2×2 Matrix Eigenvector Formulas

A shortcut for finding the eigenvectors of a

2 \times 2

matrix.

Table of contents

Introduction
Eigenvalues
Eigenvector formulas
Examples
1 2 3 4 5 6 7
Eigenvector tool
References

Introduction

A = [\begin{matrix} a & b \\ c & d \end{matrix}]

If matrix $A$ is a $2 \times 2$ matrix and $x$ is a nonzero vector and both are in $R^{2}$ or $C^{2}$ , then $x$ is an eigenvector of $A$ if $A x$ is a scalar multiple of $x$ and expressed as

A x = λ x

where

λ

is a scalar known as the eigenvalue associated with

x

Eigenvalues of $A$ are

λ_{1} = \frac{a + d + \sqrt{(a - d)^{2} + 4 b c}}{2}

and

λ_{2} = \frac{a + d - \sqrt{(a - d)^{2} + 4 b c}}{2} .

For matrix $A$ , we present two sets of general eigenvector formulas as well as a pair of special case formulas. The pair of eigenvector formulas associated with $λ_{1}$ are

x_{1} = (\begin{matrix} a - d + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix})

and

x_{1}^{'} = (\begin{matrix} 2 b \\ d - a + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) .

The pair of eigenvector formulas associated with $λ_{2}$ are

x_{2} = (\begin{matrix} a - d - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix})

and

x_{2}^{'} = (\begin{matrix} 2 b \\ d - a - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) .

When $b$ or $c$ is 0, the pair of special case formulas are

x_{λ = a} = (\begin{matrix} a - d \\ c \end{matrix}) and x_{λ = d} = (\begin{matrix} b \\ d - a \end{matrix}) .

Eigenvalues

The characteristic equation for eigenvalues of $A$ is $det (A - λ I) = 0$ and written out is

| \begin{matrix} a - λ & b \\ c & d - λ \end{matrix} | = 0

and the equation is $λ^{2} - (a + d) λ + (a d - b c) = 0$ . The characteristic polynomial $p (λ)$ is $λ^{2} - (a + d) λ + (a d - b c) .$

Eigenvalues of $A$ are determined by solving the characteristic equation and they are

λ_{1} = \frac{a + d + \sqrt{(a - d)^{2} + 4 b c}}{2}

and

λ_{2} = \frac{a + d - \sqrt{(a - d)^{2} + 4 b c}}{2} .

The original discriminant is

(a + d)^{2} - 4 (a d - b c)

Lemma

If $b = 0$ or $c = 0$ , then $λ = {a, d}$ .

Proof

When $b = 0$ or $c = 0$ the characteristic polynomial is $λ^{2} - (a + d) λ + a d$ with roots $λ = {a, d}$ .

Eigenvector formulas

General formulas for $λ_{1}$

Theorem

$λ_{1}$ is an eigenvalue of $A$ associated with eigenvectors

x_{1} = (\begin{matrix} 2 b \\ d - a + \sqrt{(d - a)^{2} + 4 b c} \end{matrix})

and

x_{1}^{'} = (\begin{matrix} a - d + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) .

Proof

Let $x = (x_{1}, x_{2})$ and $x$ is in $R^{2}$ or $C^{2}$ and

\begin{aligned} D & = (a - d)^{2} + 4 b c \\ = (d - a)^{2} + 4 b c \\ = a^{2} - 2 a d + d^{2} + 4 b c . \end{aligned}

The equation for determining eigenvectors corresponding to the eigenvalue $λ_{1}$ is $(A - λ_{1} I) x = 0$ .

\begin{matrix} (1) & (A - λ_{1} I) x = (\begin{matrix} \frac{1}{2} (a - d - \sqrt{D}) x_{1} + b x_{2} \\ c x_{1} + \frac{1}{2} (d - a - \sqrt{D}) x_{2} \end{matrix}) = 0 . \end{matrix}

Solving the first component in $(1)$ , $\frac{1}{2} (a - d - \sqrt{D}) x_{1} + b x_{2} = 0$ , we have

x_{1} = \frac{2 b}{d - a + \sqrt{D}} x_{2}

and the eigenvector is

x_{1} = (\begin{matrix} 2 b \\ d - a + \sqrt{D} \end{matrix}) .

