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

Complex Eigenvalues

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Complex Eigenvalues and Eigenvectors

A square matrix with real entries can have complex eigenvalues!
As usual, given a matrix AA, we find eigenvalues λ\lambda and corresponding eigenvectors x⃗\vec x.
Note: Every n×nn \times n matrix has exactly nn complex eigenvalues (counting any multiplicities).
Wize Tip
If AA has real entries but complex eigenvalues, then the eigenvalues come in conjugate pairs:
If λ1=a+bi\lambda_1=a+bi is an eigenvalue, then λ2=a−bi\lambda_2=a-bi is another eigenvalue.
Once you find one eigenvector x⃗1\vec x_1 corresponding to λ1\lambda_1, then the conjugate of x⃗1\vec x_1 is an eigenvector of the conjugate eigenvalue λ2\lambda_2.

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Steps

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

Example: Complex Eigenvalues and Eigenvectors

Suppose we are given a real 3×33\times3 matrix and we are told two of its eigenvalues are λ1=4\lambda_1=4 and λ2=2−3i\lambda_2=2-3i.
If their corresponding eigenvectors are x⃗1=[11]\vec x_1 = \begin{bmatrix} 1\\ 1 \end{bmatrix} and x⃗2=[2+i5]\vec x_2= \begin{bmatrix} 2+i\\ 5 \end{bmatrix}, find the missing eigenvalue and eigenvector pair.
Since λ2\lambda_2 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}
Similarly, since we know the complex eigenvector associated with λ2\lambda_2, we simply take its conjugate to find the eigenvector for λ3\lambda_3:
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} }

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].

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} \detp{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}
We can solve for λ\lambda 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 i
Then the eigenvalues are λ1=1+i\boxed{ \lambda_1 = 1+i } and its conjugate, λ2=1−i\boxed{ \lambda_2=1-i }.

Find the Eigenvectors

λ1=1+i\colorThree{\lambda_1 = 1+i}
Substituting this eigenvalue into the matrix above (A−λIA-\lambda I), our goal is to solve for x⃗1\vec x_1:
[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}

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}
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}
So an eigenvector associated with λ1=1+i\lambda_1 = 1+i is: x⃗1=[2+i1]\boxed{ \vec x_1 = \left[\begin{array}{c} 2+i\\ 1 \end{array}\right] }
λ2=1−i\colorThree{\lambda_2 = 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] }

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

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].