$$ \newcommand\A{\begin{bmatrix} a & b \\ c & d \end{bmatrix}} \newcommand\matr[1]{#1} \newcommand\sqrtD{\sqrt{D}} \renewcommand{\vec}[1]{\mathbf{#1}} $$

A shortcut for finding the eigenvectors of a $2\times2$ matrix.
Table of contents
  1. Introduction
  2. Eigenvalues
  3. Eigenvector formulas
    1. General formulas for $\lambda_1$
    2. General formulas for $\lambda_2$
    3. Special case formulas
  4. Examples
    1 2 3 4 5 6 7
  5. Eigenvector tool
  6. References

Introduction

$$ \matr A = \A $$

If matrix $\matr A$ is a $2 \times 2$ matrix and $\vec{x}$ is a nonzero vector and both are in $\mathbb{R}^2$ or $\mathbb{C}^2$, then $\vec{x}$ is an eigenvector of $\matr A$ if $\matr A\vec{x}$ is a scalar multiple of $\vec{x}$ and expressed as $$ \matr A \vec{x} = \lambda \vec{x} $$ where $\lambda$ is a scalar known as the eigenvalue associated with $\vec{x}$.

Eigenvalues of $\matr A$ are $$ \lambda_1 = \frac{a + d + \sqrt{(a-d)^2 + 4bc} }{2} $$ and $$ \lambda_2 = \frac{a + d - \sqrt{(a-d)^2 + 4bc} }{2} \text. $$

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 $\lambda_1$ are

$$ \vec{x}_1 = \begin{pmatrix} a - d + \sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} $$ and $$ \vec{x}_1' = \begin{pmatrix} 2b \\ d-a + \sqrt{(d-a)^2+4bc} \end{pmatrix} \text.$$

The pair of eigenvector formulas associated with $\lambda_2$ are

$$ \vec{x}_2 = \begin{pmatrix} a - d - \sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} $$ and $$ \vec{x}_2' = \begin{pmatrix} 2b \\ d-a - \sqrt{(d-a)^2+4bc} \end{pmatrix} \text.$$

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

$$ \vec x_{\lambda=a} = \begin{pmatrix} a-d \\ c \end{pmatrix} \hspace{2em} \mbox{ and } \hspace{2em} \vec x_{\lambda=d} = \begin{pmatrix} b \\ d-a \end{pmatrix} \text. $$

Eigenvalues

The characteristic equation for eigenvalues of $A$ is $ \text{det}(\matr A - \lambda I) = 0 $ and written out is

$$ \begin{vmatrix} a - \lambda & b \\ c & d - \lambda \end{vmatrix} = 0 $$

and the equation is $\lambda^2-(a+d)\lambda+(ad-bc)=0$. The characteristic polynomial $p(\lambda)$ is $ \lambda^2 - (a + d) \lambda + (ad - bc) \text. $

Eigenvalues of $\matr A$ are determined by solving the characteristic equation and they are $$ \lambda_1 = \frac{a + d + \sqrt{(a-d)^2 + 4bc} }{2} $$ and $$ \lambda_2 = \frac{a + d - \sqrt{(a-d)^2 + 4bc} }{2} \text. $$ The original discriminant is $(a+d)^2-4(ad-bc)$.

Lemma

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

Proof

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

Eigenvector formulas

General formulas for $\lambda_1$

Theorem

$\lambda_1$ is an eigenvalue of $A$ associated with eigenvectors $$ \vec{x}_1 = \begin{pmatrix} 2b \\ d-a + \sqrt{(d-a)^2+4bc} \end{pmatrix} $$ and $$ \vec{x}_1' = \begin{pmatrix} a-d+\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \text. $$

Proof

Let $\vec{x} = (x_1, x_2)$ and $\vec{x}$ is in $\mathbb{R}^2$ or $\mathbb{C}^2$ and

$$ \begin{aligned} D & = (a-d)^2 + 4bc \\ & = (d-a)^2 + 4bc \\ & = a^2 - 2ad + d^2 + 4bc \text. \end{aligned} $$

The equation for determining eigenvectors corresponding to the eigenvalue $\lambda_1$ is $(A-\lambda_1 I)\vec x=0$.

$$ (A-\lambda_{1}I)\vec{x} = \begin{pmatrix} \frac{1}{2}(a - d-\sqrtD)x_1 + bx_2 \\ cx_1 + \frac{1}{2}(d-a-\sqrtD)x_2 \end{pmatrix} = 0 \text. \tag{1} $$