Next, we compute the dot product of the eigenvector

x_{1}

and

A

\begin{aligned} [\begin{matrix} a & b \\ c & d \end{matrix}] (\begin{matrix} 2 b \\ d - a + \sqrt{D} \end{matrix}) & = (\begin{matrix} b (a + d + \sqrt{D}) \\ 2 b c + d (d - a + \sqrt{D}) \end{matrix}) . \end{aligned}

To verify $A x_{1} = λ_{1} x_{1}$ , we compute the scalar product of $λ_{1}$ and the eigenvector $x_{1}$ .

\begin{aligned} \frac{a + d + \sqrt{D}}{2} (\begin{matrix} 2 b \\ d - a + \sqrt{D} \end{matrix}) & = (\begin{matrix} b (a + d + \sqrt{D}) \\ \frac{1}{2} (a + d + \sqrt{D}) (d - a + \sqrt{D}) \end{matrix}) \\ = (\begin{matrix} b (a + d + \sqrt{D}) \\ \frac{1}{2} (2 d^{2} + 2 d \sqrt{D} - 2 a d + 4 b c) \end{matrix}) \\ = (\begin{matrix} b (a + d + \sqrt{D}) \\ 2 b c + d (d - a + \sqrt{D}) \end{matrix}) . \end{aligned}

Solving the second component in $(1)$ , $c x_{1} + \frac{1}{2} (d - a - \sqrt{D}) x_{2} = 0$ , we find

x_{1} = \frac{a - d + \sqrt{D}}{2 c} x_{2}

and the eigenvector is

x_{1}^{'} = (\begin{matrix} a - d + \sqrt{D} \\ 2 c \end{matrix}) .

Next, we compute the dot product of the eigenvector $x_{1}^{'}$ and $A$ .

\begin{aligned} [\begin{matrix} a & b \\ c & d \end{matrix}] (\begin{matrix} a - d + \sqrt{D} \\ 2 c \end{matrix}) & = (\begin{matrix} a (a - d + \sqrt{D}) + 2 b c \\ c (a + d + \sqrt{D}) \end{matrix}) . \end{aligned}

To verify $A x_{1}^{'} = λ_{1} x_{1}^{'}$ , we compute the scalar product of $λ_{1}$ and the eigenvector $x_{1}^{'}$ .

\begin{aligned} \frac{a + d + \sqrt{D}}{2} (\begin{matrix} a - d + \sqrt{D} \\ 2 c \end{matrix}) & = (\begin{matrix} \frac{1}{2} (a + d + \sqrt{D}) (a - d + \sqrt{D}) \\ c (a + d + \sqrt{D}) \end{matrix}) \\ = (\begin{matrix} \frac{1}{2} (2 a^{2} + 2 a \sqrt{D} - 2 a d + 4 b c) \\ c (a + d + \sqrt{D}) \end{matrix}) \\ = (\begin{matrix} a (a - d + \sqrt{D}) + 2 b c \\ c (a + d + \sqrt{D}) \end{matrix}) . \end{aligned}

◻

General formulas for $λ_{2}$

Theorem

$λ_{2}$ is an eigenvalue of $A$ associated with $x_{2}$ which are given by

x_{2} = (\begin{matrix} 2 b \\ d - a - \sqrt{(d - a)^{2} + 4 b c} \end{matrix})

and

x_{2}^{'} = (\begin{matrix} a - d - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) .

Proof

Let $x = (x_{1}, x_{2})$ and $x$ in $R^{2}$ or $C^{2}$ and

\begin{aligned} D & = (a - d)^{2} + 4 b c \\ = (d - a)^{2} + 4 b c \\ = a^{2} - 2 a d + d^{2} + 4 b c . \end{aligned}

The equation for determining eigenvectors corresponding to the eigenvalue $λ_{2}$ is $(A - λ_{2} I) x = 0$ .

\begin{matrix} (2) & (A - λ_{2} I) x = (\begin{matrix} \frac{1}{2} (a - d + \sqrt{D}) x_{1} + b x_{2} \\ c x_{1} + \frac{1}{2} (d - a + \sqrt{D}) x_{2} \end{matrix}) = 0 . \end{matrix}

Solving the first component in $(2)$ , $\frac{1}{2} (a - d + \sqrt{D}) x_{1} + b x_{2} = 0$ , we find

x_{1} = \frac{2 b}{d - a - \sqrt{D}} x_{2}

and the eigenvector is

x_{2} = (\begin{matrix} 2 b \\ d - a - \sqrt{D} \end{matrix}) .

Next, we compute the dot product of the eigenvector $x_{2}$ and $A$ .

[\begin{matrix} a & b \\ c & d \end{matrix}] (\begin{matrix} 2 b \\ d - a - \sqrt{D} \end{matrix}) = (\begin{matrix} b (a + d - \sqrt{D}) \\ 2 b c + d (d - a - \sqrt{D}) \end{matrix}) .

