Linear Algebra Examples

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

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 $[\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] - λ I_{2})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 3 - λ & 4 + 0 \\ - 2 + 0 & - 3 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 4 \\ - 2 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.2

Add $- 2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 4 \\ - 2 & - 3 - λ \end{array}]$

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

$p (λ) = (3 - λ) (- 3 - λ) - (- 2 \cdot 4)$

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 $(3 - λ) (- 3 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

$p (λ) = 3 (- 3 - λ) - λ (- 3 - λ) - (- 2 \cdot 4)$

Step 1.5.2.1.1.2

Apply the distributive property.

$p (λ) = 3 \cdot - 3 + 3 (- λ) - λ (- 3 - λ) - (- 2 \cdot 4)$

Step 1.5.2.1.1.3

Apply the distributive property.

$p (λ) = 3 \cdot - 3 + 3 (- λ) - λ \cdot - 3 - λ (- λ) - (- 2 \cdot 4)$

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 $3$ by $- 3$ .

$p (λ) = - 9 + 3 (- λ) - λ \cdot - 3 - λ (- λ) - (- 2 \cdot 4)$

Step 1.5.2.1.2.1.2

Multiply $- 1$ by $3$ .

$p (λ) = - 9 - 3 λ - λ \cdot - 3 - λ (- λ) - (- 2 \cdot 4)$

Step 1.5.2.1.2.1.3

Multiply $- 3$ by $- 1$ .

$p (λ) = - 9 - 3 λ + 3 λ - λ (- λ) - (- 2 \cdot 4)$

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 9 - 3 λ + 3 λ - 1 \cdot - 1 λ \cdot λ - (- 2 \cdot 4)$

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 $λ$ .

$p (λ) = - 9 - 3 λ + 3 λ - 1 \cdot - 1 (λ \cdot λ) - (- 2 \cdot 4)$

Step 1.5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = - 9 - 3 λ + 3 λ - 1 \cdot - 1 λ^{2} - (- 2 \cdot 4)$

Step 1.5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = - 9 - 3 λ + 3 λ + 1 λ^{2} - (- 2 \cdot 4)$

Step 1.5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = - 9 - 3 λ + 3 λ + λ^{2} - (- 2 \cdot 4)$

Step 1.5.2.1.2.2

Add $- 3 λ$ and $3 λ$ .

$p (λ) = - 9 + 0 + λ^{2} - (- 2 \cdot 4)$

Step 1.5.2.1.2.3

Add $- 9$ and $0$ .

$p (λ) = - 9 + λ^{2} - (- 2 \cdot 4)$

Step 1.5.2.1.3

Multiply $- (- 2 \cdot 4)$ .

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Step 1.5.2.1.3.1

Multiply $- 2$ by $4$ .

$p (λ) = - 9 + λ^{2} - - 8$

Step 1.5.2.1.3.2

Multiply $- 1$ by $- 8$ .

$p (λ) = - 9 + λ^{2} + 8$

Step 1.5.2.2

Add $- 9$ and $8$ .

$p (λ) = λ^{2} - 1$

Step 1.6

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

$λ^{2} - 1 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Add $1$ to both sides of the equation.

$λ^{2} = 1$

Step 1.7.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$λ = \pm \sqrt{1}$

Step 1.7.3

Any root of $1$ is $1$ .

$λ = \pm 1$

Step 1.7.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.7.4.1

First, use the positive value of the $\pm$ to find the first solution.

$λ = 1$

Step 1.7.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$λ = - 1$

Step 1.7.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$λ = 1, - 1$

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.

$N ([\begin{array}{c} 3 & 4 \\ - 2 & - 3 \end{array}] - [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Subtract the corresponding elements.

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

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Subtract $1$ from $3$ .

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

Step 3.2.2.2

Subtract $0$ from $4$ .

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

Step 3.2.2.3

Subtract $0$ from $- 2$ .

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

Step 3.2.2.4

Subtract $1$ from $- 3$ .

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

Step 3.3

Find the null space when $λ = 1$ .

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Step 3.3.1

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

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

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 $\frac{1}{2}$ 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 $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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

Step 3.3.2.1.2

Simplify $R_{1}$ .

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

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

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.3

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

$x + 2 y = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = - 1$ .

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Step 4.1

Substitute the known values into the formula.

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

Step 4.2

Simplify.

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Step 4.2.1

Add the corresponding elements.

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

Step 4.2.2

Simplify each element.

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Step 4.2.2.1

Add $3$ and $1$ .

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

Step 4.2.2.2

Add $4$ and $0$ .

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

Step 4.2.2.3

Add $- 2$ and $0$ .

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

Step 4.2.2.4

Add $- 3$ and $1$ .

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

Step 4.3

Find the null space when $λ = - 1$ .

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Step 4.3.1

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

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

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

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

Step 4.3.2.1.2

Simplify $R_{1}$ .

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

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

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

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.

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