Wize University Linear Algebra Textbook > Eigenvalues and Eigenvectors (Spectral Theory)

Complex Eigenvalues

Eigenvalues and EigenvectorsPrevious Section

Complex Eigenvalues and Eigenvectors
A square matrix with real entries can have complex eigenvalues!
As usual, given a matrix AAA, we find eigenvalues λ\lambdaλ and corresponding eigenvectors x⃗\vec xx.
Note: Every n×nn \times nn×n matrix has exactly nnn complex eigenvalues (counting any multiplicities).
Wize Tip
If AAA has real entries but complex eigenvalues, then the eigenvalues come in conjugate pairs:
If λ1=a+bi\lambda_1=a+biλ1​=a+bi  is an eigenvalue, then λ2=a−bi\lambda_2=a-biλ2​=a−bi  is another eigenvalue.
Once you find one eigenvector x⃗1\vec x_1x1​ corresponding to λ1\lambda_1λ1​, then the conjugate of x⃗1\vec x_1x1​ is an eigenvector of the conjugate eigenvalue λ2\lambda_2λ2​.

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Steps

Given a real matrix AAA, find the eigenvalues/eigenvectors as follows:
As usual, find det(A−λI){\rm det}(A-\lambda I)det(A−λI) and set it equal to zero. Solve for λ\lambdaλ.
→\rightarrow→If one of the eigenvalues is complex, you should see that its complex conjugate is also an eigenvalue.
Take each eigenvalue one at a time and solve for its corresponding eigenvector(s):
→\rightarrow→Find x⃗\vec xx such that (A−λI)x⃗=0⃗(A-\lambda I)\vec x = \vec 0 (A−λI)x=0
After finding an eigenvector for a complex eigenvalue, you can easily find the eigenvector for the conjugate eigenvalue:
→\rightarrow→Simply take the complex conjugate of the eigenvector you've already found!

Example: Complex Eigenvalues and Eigenvectors
Suppose we are given a real 3×33\times33×3 matrix and we are told two of its eigenvalues are λ1=4\lambda_1=4λ1​=4 and λ2=2−3i\lambda_2=2-3iλ2​=2−3i.
If their corresponding eigenvectors are x⃗1=[11]\vec x_1
=
\begin{bmatrix}
1\\ 1
\end{bmatrix}x1​=[11​] and x⃗2=[2+i5]\vec x_2=

\begin{bmatrix}
2+i\\ 5
\end{bmatrix}x2​=[2+i5​], find the missing eigenvalue and eigenvector pair.

Since λ2\lambda_2λ2​ is complex, its complex conjugate must also be an eigenvalue! 
Therefore, λ3=λ‾2    ⟹    λ3=2+3i\lambda_3 = \overline{\lambda}_2
\ \ \implies\ \ 
\boxed{\lambda_3 = 2 \colorFour{+}3i}λ3​=λ2​  ⟹  λ3​=2+3i​ 
Similarly, since we know the complex eigenvector associated with λ2\lambda_2λ2​, we simply take its conjugate to find the eigenvector for λ3\lambda_3λ3​:
x⃗3=x⃗2‾    ⟹    x⃗3=[2−i5]\vec x_3 = \overline{\vec x_2}
\ \ \implies \ \ 
\boxed{
\vec x_3 = 
\begin{bmatrix}
2 \colorFour{-} i\\ 5
\end{bmatrix}
}x3​=x2​​  ⟹  x3​=[2−i5​]​

Example: Complex Eigenvalues and Eigenvectors
Find the eigenvalues and eigenvectors of A=[3−51−1]A=\left[\begin{array}{rr}
3&-5\\1&-1
\end{array}\right]A=[31​−5−1​].