Solving the first component in $(1)$, $\frac{1}{2}(a - d-\sqrtD)x_1 + bx_2 = 0$, we have $$ x_1 = \frac{2b}{d-a+\sqrtD}x_2 $$ and the eigenvector is $$ \vec{x}_1 = \begin{pmatrix} 2b \\ d-a+\sqrtD \end{pmatrix} \text.$$ Next, we compute the dot product of the eigenvector $\vec{x}_1$ and $A$.

$$ \begin{aligned} \A \begin{pmatrix} 2b \\ d-a + \sqrt{D} \end{pmatrix} &= \begin{pmatrix} b(a+d+\sqrtD)\\ 2bc + d(d-a + \sqrtD) \end{pmatrix} \text. \end{aligned} $$

To verify $A\vec{x}_1 = \lambda_1 \vec{x}_1$, we compute the scalar product of $\lambda_1$ and the eigenvector $\vec{x}_1$.

$$ \begin{aligned} % \left(\frac{a+d+\sqrtD}{2} \right) \begin{pmatrix} 2b \\ d-a+\sqrtD \end{pmatrix} &= % \begin{pmatrix} b(a+d+\sqrtD) \\ \frac{1}{2}(a+d+\sqrtD)(d-a+\sqrtD) \end{pmatrix} \\[10pt] \frac{a+d+\sqrtD}{2} \begin{pmatrix} 2b \\ d-a+\sqrtD \end{pmatrix} &= \begin{pmatrix} b(a+d+\sqrtD) \\ \frac{1}{2}(a+d+\sqrtD)(d-a+\sqrtD) \end{pmatrix} \\[10pt] &= \begin{pmatrix} b(a+d+\sqrtD) \\ \frac{1}{2}(2d^2+2d\sqrtD-2ad + 4bc) \end{pmatrix} \\[10pt] &= \begin{pmatrix} b(a+d+\sqrtD) \\ 2bc + d(d-a+\sqrtD) \end{pmatrix} \text. \end{aligned} $$

Solving the second component in $(1)$, $cx_1 + \frac{1}{2}(d-a-\sqrtD)x_2 = 0$, we find

$$ x_1 = \frac{a-d+\sqrtD}{2c}x_2 $$

and the eigenvector is

$$ \vec{x}_1' = \begin{pmatrix} a-d+\sqrtD \\ 2c \end{pmatrix} \text. $$

Next, we compute the dot product of the eigenvector $\vec{x}_1'$ and $A$.

$$ \begin{aligned} \A \begin{pmatrix} a-d+\sqrtD \\ 2c \end{pmatrix} &= \begin{pmatrix} a(a-d+\sqrtD)+2bc \\ c(a+d+\sqrtD) \end{pmatrix} \text. \end{aligned} $$

To verify $A\vec{x}_1' = \lambda_1 \vec{x}_1'$, we compute the scalar product of $\lambda_1$ and the eigenvector $\vec{x}_1'$.

$$ \begin{aligned} % \left( \frac{a+d+\sqrtD}{2} \right) \begin{pmatrix} a-d+\sqrtD \\ 2c \end{pmatrix} &= \frac{a+d+\sqrtD}{2} \begin{pmatrix} a-d+\sqrtD \\ 2c \end{pmatrix} &= \begin{pmatrix} \frac{1}{2}(a+d+\sqrtD)(a-d+\sqrtD) \\ c(a+d+\sqrtD) \end{pmatrix} \\[10pt] &= \begin{pmatrix} \frac{1}{2}(2a^2+2a\sqrtD-2ad+4bc) \\ c(a+d+\sqrtD) \end{pmatrix} \\[10pt] &= \begin{pmatrix} a(a-d+\sqrtD)+2bc \\ c(a+d+\sqrtD) \end{pmatrix} \text. \end{aligned} $$

$$\square$$

General formulas for $\lambda_2$

Theorem

$\lambda_2$ is an eigenvalue of $A$ associated with $\vec{x}_2$ which are given by $$ \vec{x}_2 = \begin{pmatrix} 2b \\ d-a-\sqrt{(d-a)^2+4bc} \end{pmatrix} $$ and $$ \vec{x}_2' = \begin{pmatrix} a-d-\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \text. $$

Proof

Let $\vec{x} = (x_1,x_2)$ and $\vec{x}$ in $\mathbb{R}^2$ or $\mathbb{C}^2$ and

$$ \begin{aligned} D & = (a-d)^2 + 4bc \\ & = (d-a)^2 + 4bc \\ & = a^2 - 2ad + d^2 + 4bc \text. \end{aligned} $$

The equation for determining eigenvectors corresponding to the eigenvalue $\lambda_2$ is $(A-\lambda_2 I)\vec x=0$.

