How to calculate eigenvectors

Eigenvectors are concepts used in linear algebra to analyze the properties of square matrices. They are crucial for understanding how a matrix transformation affects the direction of vectors along them. Eigenvalues provide scaling information for these transformations, finding use in physics, engineering, computer graphics, and other fields involving matrices to represent systems and transformations.

Formal definition

An eigenvector of a square matrix is a non-zero vector that becomes equal to a scaled version of the same vector when multiplied by the matrix.

Consider a square matrix AA of size n×nn \times n, and if x is a non-zero vector, the product of AA and xx can be represented as λx\lambda x, where xx is the eigenvector, and λ\lambda is the scalar value known as the eigenvalue associated with matrix AA

As an example, we can see below that we have an eigenvector x=[12]x=\begin{bmatrix} 1 \\ 2 \end{bmatrix} and a square matrix A=[308−1] A=\begin{bmatrix} 3 & 0 \\ 8 & -1 \end{bmatrix} where the result of AxAx gives us the vector xx back but with a scaling of λ=3 \lambda =3 .

Example of an eigenvector with its eigen value
Example of an eigenvector with its eigen value

Calculating eigenvectors

To find the eigenvectors of a square matrix, follow these steps:

  1. Given a square matrix AA of size n×nn \times n , start by finding the eigenvalues of the matrix. To do this, solve the characteristic equationThe characteristic equation is a polynomial equation used to find eigenvalues of a matrix.:

Note: Here, II represents the identity matrix of the same size as AA, and λ \lambda is the unknown eigenvalue.

  1. After obtaining the eigenvalues (λ1,λ2,...,λn)(\lambda_{1},\lambda_{2},...,\lambda_{n}), each eigenvalue corresponds to a specific eigenvector.

  2. For each eigenvalue λi\lambda_{i}, solve the equation (A−λiI)xi=0(A-\lambda_{i} I)x_{i} = 0, where xix_{i}is the eigenvector corresponding to λi\lambda_{i}. This equation can be rewritten as:

  1. Solve the system of linear equations (A−λiI)xi=0 (A-\lambda_{i} I)x_{i} = 0 to find the non-zero vector xix_{i}. Keep in mind that the eigenvector is unique only up to a scalar multiple, so the resulting eigenvector may be normalized.

  2. Repeat steps 3 and 4 for each eigenvalue to find all the corresponding eigenvectors.

Note: It's important to remember that not all square matrices have distinct eigenvalues, and in some cases, the process of finding eigenvalues and eigenvectors might involve more complex calculations or numerical methods, especially for larger matrices. However, for (2×2)(2 \times 2) and (3×3)(3 \times 3) matrices, the above steps are generally sufficient to find eigenvalues and eigenvectors.

Example

We will see a basic example of finding eigenvectors and eigenvalues given a square matrix AA.

  • Question: Find the eigenvectors of the matrix A=[308−1] A=\begin{bmatrix} 3 & 0 \\ 8 & -1 \end{bmatrix} . Note that this matrix is similar to the one we saw above. Let's calculate its eigenvectors using the above steps.

  • Answer: Following the first step, we must set up the characteristic equation and solve for eigenvalues.

Next, we need to solve the equation (A−λiI)xi=0(A-\lambda_{i} I)x_{i} = 0 for each λi\lambda_{i}.

We will start solving with λ1=3\lambda_{1} =3 first, and then we'll solve for λ2=−1\lambda_{2}=-1

Solving for eigenvector x1x_{1}

Suppose x1=[xy]x_{1}= \begin{bmatrix} x \\ y \end{bmatrix}then we can find it using the (A−λ1I)x1=0(A-\lambda_{1} I)x_{1} = 0 formula.

We can introduce a free variableIn linear algebra, a free variable is a variable in a system of linear equations that can take any value and does not depend on other variables in the system. ss and let s=xs=x then y=2sy=2s. Now we can get our eigenvector in matrix form x1=[12]sx_{1} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}s .

Solving for eigenvector x2x_{2}

Suppose x2=[xy]x_{2}= \begin{bmatrix} x \\ y \end{bmatrix}then we can find it using the (A−λ2I)x2=0(A-\lambda_{2} I)x_{2} = 0 formula.

We can introduce a free variable ss and let s=ys=y. Now we can get our second eigenvector in matrix form x2=[01]sx_{2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}s .

We can summarize the results as follows:

  • λ1=3,x1=[12]\lambda_{1} = 3 , x_{1}= \begin{bmatrix} 1 \\ 2 \end{bmatrix}

  • λ2=−1,x2=[01]\lambda_{2} = -1 , x_{2}= \begin{bmatrix} 0 \\ 1 \end{bmatrix}

Quiz

Now that you know how to find eigenvectors and eigenvalues, try this quiz to test your understanding.

Eigenvectors and eigenvalues

Q

Find eigenvectors and corresponding eigen values of the matrix A=[−1320]A=\begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix}

A)

x1=[11],λ1=2x_{1}=\begin{bmatrix} 1 \\ 1 \end{bmatrix},\lambda_{1}=2

x2=[−321],λ2=−3x_{2}=\begin{bmatrix} \frac{-3}{2} \\ 1 \end{bmatrix},\lambda_{2}=-3

B)

x1=[05],λ1=2x_{1}=\begin{bmatrix} 0 \\ 5 \end{bmatrix},\lambda_{1}=2

x2=[521],λ2=10x_{2}=\begin{bmatrix} \frac{5}{2} \\ 1 \end{bmatrix},\lambda_{2}=10

C)

x1=[11],λ1=6x_{1}=\begin{bmatrix} 1 \\ 1 \end{bmatrix},\lambda_{1}=6

x2=[−121],λ2=−3x_{2}=\begin{bmatrix} \frac{-1}{2} \\ 1 \end{bmatrix},\lambda_{2}=-3

Conclusion

Understanding eigenvectors and eigenvalues is crucial in various fields, including linear algebra, physics, and data analysis. These concepts help us analyze the behavior of matrices and systems, enabling us to find important patterns and solutions.

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