To verify $A x_{2} = λ_{2} x_{2}$ , we compute the scalar product of $λ_{2}$ and the eigenvector $x_{2}$ .

\begin{aligned} (\frac{a + d - \sqrt{D}}{2}) (\begin{matrix} 2 b \\ d - a - \sqrt{D} \end{matrix}) & = (\begin{matrix} b (a + d - \sqrt{D}) \\ \frac{1}{2} (a + d - \sqrt{D}) (d - a - \sqrt{D}) \end{matrix}) \\ = (\begin{matrix} b (a + d - \sqrt{D}) \\ \frac{1}{2} (2 d^{2} - 2 d \sqrt{D} - 2 a d + 4 b c) \end{matrix}) \\ = (\begin{matrix} b (a + d - \sqrt{D}) \\ 2 b c + d (d - a - \sqrt{D}) \end{matrix}) . \end{aligned}

Solving the second component in $(2)$ , $c x_{1} + \frac{1}{2} (d - a + \sqrt{D}) x_{2} = 0$ , we find

x_{1} = \frac{a - d - \sqrt{D}}{2 c} x_{2}

and the eigenvector is

x_{2}^{'} = (\begin{matrix} a - d - \sqrt{D} \\ 2 c \end{matrix}) .

Next, we compute the dot product of the eigenvector $x_{2}^{'}$ and $A$ .

[\begin{matrix} a & b \\ c & d \end{matrix}] (\begin{matrix} a - d - \sqrt{D} \\ 2 c \end{matrix}) = (\begin{matrix} a (a - d - \sqrt{D}) + 2 b c \\ c (a + d - \sqrt{D}) \end{matrix}) .

To verify $A x_{2}^{'} = λ_{2} x_{2}^{'}$ , we compute the scalar product of $λ_{2}$ and the eigenvector $x_{2}^{'}$ .

\begin{aligned} (\frac{a + d - \sqrt{D}}{2}) (\begin{matrix} a - d - \sqrt{D} \\ 2 c \end{matrix}) & = (\begin{matrix} \frac{1}{2} (a + d - \sqrt{D}) (a - d - \sqrt{D}) \\ c (a + d - \sqrt{D}) \end{matrix}) \\ = (\begin{matrix} \frac{1}{2} (2 a^{2} - 2 a d - 2 a \sqrt{D} + 4 b c) \\ c (a + d - \sqrt{D}) \end{matrix}) \\ = (\begin{matrix} a (a - d - \sqrt{D}) + 2 b c \\ c (a + d - \sqrt{D}) \end{matrix}) . \end{aligned}

◻

Special case formulas

A set of special case eigenvector formulas exist when $b$ or $c$ is 0.

Theorem

When $b = 0$ then $λ = a$ belongs to the eigenvector $x_{λ = a} = (\begin{matrix} a - d \\ c \end{matrix}) .$

Proof

When $b = 0$ and $λ = a$ , the equation to solve for the eigenvector is $(A - a I) x = 0$ and written out is

(\begin{matrix} 0 \\ c x_{1} + (d - a) x_{2} \end{matrix}) = 0 .

Solving the second component we find

x_{1} = \frac{a - d}{c} x_{2}

and the eigenvector is

x_{λ = a} = (\begin{matrix} a - d \\ c \end{matrix}) .

We confirm $x_{λ = a}$ is an eigenvector by showing $A x = λ x$ and we have

[\begin{matrix} a & 0 \\ c & d \end{matrix}] (\begin{matrix} a - d \\ c \end{matrix}) = (\begin{matrix} a (a - d) \\ a c \end{matrix}) = a (\begin{matrix} a - d \\ c \end{matrix}) .

◻

Theorem

If $c = 0$ then $λ = d$ belongs to the eigenvector $x_{λ = d} = (\begin{matrix} b \\ d - a \end{matrix}) .$

Proof

When $c = 0$ and $λ = d$

(A - d I) x = (\begin{matrix} (a - d) x_{1} + b x_{2} \\ 0 \end{matrix}) = 0

Solving the first component we find $(d - a) x_{1} = b x_{2}$ and the eigenvector is

x_{λ = d} = (\begin{matrix} b \\ d - a \end{matrix}) .

We confirm $x_{λ = d}$ is an eigenvector by showing $A x = λ x$ and we have

[\begin{matrix} a & b \\ 0 & d \end{matrix}] (\begin{matrix} b \\ d - a \end{matrix}) = (\begin{matrix} b d \\ d (d - a) \end{matrix}) = d (\begin{matrix} b \\ d - a \end{matrix}) .