$$ (A-\lambda_2 I)\vec x = \begin{equation} \begin{pmatrix} \frac{1}{2}(a-d+\sqrtD)x_1 + bx_2 \\ cx_1 + \frac{1}{2}(d-a+\sqrtD)x_2 \end{pmatrix} = 0 \text. \tag{2} \end{equation} $$

Solving the first component in $(2)$, $\frac{1}{2}(a-d+\sqrtD)x_1 + bx_2 = 0$, we find

$$ x_1 = \frac{2b}{d-a-\sqrtD}x_2 $$

and the eigenvector is

$$ \vec{x}_2 = \begin{pmatrix} 2b \\ d-a-\sqrtD \end{pmatrix} \text.$$

Next, we compute the dot product of the eigenvector $\vec{x}_2$ and $A$.

$$ \A \begin{pmatrix} 2b \\ d-a-\sqrtD \end{pmatrix} = \begin{pmatrix} b(a+d-\sqrtD) \\ 2bc + d(d-a-\sqrtD) \end{pmatrix} \text. $$

To verify $A\vec{x}_2 = \lambda_2 \vec{x}_2$, we compute the scalar product of $\lambda_2$ and the eigenvector $\vec{x}_2$.

$$ \begin{aligned} \left( \frac{a+d-\sqrtD}{2} \right) \begin{pmatrix} 2b \\ d-a-\sqrtD \end{pmatrix} &= \begin{pmatrix} b(a+d-\sqrtD) \\ \frac{1}{2}(a+d-\sqrtD)(d-a-\sqrtD) \end{pmatrix} \\[10pt] &= \begin{pmatrix} b(a+d-\sqrtD) \\ \frac{1}{2}(2d^2-2d\sqrtD-2ad+4bc) \end{pmatrix}\\[10pt] &= \begin{pmatrix} b(a+d-\sqrtD) \\ 2bc+d(d-a-\sqrtD) \end{pmatrix} \text. \end{aligned} $$

Solving the second component in $(2)$, $cx_1 + \frac{1}{2}(d-a+\sqrtD)x_2 = 0$, we find

$$ x_1 = \frac{a-d-\sqrtD}{2c}x_2 $$

and the eigenvector is

$$ \vec{x}_2' = \begin{pmatrix} a-d-\sqrtD \\ 2c \end{pmatrix} \text. $$

Next, we compute the dot product of the eigenvector $\vec{x}_2'$ and $A$.

$$ \A \begin{pmatrix} a-d-\sqrtD \\ 2c \end{pmatrix} = \begin{pmatrix} a(a-d-\sqrtD)+2bc \\ c(a+d-\sqrtD) \end{pmatrix} \text. $$

To verify $A\vec{x}_2' = \lambda_2 \vec{x}_2'$, we compute the scalar product of $\lambda_2$ and the eigenvector $\vec{x}_2'$.

$$ \begin{aligned} \left( \frac{a+d-\sqrtD}{2} \right) \begin{pmatrix} a-d-\sqrtD \\ 2c \end{pmatrix} &= \begin{pmatrix} \frac{1}{2}(a+d-\sqrtD)(a-d-\sqrtD) \\ c(a+d-\sqrtD) \end{pmatrix}\\[10pt] &= \begin{pmatrix} \frac{1}{2}(2a^2-2ad-2a\sqrtD+4bc) \\ c(a+d-\sqrtD) \end{pmatrix}\\[10pt] &= \begin{pmatrix} a(a-d-\sqrtD)+2bc \\ c(a+d-\sqrtD) \end{pmatrix} \text. \end{aligned} $$

$$\square$$

Special case formulas

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

Theorem

When $b=0$ then $\lambda=a$ belongs to the eigenvector $\vec x_{\lambda=a} = \begin{pmatrix} a-d \\ c \end{pmatrix} \text.$

Proof

When $b=0$ and $\lambda=a$, the equation to solve for the eigenvector is $(A-aI)\vec{x}=0$ and written out is

$$ \begin{pmatrix} 0 \\ cx_1 + (d-a)x_2 \end{pmatrix} = 0 \text. $$

Solving the second component we find

$$ x_1 = \frac{a-d}{c} x_2 $$

and the eigenvector is

$$ \vec{x}_{\lambda=a} = \begin{pmatrix} a-d \\ c \end{pmatrix} \text. $$

We confirm $\vec{x}_{\lambda=a}$ is an eigenvector by showing $A\vec{x} = \lambda\vec{x}$ and we have

$$ \begin{bmatrix} a & 0 \\ c & d \end{bmatrix} \begin{pmatrix} a-d \\ c \end{pmatrix} = \begin{pmatrix} a(a-d) \\ ac \end{pmatrix} = a \begin{pmatrix} a-d \\ c \end{pmatrix} \text. $$

$$\square$$
Theorem

If $c=0$ then $\lambda=d$ belongs to the eigenvector $\vec x_{\lambda=d} = \begin{pmatrix} b \\ d-a \end{pmatrix} \text.$

