3$

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 $0.8$ by $0.7$ .

$p (λ) = 0.56 + 0.8 (- λ) - λ \cdot 0.7 - λ (- λ) - 0.2 \cdot 0.3$

Step 1.5.2.1.2.1.2

Multiply $- 1$ by $0.8$ .

$p (λ) = 0.56 - 0.8 λ - λ \cdot 0.7 - λ (- λ) - 0.2 \cdot 0.3$

Step 1.5.2.1.2.1.3

Multiply $0.7$ by $- 1$ .

$p (λ) = 0.56 - 0.8 λ - 0.7 λ - λ (- λ) - 0.2 \cdot 0.3$

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0.56 - 0.8 λ - 0.7 λ - 1 \cdot - 1 λ \cdot λ - 0.2 \cdot 0.3$

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 $λ$ .

$p (λ) = 0.56 - 0.8 λ - 0.7 λ - 1 \cdot - 1 (λ \cdot λ) - 0.2 \cdot 0.3$

Step 1.5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0.56 - 0.8 λ - 0.7 λ - 1 \cdot - 1 λ^{2} - 0.2 \cdot 0.3$

Step 1.5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0.56 - 0.8 λ - 0.7 λ + 1 λ^{2} - 0.2 \cdot 0.3$

Step 1.5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0.56 - 0.8 λ - 0.7 λ + λ^{2} - 0.2 \cdot 0.3$

Step 1.5.2.1.2.2

Subtract $0.7 λ$ from $- 0.8 λ$ .

$p (λ) = 0.56 - 1.5 λ + λ^{2} - 0.2 \cdot 0.3$

Step 1.5.2.1.3

Multiply $- 0.2$ by $0.3$ .

$p (λ) = 0.56 - 1.5 λ + λ^{2} - 0.06$

Step 1.5.2.2

Subtract $0.06$ from $0.56$ .

$p (λ) = - 1.5 λ + λ^{2} + 0.5$

Step 1.5.2.3

Reorder $- 1.5 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 1.5 λ + 0.5$

Step 1.6

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

$λ^{2} - 1.5 λ + 0.5 = 0$

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 $a = 1$ , $b = - 1.5$ , and $c = 0.5$ into the quadratic formula and solve for $λ$ .

$\frac{1.5 \pm \sqrt{{(- 1.5)}^{2} - 4 \cdot (1 \cdot 0.5)}}{2 \cdot 1}$

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 $- 1.5$ to the power of $2$ .

$λ = \frac{1.5 \pm \sqrt{2.25 - 4 \cdot 1 \cdot 0.5}}{2 \cdot 1}$

Step 1.7.3.1.2

Multiply $- 4 \cdot 1 \cdot 0.5$ .

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Step 1.7.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{1.5 \pm \sqrt{2.25 - 4 \cdot 0.5}}{2 \cdot 1}$

Step 1.7.3.1.2.2

Multiply $- 4$ by $0.5$ .

$λ = \frac{1.5 \pm \sqrt{2.25 - 2}}{2 \cdot 1}$

Step 1.7.3.1.3

Subtract $2$ from $2.25$ .

$λ = \frac{1.5 \pm \sqrt{0.25}}{2 \cdot 1}$

Step 1.7.3.1.4

Rewrite $0.25$ as ${0.5}^{2}$ .

$λ = \frac{1.5 \pm \sqrt{{0.5}^{2}}}{2 \cdot 1}$

Step 1.7.3.1.5

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

$λ = \frac{1.5 \pm 0.5}{2 \cdot 1}$

Step 1.7.3.2

Multiply $2$ by $1$ .

$λ = \frac{1.5 \pm 0.5}{2}$

Step 1.7.3.3

Simplify $\frac{1.5 \pm 0.5}{2}$ .

$λ = \frac{3 \pm 1}{4}$

Step 1.7.4

The final answer is the combination of both solutions.

$λ = 1, \frac{1}{2}$

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.

$N ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] - [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Subtract the corresponding elements.

$[\begin{array}{c} 0.8 - 1 & 0.3 - 0 \\ 0.2 - 0 & 0.7 - 1 \end{array}]$

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Subtract $1$ from $0.8$ .

$[\begin{array}{c} - 0.2 & 0.3 - 0 \\ 0.2 - 0 & 0.7 - 1 \end{array}]$

Step 3.2.2.2

Subtract $0$ from $0.3$ .

$[\begin{array}{c} - 0.2 & 0.3 \\ 0.2 - 0 & 0.7 - 1 \end{array}]$

Step 3.2.2.3

Subtract $0$ from $0.2$ .

$[\begin{array}{c} - 0.2 & 0.3 \\ 0.2 & 0.7 - 1 \end{array}]$

Step 3.2.2.4

Subtract $1$ from $0.7$ .

