Eigenvalues and eigenvectors are special numbers and vectors associated with a matrix. They describe directions that do not change direction when a matrix transforms them—they only get stretched, compressed, or flipped.
Solved Examples
Example 1: Find the eigenvectors of the matrix A =
Solution:
The eigen values of the matrix is found using,
|A - λI| = 0\begin{bmatrix}1-λ & 1 & 0\\0 & 1-λ & 1\\0 & 0 & 1-λ\end{bmatrix} = 0
(1 - λ)3 = 0Thus, the eigen values are, λ = 1, 1, 1
As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 1, using (A - λI)v = O
\begin{bmatrix}1-1 & 1 & 0\\0 & 1-1 & 1\\0 & 0 & 1-1\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}0\\ 0\\0\end{bmatrix}
\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}.\begin{bmatrix}a\\ b\\c\end{bmatrix} = \begin{bmatrix}0\\ 0\\0\end{bmatrix}
solving the above equation we get,
- a = K
- y = 0
- z = 0
Then the eigenvector is,
\begin{bmatrix}a\\ b\\c\end{bmatrix}= \begin{bmatrix}k\\ 0\\0\end{bmatrix} = k\begin{bmatrix}1\\ 0\\0\end{bmatrix}
Example 2: Find the eigenvectors of the matrix A =
Solution:
The eigen values of the matrix is found using,
|A - λI| = 0\begin{bmatrix}5-λ & 0\\0 & 5-λ \end{bmatrix} = 0
(5 - λ)2 = 0Thus, the eigen values are,
λ = 5, 5As the all the eigenvalues are equal we have three identical eigenvectors. We will find the eigenvectors for λ = 5, using
(A - λI)v = O\begin{bmatrix}5-5 & 0 \\ 0 & 5-5\end{bmatrix}.\begin{bmatrix}a\\ b\end{bmatrix} = \begin{bmatrix}0\\ 0\end{bmatrix} Simplifying the above we get,
a = 1, b = 0
a = 0, b = 1Then the eigenvector is,
\begin{bmatrix}a\\ b\end{bmatrix}= \begin{bmatrix}1\\ 0\end{bmatrix} , \begin{bmatrix}0\\ 1\end{bmatrix}
Example 3: Given matrix
Solution:
1) Find Eigenvalues:
The eigenvalues are found by solving the characteristic equation det(A−λI) = 0.
A−λI= \begin{pmatrix} 4−λ & 1 \\ 2 & 3−λ \end{pmatrix} Determinant: (4−λ)(3−λ)−2⋅1 = 0.
λ2−7λ+10 = 0.
Solving for λ: λ = 5 or λ = 2.
2) Find Eigenvectors:
For λ = 5: A−5I =
\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}. Solving (A−5I)x = 0 gives eigenvector
x= k\begin{pmatrix} 1 \\ 1 \end{pmatrix} where k is a scalar.For λ=2: A−2I
= \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} .Solving (A−2I)x = 0, gives eigenvector
x= k\begin{pmatrix} -1 \\ 2 \end{pmatrix} where k is a scalar.
Example 4: In mechanical engineering, eigenvalues can be used to determine natural frequencies of a system. Consider a 2-DOF mass-spring system with mass matrix
Find the natural frequencies.
Solution:
Form the generalized eigenvalue problem: det(K−λM)=0
Find Eigenvalues:
K−λM
= \begin{pmatrix} 5−2λ & -1 \\ -1 & 5−λ \end{pmatrix} Determinant: (5−2λ)(5−λ) − { (-1)(−1) } = 0
2λ2−15λ +24=0
Solving for λ: λ = 5.186 or λ = 2.313
Natural Frequencies:
ω1 ≈ 2.2775, ω2 ≈ 1.5211
Practice Problems
1. Given matrix
- Find the eigenvalues.
- Find the corresponding eigenvectors.
2. An electrical circuit has an impedance matrix
3. Consider a mass-spring system with mass matrix
4. Given the covariance matrix
- Find the eigenvalues.
- Find the eigenvectors.
- Use the eigenvectors to determine the principal components.
5. A communication channel is represented by the matrix