Find the Eigenvalues
Solve the equation:
det(A−λI)=0∣3−λ−51−1−λ∣=0(3−λ)(−1−λ)−(−5)(1)=0(λ2−2λ−3)+5=0λ2−2λ+2=0\begin{array}{rcl}
\det{A-\lambda I}&=&0\\[1em]

\colorFour{
\left|\begin{array}{cc}3-\lambda & -5\\1&-1-\lambda\end{array}\right|}
&=&0\\[1em]

(3-\lambda)(-1-\lambda)-(-5)(1)&=&0\\[1em]

(\lambda^2-2\lambda-3)+5&=&0\\[1em]

\lambda^2-2\lambda+2&=&0
\end{array}det(A−λI)​3−λ1​−5−1−λ​​(3−λ)(−1−λ)−(−5)(1)(λ2−2λ−3)+5λ2−2λ+2​=====​00000​
We can solve for λ\lambdaλ using the quadratic formula:
λ  =  −(−2)±(−2)2−4(1)(2)2(1)  =  2±−42  =  2±2i2  =  1±i\lambda
\ \ =\ \ \dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}
\ \ =\ \ \dfrac{2 \pm \sqrt{-4}}{2}
\ \ =\ \ \dfrac{2 \pm 2i}{2}
\ \ =\ \ 1\pm iλ  =  2(1)−(−2)±(−2)2−4(1)(2)​​  =  22±−4​​  =  22±2i​  =  1±i
Then the eigenvalues are λ1=1+i\boxed{
\lambda_1 = 1+i
}λ1​=1+i​ and its conjugate, λ2=1−i\boxed{
\lambda_2=1-i
}λ2​=1−i​.
Find the Eigenvectors
λ1=1+i\colorThree{\lambda_1 = 1+i}λ1​=1+i
Substituting this eigenvalue into the matrix above (A−λIA-\lambda IA−λI), our goal is to solve for x⃗1\vec x_1x1​:
[3−(1+i)−51−1−(1+i)]x⃗1=[00]\colorFour{
\begin{bmatrix}
3-\colorThree{(1+i)} & -5\\
1 & -1-\colorThree{(1+i)}
\end{bmatrix}}
\vec x_1 
= 
\begin{bmatrix}
0 \\0
\end{bmatrix}[3−(1+i)1​−5−1−(1+i)​]x1​=[00​]

Solve by writing the augmented matrix of this system and row-reducing:
[2−i−501−2−i0] R2↔R1⟶[1−2−i02−i−50] R2−(2−i)R1Note:  (−2−i)(2−i)=−4−2i+2i+i2=−5⟶[1−2−i0000]\begin{array}{ll}

&\left[\begin{array}{cc|r}
2-i& -5 &0\\
1&-2-i& 0\\
\end{array}\right]

\begin{array}{c}
~\\
R_2 \leftrightarrow R_1\\
\end{array}
\\[2em]

\longrightarrow
&\left[\begin{array}{cc|r}
1&-2-i& 0\\
2-i& -5 &0\\
\end{array}\right]

\begin{array}{c}
~\\
R_2 - (2-i)R_1\\
\end{array}
\\[2em]
&\text{Note: }\ 
(-2-i)(2-i) = -4-\cancel{2i}+\cancel{2i}+i^2=-5
\\[1em]

\longrightarrow
&\left[\begin{array}{cc|r}
1&-2-i& 0\\
0& 0 &0\\
\end{array}\right]

\end{array}⟶⟶​[2−i1​−5−2−i​00​] R2​↔R1​​[12−i​−2−i−5​00​] R2​−(2−i)R1​​Note:  (−2−i)(2−i)=−4−2i+2i+i2=−5[10​−2−i0​00​]​
Including a parameter for the second variable (since there is no leading 1 in the second column), we get the equations:
{1x1+(−2−i)x2=0x2=t  ⟹  {x1=(2+i)tx2=t  ⟹  x⃗1=t[2+i1]\left\{\begin{array}{rcl}
1x_1+(-2-i)x_2&=&0\\
x_2&=&t
\end{array}\right.

\quad \implies \quad

\left\{\begin{array}{rcl}
x_1&=&(2+i)t\\
x_2&=&t
\end{array}\right.