◻

Examples

Example 1

Let $A = [\begin{matrix} 1 & 3 \\ 4 & 5 \end{matrix}]$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} - 6 λ - 7$ and the eigenvalues are $λ_{1} = 7, λ_{2} = - 1$ . The eigenvectors of $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - 5) + \sqrt{16 + 4 (3) (4)} \\ 2 (4) \end{matrix}) = 4 (\begin{matrix} 1 \\ 2 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (3) \\ (5 - 1) + \sqrt{16 + 4 (3) (4)} \end{matrix}) = 6 (\begin{matrix} 1 \\ 2 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - 5) - \sqrt{16 + 4 (3) (4)} \\ 2 (4) \end{matrix}) = 4 (\begin{matrix} - 3 \\ 2 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (3) \\ (5 - 1) - \sqrt{16 + 4 (3) (4)} \end{matrix}) = - 2 (\begin{matrix} - 3 \\ 2 \end{matrix}) \end{aligned}

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Example 2

Let $A = [\begin{matrix} 1 & i \\ - i & - 1 \end{matrix}]$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} - 2$ and the eigenvalues are $λ_{1} = \sqrt{2}, λ_{2} = - \sqrt{2}$ . The eigenvectors of $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - (- 1)) + \sqrt{(1 - (- 1))^{2} + 4 (i) (- i)} \\ 2 (- i) \end{matrix}) = 2 (\begin{matrix} (1 + \sqrt{2}) i \\ 1 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (i) \\ (- 1 - 1) + \sqrt{(- 1 - 1)^{2} + 4 (i) (- i)} \end{matrix}) = 2 (\begin{matrix} (1 + \sqrt{2}) i \\ 1 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - (- 1)) - \sqrt{(1 - (- 1))^{2} + 4 (i) (- i)} \\ 2 (- i) \end{matrix}) = 2 (\begin{matrix} (1 - \sqrt{2}) i \\ 1 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (i) \\ (- 1 - 1) - \sqrt{(- 1 - 1)^{2} + 4 (i) (- i)} \end{matrix}) = 2 (\begin{matrix} (1 - \sqrt{2}) i \\ 1 \end{matrix}) \end{aligned}

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Example 3

Let $A = [\begin{matrix} 1 & 3 \\ 0 & 5 \end{matrix}]$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} - 6 λ + 5$ and the eigenvalues are $λ_{1} = 5, λ_{2} = 1$ . The eigenvectors of $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - 5) + \sqrt{(1 - 5)^{2} + 4 (3) (0)} \\ 2 (0) \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ x_{λ = d} & = (\begin{matrix} b \\ d - a \end{matrix}) = (\begin{matrix} 3 \\ 4 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (3) \\ (5 - 1) + \sqrt{16} \end{matrix}) = 2 (\begin{matrix} 3 \\ 4 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - 5) - \sqrt{(1 - 5)^{2} + 4 (3) (0)} \\ 2 (0) \end{matrix}) = - 8 (\begin{matrix} 1 \\ 0 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (3) \\ (5 - 1) - \sqrt{16} \end{matrix}) = 6 (\begin{matrix} 1 \\ 0 \end{matrix}) \end{aligned}

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Example 4

Let $A = [\begin{matrix} 1 & 0 \\ 4 & 3 \end{matrix}]$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} - 4 λ + 3$ and the eigenvalues are $λ_{1} = 3, λ_{2} = 1$ . The eigenvectors of $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - 3) + \sqrt{(1 - 3)^{2}} \\ 2 (4) \end{matrix}) = 8 (\begin{matrix} 0 \\ 1 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (0) \\ (3 - 1) + \sqrt{(3 - 1)^{2}} \end{matrix}) = 4 (\begin{matrix} 0 \\ 1 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - 3) - \sqrt{(1 - 3)^{2}} \\ 2 (4) \end{matrix}) = 4 (\begin{matrix} - 1 \\ 2 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (0) \\ (3 - 1) - \sqrt{(3 - 1)^{2}} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ x_{λ = a} & = (\begin{matrix} a - d \\ c \end{matrix}) = 2 (\begin{matrix} - 1 \\ 2 \end{matrix}) \end{aligned}

