How to Find Basis of Eigenspace?

To find the basis for eigenspaces, we follow three steps.

Step 1: we find the eigenvalues of the matrix.

Step 2: we find the corresponding eigenvectors for each eigenvalue and check for linear independence among them.

Step 3: we determine the basis for each eigenspace using the set of linearly independent eigenvectors.

In the examples provided below, you will gain a clear understanding of how to find the basis for eigenspaces using different methods.

Example for Finding Basis for Eigenspaces

Example : Find basis of eigenspaces for matrix A = [Tex]\begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix} [/Tex]

Solution:

Step 1: Find eigvalues of matrix A

det (A – λI) = 0

As (A – λI) = [Tex]\begin{bmatrix} 2 – \lambda & 1 \\ 2 & 3 – \lambda \end{bmatrix} [/Tex]

Thus, det (A – λI) = (2−λ) (3−λ) – 2×1 = λ² − 5λ + 6 – 2 = 0

λ² − 5λ + 4 = 0

⇒ λ² − 4λ – λ + 4 = 0

⇒ λ(λ – 4) – (λ – 4) = 0

⇒ (λ−1) (λ−4) = 0​

We get, λ₁ = 1 and λ₂ = 4 as the eigenvalues.

Step 2 : Find the eigenvector for eigenvalues λ ₁ = 1 and λ ₂ = 4

For λ₁ = 1

(A – λI) = [Tex]\begin{bmatrix} 2 – 1 & 1 \\ 2 & 3 – 1 \end{bmatrix} [/Tex]

= [Tex]\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} [/Tex]

Now, you solve to get eigenvector x₁, (A – λ₁ I)x₁ = 0

[Tex]\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} [/Tex]

x + y = 0 . . . (i)

2x + 2y = 0 . . . (ii)

On solving both the eq (i) and eq (ii), we get x = −y

This equation holds true for any value of y, so we can take any non-zero value for y.

We take, y =1, then x = −1.

Therefore, the eigenvector for λ₁ = 1 is x₁ = [Tex]\begin{bmatrix} -1 \\ 1 \end{bmatrix} [/Tex]

Now, In the similar way, you can find for λ₂ = 4 ​

(A – λ₂I) = [Tex]\begin{bmatrix} 2 – 4 & 1 \\ 1 & 3 – 4 \end{bmatrix} [/Tex] = [Tex]\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix} [/Tex]

(A – λ₂I) x₂ = 0

[Tex]\begin{bmatrix} -2 & 1 \\ 1 & -1 \end{bmatrix} [/Tex] [Tex]\begin{bmatrix} x \\ y \end{bmatrix} [/Tex] = [Tex]\begin{bmatrix} 0 \\ 0 \end{bmatrix} [/Tex]

−2x + y = 0 . . . (iii)

x − y = 0 . . . (iv)

To solve eq (iii) and eq (iv) , follow similar approach used to solve eq1 and eq 2.

After solving eq (iii) and eq (iv) , we get

[Tex]\begin{bmatrix} 1 \\ 1 \end{bmatrix} [/Tex] as , you get y = 1 and x = 1

Therefore, the eigenvector λ₂ = 4, x₂ = [Tex]\begin{bmatrix} 1 \\ 1 \end{bmatrix} [/Tex]

Step 3: Now, you need to find basis for each eigenspace.

basis of λ₁ = [Tex]\begin{bmatrix} 1 \\ -1 \end{bmatrix} [/Tex]

basis of λ₂ = [Tex]\begin{bmatrix} 1 \\ 1 \end{bmatrix} [/Tex]

Therefore, the basis of the eigenspaces for matrix A [Tex]\begin{bmatrix} 1 \\ -1 \end{bmatrix} [/Tex] , [Tex]\begin{bmatrix} 1 \\ 1 \end{bmatrix} [/Tex]

Example 2: Find basis of eigenspaces for matrix A = [Tex]\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} [/Tex]

Solution:

Step 1 : Find eigenvalues of matrix A

det(A − λI) = 0

As, (A – λI) = [Tex]\begin{bmatrix} 2 – \lambda & 1 \\ 1 & 2 – \lambda \end{bmatrix} [/Tex] = 0

Thus, det (A – λI) = (2 − λ) ² − 1 = 0

⇒ λ² − 4λ + 3 = 0

⇒ (λ − 1) (λ − 3) = 0

We get, λ1 = 1 and λ2 = 3 as the eigenvalues.

Step 2: Find the eigenvector for eigenvalues λ 1 = 1 and λ 2 = 3

For λ₁ = 1

(A − λI) = [Tex]\begin{bmatrix} 2 – 1 & 1 \\ 1 & 2 – 1 \end{bmatrix} [/Tex] = [Tex]\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} [/Tex]

Now, you solve to get eigenvector x₁, (A – λ₁ I)x₁ = 0

[Tex]\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} [/Tex] [Tex]\begin{bmatrix} x \\ y \end{bmatrix} [/Tex] = [Tex]\begin{bmatrix} 0 \\ 0 \end{bmatrix} [/Tex]

x + y = 0 . . . (i)

x + y = 0 . . . (ii)

On solving both the eq (i) and eq (ii), we get x = -y

This equation holds true for any value of y, so we can take any non-zero value for y.

We take, y =1, then x = −1.

Therefore, the eigenvector for λ₁ = 1 is x₁ = [Tex]\begin{bmatrix} -1 \\ 1 \end{bmatrix} [/Tex]

Now, In the similar way, you can find for λ₂ = 3

(A – λ₂I) = \begin{pmatrix} 2 – 3 & 1 \\ 1 & 2 – 3 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

(A – λ₂I) x₂ = 0

[Tex]\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} [/Tex] [Tex]\begin{bmatrix} x \\ y \end{bmatrix} [/Tex] = [Tex]\begin{bmatrix} 0 \\ 0 \end{bmatrix} [/Tex]

−x + y = 0 . . . (iii)

x – y = 0 . . . (iv)

To solve eq (iii) and eq (iv) , follow similar approach used to solve eq1 and eq 2.

After solving eq (iii) and eq (iv) , we get

[Tex]\begin{bmatrix} 1 \\ 1 \end{bmatrix} [/Tex]as , you get y = 1 and x = 1

Therefore, the eigenvector λ₂ = 3, x₂ = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

Step 3: Now, you need to find basis for each eigenspace.

basis of λ₁ = [Tex] \begin{bmatrix} -1 \\ 1 \end{bmatrix} [/Tex]

basis of λ₂ = [Tex] \begin{bmatrix} 1 \\ 1 \end{bmatrix} [/Tex]

Therefore, the basis of the eigenspaces for matrix A [Tex] \begin{bmatrix} -1 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \end{bmatrix}[/Tex].

Eigenspaces are a fundamental concept in linear algebra. When you apply a linear transformation to a vector, some vectors get stretched or compressed but don’t change direction. Basis of an eigenspace consists of a set of eigenvectors associated with a specific eigenvalue. These eigenvectors form the building blocks or foundation of the eigenspace.

But how do we find the basis for these crucial spaces? This article breaks down the process into simple steps, guiding you through the concept of eigenspaces and providing practical methods to identify their bases. By the end, you’ll have a clear understanding of how to find the basis of eigenspaces.