\quad \implies \quad

\vec x_1 = 
t \begin{bmatrix}
2+i\\
1
\end{bmatrix}{1x1​+(−2−i)x2​x2​​==​0t​⟹{x1​x2​​==​(2+i)tt​⟹x1​=t[2+i1​]
So an eigenvector associated with λ1=1+i\lambda_1 = 1+iλ1​=1+i is: x⃗1=[2+i1]\boxed{
\vec x_1
=
\left[\begin{array}{c}
2+i\\
1
\end{array}\right]
}x1​=[2+i1​]​
λ2=1−i\colorThree{\lambda_2 = 1-i}λ2​=1−i
Since the second eigenvalue is the conjugate of the first, its eigenvector is simply the conjugate of the first eigenvector!
Therefore, negating all of the imaginary parts, we have: x⃗2=[2−i1]\boxed{
\vec x_2
=
\left[\begin{array}{c}
2-i\\
1
\end{array}\right]
}x2​=[2−i1​]​

For practice, check your work by applying the definition of eigenvalues/eigenvectors: Ax⃗=λx⃗A\vec x = \lambda \vec xAx=λx.

Practice: Complex Eigenvalues and Eigenvectors
Find the eigenvalues and eigenvectors of A=[−6−2012]A=\left[\begin{array}{rr}
-6&-20\\1&2
\end{array}\right]A=[−61​−202​].

A) λ1=−2−iλ2=−2−ix⃗1=[2−i1]x⃗2=[2+i1]\begin{array}{c|c}
\lambda_1=-2-i \quad & \quad 
\lambda_2=-2-i \\[0.5em]

\vec x_1 = \begin{bmatrix} 2-i\\1 \end{bmatrix}
& \quad
\vec x_2 = \begin{bmatrix} 2+i\\1 \end{bmatrix}
\end{array}λ1​=−2−ix1​=[2−i1​]​λ2​=−2−ix2​=[2+i1​]​

B) λ1=−1−2iλ2=−1+2ix⃗1=[2−i1]x⃗2=[2+i1]\begin{array}{c|c}
\lambda_1=-1-2i \quad & \quad 
\lambda_2=-1+2i \\[0.5em]

\vec x_1 = \begin{bmatrix} 2-i\\1 \end{bmatrix}
& \quad
\vec x_2 = \begin{bmatrix} 2+i\\1 \end{bmatrix}
\end{array}λ1​=−1−2ix1​=[2−i1​]​λ2​=−1+2ix2​=[2+i1​]​

C) λ1=−2+2iλ2=−2−2ix⃗1=[14+2i]x⃗2=[14−2i]\begin{array}{c|c}
\lambda_1=-2+2i \quad & \quad 
\lambda_2=-2-2i \\[0.5em]

\vec x_1 = \begin{bmatrix} 1\\4+2i \end{bmatrix}
& \quad
\vec x_2 = \begin{bmatrix} 1\\4-2i \end{bmatrix}
\end{array}λ1​=−2+2ix1​=[14+2i​]​λ2​=−2−2ix2​=[14−2i​]​

D) λ1=−2+2iλ2=−2−2ix⃗1=[−4+2i1]x⃗2=[−4−2i1]\begin{array}{c|c}
\lambda_1=-2+2i \quad & \quad 
\lambda_2=-2-2i \\[0.5em]

\vec x_1 = \begin{bmatrix} -4+2i\\1 \end{bmatrix}
& \quad
\vec x_2 = \begin{bmatrix} -4-2i\\1 \end{bmatrix}
\end{array}λ1​=−2+2ix1​=[−4+2i1​]​λ2​=−2−2ix2​=[−4−2i1​]​

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Extra Practice

Practice: Complex Eigenvalues and Eigenvectors
Find the eigenvalues and eigenvectors of A=[−6−2012]A=\left[\begin{array}{rr}
-6&-20\\1&2
\end{array}\right]A=[−61​−202​].

Wize University Linear Algebra Textbook > Eigenvalues and Eigenvectors (Spectral Theory)

Complex Eigenvalues

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