Proof

When $c=0$ and $\lambda = d$

$$ (\matr A - dI)\vec x = \begin{pmatrix} (a-d)x_1 + bx_2 \\ 0 \end{pmatrix} = 0 $$

Solving the first component we find $(d-a)x_1 = bx_2$ and the eigenvector is

$$ \vec{x}_{\lambda=d} = \begin{pmatrix} b \\ d-a \end{pmatrix} \text. $$

We confirm $\vec{x}_{\lambda=d}$ is an eigenvector by showing $A\vec{x} = \lambda\vec{x}$ and we have

$$ \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} \begin{pmatrix} b \\ d-a \end{pmatrix} = \begin{pmatrix} bd \\ d(d-a) \end{pmatrix} = d \begin{pmatrix} b \\ d-a \end{pmatrix} \text. $$

$$\square$$

Examples

Example 1

Let $\matr A = \begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix}$. The characteristic polynomial of $\matr A$ is $p(\lambda) = \lambda^2-6\lambda-7$ and the eigenvalues are $\lambda_1 = 7,\hspace{3pt} \lambda_2 = -1$. The eigenvectors of $\matr A$ are

$$ \begin{aligned} \vec{x}_1 &= \begin{pmatrix} (a-d)+\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\[1em] &= \begin{pmatrix} (1-5)+\sqrt{16+4(3)(4)} \\ 2(4) \end{pmatrix} = 4\begin{pmatrix} 1 \\ 2 \end{pmatrix} \\[20pt] %
\vec{x}_1 &= \begin{pmatrix} 2b \\ (d-a)+\sqrt{(d-a)^2+4bc} \end{pmatrix} \\[1em] &= \begin{pmatrix} 2(3) \\ (5-1)+\sqrt{16+4(3)(4)} \end{pmatrix} = 6\begin{pmatrix} 1 \\ 2 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} (a-d)-\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\[1em] &= \begin{pmatrix} (1-5)-\sqrt{16+4(3)(4)} \\ 2(4) \end{pmatrix} = 4\begin{pmatrix} -3 \\ 2\end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} 2b \\ (d-a)-\sqrt{(d-a)^2+4bc} \end{pmatrix} \\[1em] &= \begin{pmatrix} 2(3) \\ (5-1)-\sqrt{16+4(3)(4)} \end{pmatrix} = -2\begin{pmatrix} -3 \\ 2 \end{pmatrix} \end{aligned} $$ Wolfram|Alpha

Example 2

Let $\matr A = \begin{bmatrix} 1 & i \\ -i & -1 \end{bmatrix}$. The characteristic polynomial of $\matr A$ is $p(\lambda) = \lambda^2 -2$ and the eigenvalues are $\lambda_1 = \sqrt{2}, \lambda_2=-\sqrt{2}$. The eigenvectors of $\matr A$ are

$$ \begin{aligned} \vec{x}_1 &= \begin{pmatrix} (a-d)+\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-(-1))+\sqrt{(1-(-1))^2+4(i)(-i)} \\ 2(-i) \end{pmatrix} = 2\begin{pmatrix} (1+\sqrt{2})i \\ 1 \end{pmatrix} \\[20pt] % \vec{x}_1 &= \begin{pmatrix} 2b \\ (d-a)+\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(i) \\ (-1-1)+\sqrt{(-1-1)^2+4(i)(-i)} \end{pmatrix} = 2\begin{pmatrix} (1+\sqrt{2})i \\ 1 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} (a-d)-\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-(-1))-\sqrt{(1-(-1))^2+4(i)(-i)} \\ 2(-i) \end{pmatrix} = 2\begin{pmatrix} (1-\sqrt{2})i \\ 1 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} 2b \\ (d-a)-\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(i) \\ (-1-1)-\sqrt{(-1-1)^2+4(i)(-i)} \end{pmatrix} = 2\begin{pmatrix} (1-\sqrt{2})i \\ 1 \end{pmatrix} \end{aligned} $$ Wolfram|Alpha

