Linear Algebra Examples

$[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}]$

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 $[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] - λ I_{2})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 5 - λ & 0 + 0 \\ 1 + 0 & 5 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 5 - λ & 0 \\ 1 + 0 & 5 - λ \end{array}]$

Step 1.4.3.2

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 5 - λ & 0 \\ 1 & 5 - λ \end{array}]$

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

$p (λ) = (5 - λ) (5 - λ) - 1 \cdot 0$

Step 1.5.2

Simplify the determinant.

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Step 1.5.2.1

Expand $(5 - λ) (5 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1

Apply the distributive property.

$p (λ) = 5 (5 - λ) - λ (5 - λ) - 1 \cdot 0$

Step 1.5.2.1.2

Apply the distributive property.

$p (λ) = 5 \cdot 5 + 5 (- λ) - λ (5 - λ) - 1 \cdot 0$

Step 1.5.2.1.3

Apply the distributive property.

$p (λ) = 5 \cdot 5 + 5 (- λ) - λ \cdot 5 - λ (- λ) - 1 \cdot 0$

Step 1.5.2.2

Simplify and combine like terms.

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Step 1.5.2.2.1

Simplify each term.

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Step 1.5.2.2.1.1

Multiply $5$ by $5$ .

$p (λ) = 25 + 5 (- λ) - λ \cdot 5 - λ (- λ) - 1 \cdot 0$

Step 1.5.2.2.1.2

Multiply $- 1$ by $5$ .

$p (λ) = 25 - 5 λ - λ \cdot 5 - λ (- λ) - 1 \cdot 0$

Step 1.5.2.2.1.3

Multiply $5$ by $- 1$ .

$p (λ) = 25 - 5 λ - 5 λ - λ (- λ) - 1 \cdot 0$

Step 1.5.2.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 25 - 5 λ - 5 λ - 1 \cdot - 1 λ \cdot λ - 1 \cdot 0$

Step 1.5.2.2.1.5

Multiply $λ$ by $λ$ by adding the exponents.

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Step 1.5.2.2.1.5.1

Move $λ$ .

$p (λ) = 25 - 5 λ - 5 λ - 1 \cdot - 1 (λ \cdot λ) - 1 \cdot 0$

Step 1.5.2.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 25 - 5 λ - 5 λ - 1 \cdot - 1 λ^{2} - 1 \cdot 0$

Step 1.5.2.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 25 - 5 λ - 5 λ + 1 λ^{2} - 1 \cdot 0$

Step 1.5.2.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 25 - 5 λ - 5 λ + λ^{2} - 1 \cdot 0$

Step 1.5.2.2.2

Subtract $5 λ$ from $- 5 λ$ .

$p (λ) = 25 - 10 λ + λ^{2} - 1 \cdot 0$

Step 1.5.2.3

Subtract $0$ from $25 - 10 λ + λ^{2}$ .

$p (λ) = 25 - 10 λ + λ^{2}$

Step 1.5.2.4

Move $25$ .

$p (λ) = - 10 λ + λ^{2} + 25$

Step 1.5.2.5

Reorder $- 10 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 10 λ + 25$

Step 1.6

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

$λ^{2} - 10 λ + 25 = 0$

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 $25$ as $5^{2}$ .

$λ^{2} - 10 λ + 5^{2} = 0$

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.

$10 λ = 2 \cdot λ \cdot 5$

Step 1.7.1.3

Rewrite the polynomial.

$λ^{2} - 2 \cdot λ \cdot 5 + 5^{2} = 0$

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 $b = 5$ .

${(λ - 5)}^{2} = 0$

Step 1.7.2

Set the $λ - 5$ equal to $0$ .

$λ - 5 = 0$

Step 1.7.3

Add $5$ to both sides of the equation.

$λ = 5$

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 $λ = 5$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] - 5 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

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

$[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - 5 \cdot 1 & - 5 \cdot 0 \\ - 5 \cdot 0 & - 5 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 5$ by $1$ .

$[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - 5 & - 5 \cdot 0 \\ - 5 \cdot 0 & - 5 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 5$ by $0$ .

$[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - 5 & 0 \\ - 5 \cdot 0 & - 5 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 5$ by $0$ .

$[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - 5 & 0 \\ 0 & - 5 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 5$ by $1$ .

$[\begin{array}{c} 5 & 0 \\ 1 & 5 \end{array}] + [\begin{array}{c} - 5 & 0 \\ 0 & - 5 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 5 - 5 & 0 + 0 \\ 1 + 0 & 5 - 5 \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $5$ from $5$ .

$[\begin{array}{c} 0 & 0 + 0 \\ 1 + 0 & 5 - 5 \end{array}]$

Step 3.2.3.2

Add $0$ and $0$ .

$[\begin{array}{c} 0 & 0 \\ 1 + 0 & 5 - 5 \end{array}]$

Step 3.2.3.3

Add $1$ and $0$ .

$[\begin{array}{c} 0 & 0 \\ 1 & 5 - 5 \end{array}]$

Step 3.2.3.4

Subtract $5$ from $5$ .

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

Step 3.3

Find the null space when $λ = 5$ .

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Step 3.3.1

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

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

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

Step 3.3.3

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

$x = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

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

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