Examples

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}]$

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 $[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}]$ for $A$ .

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

Step 1.3.2

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

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- 1$ by $1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $2$ and $0$ .

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

Step 1.4.3.2

Add $- 6$ and $0$ .

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

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

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

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 $(- 1 - λ) (6 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

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

Step 1.5.2.1.1.2

Apply the distributive property.

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

Step 1.5.2.1.1.3

Apply the distributive property.

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

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 $- 1$ by $6$ .

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

Step 1.5.2.1.2.1.2

Multiply $- 1 (- λ)$ .

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Step 1.5.2.1.2.1.2.1

Multiply $- 1$ by $- 1$ .

$p (λ) = - 6 + 1 λ - λ \cdot 6 - λ (- λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.2.2

Multiply $λ$ by $1$ .

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

Step 1.5.2.1.2.1.3

Multiply $6$ by $- 1$ .

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

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 6 + λ - 6 λ - 1 \cdot - 1 λ \cdot λ - (- 6 \cdot 2)$

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 $λ$ .

$p (λ) = - 6 + λ - 6 λ - 1 \cdot - 1 (λ \cdot λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 1.5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 1.5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.2.1.2.2

Subtract $6 λ$ from $λ$ .

$p (λ) = - 6 - 5 λ + λ^{2} - (- 6 \cdot 2)$

Step 1.5.2.1.3

Multiply $- (- 6 \cdot 2)$ .

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Step 1.5.2.1.3.1

Multiply $- 6$ by $2$ .

$p (λ) = - 6 - 5 λ + λ^{2} - - 12$

Step 1.5.2.1.3.2

Multiply $- 1$ by $- 12$ .

$p (λ) = - 6 - 5 λ + λ^{2} + 12$

Step 1.5.2.2

Add $- 6$ and $12$ .

$p (λ) = - 5 λ + λ^{2} + 6$

Step 1.5.2.3

Reorder $- 5 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 5 λ + 6$

Step 1.6

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

$λ^{2} - 5 λ + 6 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Factor $λ^{2} - 5 λ + 6$ using the AC method.

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Step 1.7.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 1.7.1.2

Write the factored form using these integers.

$(λ - 3) (λ - 2) = 0$

Step 1.7.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$λ - 3 = 0$

$λ - 2 = 0$

Step 1.7.3

Set $λ - 3$ equal to $0$ and solve for $λ$ .

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Step 1.7.3.1

Set $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 1.7.3.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 1.7.4

Set $λ - 2$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set $λ - 2$ equal to $0$ .

$λ - 2 = 0$

Step 1.7.4.2

Add $2$ to both sides of the equation.

$λ = 2$

Step 1.7.5

The final solution is all the values that make $(λ - 3) (λ - 2) = 0$ true.

$λ = 3, 2$

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 $λ = 3$ .

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Step 3.1

Substitute the known values into the formula.

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

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 $- 3$ by each element of the matrix.

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 3$ by $1$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 3 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 3$ by $0$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 3 & 0 \\ - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 3$ by $0$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 3 & 0 \\ 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 3$ by $1$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 3 & 0 \\ 0 & - 3 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} - 1 - 3 & 2 + 0 \\ - 6 + 0 & 6 - 3 \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $3$ from $- 1$ .

$[\begin{array}{c} - 4 & 2 + 0 \\ - 6 + 0 & 6 - 3 \end{array}]$

Step 3.2.3.2

Add $2$ and $0$ .

$[\begin{array}{c} - 4 & 2 \\ - 6 + 0 & 6 - 3 \end{array}]$

Step 3.2.3.3

Add $- 6$ and $0$ .

$[\begin{array}{c} - 4 & 2 \\ - 6 & 6 - 3 \end{array}]$

Step 3.2.3.4

Subtract $3$ from $6$ .

$[\begin{array}{c} - 4 & 2 \\ - 6 & 3 \end{array}]$

Step 3.3

Find the null space when $λ = 3$ .

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Step 3.3.1

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

$[\begin{array}{c} - 4 & 2 & 0 \\ - 6 & 3 & 0 \end{array}]$

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 $- \frac{1}{4}$ 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 $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 2 & - \frac{1}{4} \cdot 0 \\ - 6 & 3 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 \\ - 6 & 3 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.3

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

$x - \frac{1}{2} y = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{y}{2} \\ y \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} \frac{1}{2} \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${y [\begin{array}{c} \frac{1}{2} \\ 1 \end{array}] | y \in R}$

Step 3.3.7

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

${[\begin{array}{c} \frac{1}{2} \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue $λ = 2$ .

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Step 4.1

Substitute the known values into the formula.

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

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 $- 2$ by each element of the matrix.

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 2 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 2 & 0 \\ - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 2 & 0 \\ 0 & - 2 \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - 2 & 0 \\ 0 & - 2 \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} - 1 - 2 & 2 + 0 \\ - 6 + 0 & 6 - 2 \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract $2$ from $- 1$ .

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

Step 4.2.3.2

Add $2$ and $0$ .

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

Step 4.2.3.3

Add $- 6$ and $0$ .

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

Step 4.2.3.4

Subtract $2$ from $6$ .

$[\begin{array}{c} - 3 & 2 \\ - 6 & 4 \end{array}]$

Step 4.3

Find the null space when $λ = 2$ .

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Step 4.3.1

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

$[\begin{array}{c} - 3 & 2 & 0 \\ - 6 & 4 & 0 \end{array}]$

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 2 & - \frac{1}{3} \cdot 0 \\ - 6 & 4 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & 0 \\ - 6 & 4 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & 0 \\ - 6 + 6 \cdot 1 & 4 + 6 (- \frac{2}{3}) & 0 + 6 \cdot 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & 0 \\ 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x - \frac{2}{3} y = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{2 y}{3} \\ y \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} \frac{2}{3} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${y [\begin{array}{c} \frac{2}{3} \\ 1 \end{array}] | y \in R}$

Step 4.3.7

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

${[\begin{array}{c} \frac{2}{3} \\ 1 \end{array}]}$

Step 5

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

${[\begin{array}{c} \frac{1}{2} \\ 1 \end{array}], [\begin{array}{c} \frac{2}{3} \\ 1 \end{array}]}$

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