$[\begin{array}{c} - 0.2 & 0.3 \\ 0.2 & - 0.3 \end{array}]$

Step 3.3

Find the null space when $λ = 1$ .

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Step 3.3.1

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

$[\begin{array}{c} - 0.2 & 0.3 & 0 \\ 0.2 & - 0.3 & 0 \end{array}]$

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 $\frac{1}{- 0.2}$ 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 $\frac{1}{- 0.2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{- 0.2}{- 0.2} & \frac{0.3}{- 0.2} & \frac{0}{- 0.2} \\ 0.2 & - 0.3 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1.5 & 0 \\ 0.2 & - 0.3 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} - 0.2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 0.2 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1.5 & 0 \\ 0.2 - 0.2 \cdot 1 & - 0.3 - 0.2 \cdot - 1.5 & 0 - 0.2 \cdot 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1.5 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 3.3.3

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

$x - 1.5 y = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1.5 y \\ y \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} 1.5 \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${y [\begin{array}{c} 1.5 \\ 1 \end{array}] | y \in R}$

Step 3.3.7

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

${[\begin{array}{c} 1.5 \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue $λ = \frac{1}{2}$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] - \frac{1}{2} [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 4.2

Simplify.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $- \frac{1}{2}$ by each element of the matrix.

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} \cdot 1 & - \frac{1}{2} \cdot 0 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} & - \frac{1}{2} \cdot 0 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- \frac{1}{2} \cdot 0$ .

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Step 4.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} & 0 (\frac{1}{2}) \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2.2

Multiply $0$ by $\frac{1}{2}$ .

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} & 0 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- \frac{1}{2} \cdot 0$ .

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Step 4.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} & 0 \\ 0 (\frac{1}{2}) & - \frac{1}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3.2

Multiply $0$ by $\frac{1}{2}$ .

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} & 0 \\ 0 & - \frac{1}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.3 \\ 0.2 & 0.7 \end{array}] + [\begin{array}{c} - \frac{1}{2} & 0 \\ 0 & - \frac{1}{2} \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 0.8 - \frac{1}{2} & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

To write $0.8$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} 0.8 \cdot \frac{2}{2} - \frac{1}{2} & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.2

Combine $0.8$ and $\frac{2}{2}$ .

$[\begin{array}{c} \frac{0.8 \cdot 2}{2} - \frac{1}{2} & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{0.8 \cdot 2 - 1}{2} & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.4

Simplify the numerator.

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Step 4.2.3.4.1

Multiply $0.8$ by $2$ .

$[\begin{array}{c} \frac{1.6 - 1}{2} & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.4.2

Subtract $1$ from $1.6$ .

$[\begin{array}{c} \frac{0.6}{2} & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.5

Divide $0.6$ by $2$ .

$[\begin{array}{c} 0.3 & 0.3 + 0 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.6

Add $0.3$ and $0$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 + 0 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.7

Add $0.2$ and $0$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & 0.7 - \frac{1}{2} \end{array}]$

Step 4.2.3.8

To write $0.7$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & 0.7 \cdot \frac{2}{2} - \frac{1}{2} \end{array}]$

Step 4.2.3.9

Combine $0.7$ and $\frac{2}{2}$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & \frac{0.7 \cdot 2}{2} - \frac{1}{2} \end{array}]$

Step 4.2.3.10

Combine the numerators over the common denominator.

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & \frac{0.7 \cdot 2 - 1}{2} \end{array}]$

Step 4.2.3.11

Simplify the numerator.

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Step 4.2.3.11.1

Multiply $0.7$ by $2$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & \frac{1.4 - 1}{2} \end{array}]$

Step 4.2.3.11.2

Subtract $1$ from $1.4$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & \frac{0.4}{2} \end{array}]$

Step 4.2.3.12

Divide $0.4$ by $2$ .

$[\begin{array}{c} 0.3 & 0.3 \\ 0.2 & 0.2 \end{array}]$

Step 4.3

Find the null space when $λ = \frac{1}{2}$ .

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Step 4.3.1

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

$[\begin{array}{c} 0.3 & 0.3 & 0 \\ 0.2 & 0.2 & 0 \end{array}]$

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 $\frac{1}{0.3}$ 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 $\frac{1}{0.3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.3}{0.3} & \frac{0.3}{0.3} & \frac{0}{0.3} \\ 0.2 & 0.2 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 \\ 0.2 & 0.2 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} - 0.2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 0.2 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 0 \\ 0.2 - 0.2 \cdot 1 & 0.2 - 0.2 \cdot 1 & 0 - 0.2 \cdot 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

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

Step 4.3.3

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

$x + y = 0$

$0 = 0$

Step 4.3.4

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

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

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.

${[\begin{array}{c} 1.5 \\ 1 \end{array}], [\begin{array}{c} - 1 \\ 1 \end{array}]}$