Example 3

Let $\matr A = \begin{bmatrix} 1 & 3 \\ 0 & 5 \end{bmatrix}$. The characteristic polynomial of $\matr A$ is $p(\lambda) = \lambda^2-6\lambda+5$ and the eigenvalues are $\lambda_1 = 5, \lambda_2 = 1$. The eigenvectors of $\matr A$ are

$$ \begin{aligned} \vec{x}_1 &= \begin{pmatrix} (a-d)+\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-5)+\sqrt{(1-5)^2+4(3)(0)} \\ 2(0) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\[20pt] % \vec{x}_{\lambda=d} &= \begin{pmatrix} b \\ d-a \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \\[20pt] % \vec{x}_1 &= \begin{pmatrix} 2b \\ (d-a)+\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(3) \\ (5-1)+\sqrt{16} \end{pmatrix} = 2\begin{pmatrix} 3 \\ 4 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} (a-d)-\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-5)-\sqrt{(1-5)^2+4(3)(0)} \\ 2(0) \end{pmatrix} = -8\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} 2b \\ (d-a)-\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(3) \\ (5-1)-\sqrt{16} \end{pmatrix} = 6\begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{aligned} $$ Wolfram|Alpha

Example 4

Let $\matr A = \begin{bmatrix} 1 & 0 \\ 4 & 3 \end{bmatrix}$. The characteristic polynomial of $\matr A$ is $p(\lambda) = \lambda^2-4\lambda+3$ and the eigenvalues are $\lambda_1=3, \lambda_2=1$. The eigenvectors of $\matr A$ are

$$ \begin{aligned} \vec{x}_1 &= \begin{pmatrix} (a-d)+\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-3)+\sqrt{(1-3)^2} \\ 2(4) \end{pmatrix} = 8\begin{pmatrix} 0 \\ 1 \end{pmatrix} \\[20pt] % \vec{x}_1 &= \begin{pmatrix} 2b \\ (d-a)+\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(0) \\ (3-1)+\sqrt{(3-1)^2} \end{pmatrix} = 4\begin{pmatrix} 0 \\ 1 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} (a-d)-\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-3)-\sqrt{(1-3)^2} \\ 2(4) \end{pmatrix} = 4\begin{pmatrix} -1 \\ 2 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} 2b \\ (d-a)-\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(0) \\ (3-1)-\sqrt{(3-1)^2} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\[20pt] % \vec{x}_{\lambda=a} &= \begin{pmatrix} a-d \\ c \end{pmatrix} = 2\begin{pmatrix} -1 \\ 2 \end{pmatrix} \end{aligned} $$ Wolfram|Alpha

Example 5

Let $\matr A = \begin{pmatrix} 1 & 0 \\ 4 & i \end{pmatrix}$. The characteristic polynomial of $\matr A$ is $p(\lambda) = \lambda^2-(1+i)\lambda+i$ and the eigenvalues are $\lambda_1 = 1, \lambda_2 = i$. The eigenvectors of $\matr A$ are

$$ \begin{aligned} \vec{x}_1 &= \begin{pmatrix} (a-d)+\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-i)+\sqrt{(1-i)^2+4(0)(4)} \\ 2(4) \end{pmatrix} = \begin{pmatrix} 1-i+1-i \\ 8 \end{pmatrix} = 2\begin{pmatrix} 1-i \\ 4 \end{pmatrix} \\[20pt] % \vec{x}_1 &= \begin{pmatrix} 2b \\ (d-a)+\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(0) \\ (i-1)+\sqrt{(i-1)^2+4(0)(4)} \end{pmatrix} = \begin{pmatrix} 0 \\ i-1+1-i \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\[20pt] % \vec{x}_{\lambda=a} &= \begin{pmatrix} a-d \\ c \end{pmatrix} = \begin{pmatrix} 1-i \\ 4 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} (a-d)-\sqrt{(a-d)^2+4bc} \\ 2c \end{pmatrix} \\ &= \begin{pmatrix} (1-i)-\sqrt{(1-i)^2+4(0)(4)} \\ 2(4) \end{pmatrix} = \begin{pmatrix} 1-i-1+i \\ 8 \end{pmatrix} = 8\begin{pmatrix} 0 \\ 1 \end{pmatrix} \\[20pt] % \vec{x}_2 &= \begin{pmatrix} 2b \\ (d-a)-\sqrt{(d-a)^2+4bc} \end{pmatrix} \\ &= \begin{pmatrix} 2(0) \\ (i-1)-\sqrt{(i-1)^2+4(0)(4)} \end{pmatrix} = \begin{pmatrix} 0 \\ i-1-1+i \end{pmatrix} = 2(i-1)\begin{pmatrix} 0 \\ 1 \end{pmatrix} \\[20pt] % &\sqrt{(i-1)^2} = \sqrt{-2i} = 1-i \end{aligned} $$ WolframAlpha