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Example 5

Let $A = (\begin{matrix} 1 & 0 \\ 4 & i \end{matrix})$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} - (1 + i) λ + i$ and the eigenvalues are $λ_{1} = 1, λ_{2} = i$ . The eigenvectors of $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - i) + \sqrt{(1 - i)^{2} + 4 (0) (4)} \\ 2 (4) \end{matrix}) = (\begin{matrix} 1 - i + 1 - i \\ 8 \end{matrix}) = 2 (\begin{matrix} 1 - i \\ 4 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (0) \\ (i - 1) + \sqrt{(i - 1)^{2} + 4 (0) (4)} \end{matrix}) = (\begin{matrix} 0 \\ i - 1 + 1 - i \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ x_{λ = a} & = (\begin{matrix} a - d \\ c \end{matrix}) = (\begin{matrix} 1 - i \\ 4 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (1 - i) - \sqrt{(1 - i)^{2} + 4 (0) (4)} \\ 2 (4) \end{matrix}) = (\begin{matrix} 1 - i - 1 + i \\ 8 \end{matrix}) = 8 (\begin{matrix} 0 \\ 1 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (0) \\ (i - 1) - \sqrt{(i - 1)^{2} + 4 (0) (4)} \end{matrix}) = (\begin{matrix} 0 \\ i - 1 - 1 + i \end{matrix}) = 2 (i - 1) (\begin{matrix} 0 \\ 1 \end{matrix}) \\ \sqrt{(i - 1)^{2}} = \sqrt{- 2 i} = 1 - i \end{aligned}

WolframAlpha

Example 6

Let $A = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} + 1$ and the eigenvalues are $λ_{1} = i, λ_{2} = - i$ . The eigenvectors are $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (0 - 0) + \sqrt{(0 - 0)^{2} + 4 (- 1) (1)} \\ 2 (1) \end{matrix}) = 2 (\begin{matrix} i \\ 1 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (- 1) \\ (0 - 0) + \sqrt{(0 - 0) + 4 (- 1) (1)} \end{matrix}) = 2 i (\begin{matrix} i \\ 1 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (0 - 0) - \sqrt{(0 - 0)^{2} + 4 (- 1) (1)} \\ 2 (1) \end{matrix}) = 2 (\begin{matrix} - i \\ 1 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (- 1) \\ (0 - 0) - \sqrt{(0 - 0) + 4 (- 1) (1)} \end{matrix}) = - 2 i (\begin{matrix} - i \\ 1 \end{matrix}) \end{aligned}

Wolfram|Alpha

Example 7

Let $A = (\begin{matrix} 4 & - 5 \\ 2 & - 2 \end{matrix})$ . The characteristic polynomial of $A$ is $p (λ) = λ^{2} - 2 λ + 2$ and the eigenvalues are $λ_{1} = 1 + i, λ_{2} = 1 - i$ . The eigenvectors are $A$ are

\begin{aligned} x_{1} & = (\begin{matrix} (a - d) + \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (4 - (- 2)) + \sqrt{(4 - (- 2))^{2} + 4 (- 5) (2)} \\ 2 (2) \end{matrix}) = 2 (\begin{matrix} 3 + i \\ 2 \end{matrix}) \\ x_{1} & = (\begin{matrix} 2 b \\ (d - a) + \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (- 5) \\ (- 2 - 4) + \sqrt{(- 2 - 4)^{2} + 4 (- 5) (2)} \end{matrix}) = - 10 (\begin{matrix} 3 + i \\ 2 \end{matrix}) \\ x_{2} & = (\begin{matrix} (a - d) - \sqrt{(a - d)^{2} + 4 b c} \\ 2 c \end{matrix}) \\ = (\begin{matrix} (4 - (- 2)) - \sqrt{(4 - (- 2))^{2} + 4 (- 5) (2)} \\ 2 (2) \end{matrix}) = 2 (\begin{matrix} 3 - i \\ 2 \end{matrix}) \\ x_{2} & = (\begin{matrix} 2 b \\ (d - a) - \sqrt{(d - a)^{2} + 4 b c} \end{matrix}) \\ = (\begin{matrix} 2 (- 5) \\ (- 2 - 4) - \sqrt{(- 2 - 4)^{2} + 4 (- 5) (2)} \end{matrix}) = - 10 (\begin{matrix} 3 - i \\ 2 \end{matrix}) \end{aligned}

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Eigenvector tool

Eigenvector Tool

The eigenvector tool only utilizes the previously defined formulas to calculate eigenvectors. Additionally a link to WolframAlpha that solves the same matrix is provided to confirm the results.

eigenvectorslinear algebra

Introduction

Eigenvalues

Eigenvector formulas

General formulas for λ1

General formulas for λ2

Special case formulas

Examples

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Eigenvector tool

General formulas for $λ_{1}$

General formulas for $λ_{2}$