How to Find Basis for Eigenspaces

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 find their basis. By the end, you'll have a clear understanding of how to find the basis of eigenspaces.

What are Eigenspaces?

Eigenspaces are fundamental concepts in linear algebra, particularly in the study of eigenvalues and eigenvectors. When you have a square matrix A, an eigenvector v is a non-zero vector such that when you multiply it by the matrix A, you get a new vector that is a scalar multiple of v. That scalar multiple is called the eigenvalue corresponding to that eigenvector.

Formally, if A is an n × n matrix, v is an eigenvector of A, and λ is the corresponding eigenvalue, then:

A⋅v = λ⋅v

The eigenspace corresponding to an eigenvalue λ is the set of all eigenvectors associated with that eigenvalue, along with the zero vector.

Mathematically, it is defined as:

E_λ = {v ∈ Rⁿ: A⋅v = λ⋅v}

Where Rⁿ is the vector space over the real numbers of dimension n.

Note: Eigenspaces are vector subspaces of the original vector space.

What is Basis of Eigenspace?

Basis of an eigenspace is a set of linearly independent eigenvectors corresponding to a specific eigenvalue.

For a square matrix A and an eigenvalue λ. The eigenspace corresponding to λ is denoted by E_λ​ and is defined as the set of all eigenvectors associated with λ, along with the zero vector.

If v₁, v₂, . . . ,v_k​ are linearly independent eigenvectors corresponding to the eigenvalue λ\lambdaλ, then these vectors form a basis for the eigenspace E_λ​. This means that any vector in E_λ can be expressed as a linear combination of these eigenvectors.

Mathematically, if v₁, v₂, . . . ,v_k​ are eigenvectors corresponding to the eigenvalue λ, then any vector v in E_λ​ can be expressed as:

v = c₁ ⋅ v₁ + c₂ ⋅ v₂ + . . . + c_k ⋅ v_k

Where c₁, c₂, . . . , c_k​ are scalars.

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 on How to Find Basis for Eigenspaces

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

Solution:

Step 1: Find eigvalues of matrix A

det (A - λI) = 0

As (A - λI) = \begin{bmatrix} 2 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix}

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) = \begin{bmatrix} 2 - 1 & 1 \\ 2 & 3 - 1 \end{bmatrix}

= \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}

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

\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

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 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

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

(A - λ₂I) = \begin{bmatrix} 2 - 4 & 1 \\ 1 & 3 - 4 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}

(A - λ₂I) x₂ = 0

\begin{bmatrix} -2 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

−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

\begin{bmatrix} 1 \\ 1 \end{bmatrix} as , you get y = 1 and x = 1

Therefore, the eigenvector λ₂ = 4, x₂ = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

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

basis of λ₁ = \begin{bmatrix} 1 \\ -1 \end{bmatrix}

basis of λ₂ = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

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

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

Solution:

Step 1 : Find eigenvalues of matrix A

det(A − λI) = 0

As, (A - λI) = \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} = 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) = \begin{bmatrix} 2 - 1 & 1 \\ 1 & 2 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}

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

\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

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 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

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

\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

−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

\begin{bmatrix} 1 \\ 1 \end{bmatrix} 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 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

basis of λ₂ = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

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

More Solved Examples on How to find Basis for Eigenspaces

Example 1: Find basis of eigenspaces for matrix A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}

Solution:

Step 1: Find eigenvalues of matrix A.

det(A − λI) = 0

As, (A - λI) = \begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix} = 0

Thus, det (A - λI) = (4 − λ) (3 − λ) − 2×1 = 0

λ² − 7λ + 10 = 0

⇒ λ(λ - 2) - (λ - 5) = 0

⇒ (λ − 2) (λ − 5) = 0

We get, λ₁ = 2 and λ₂ = 5 as the eigenvalues.

Step 2 : Find the eigenvector for eigenvalues λ₁ = 2 and λ₂ = 5

For λ₁ = 2

(A − λI) = \begin{bmatrix} 4 - 2 & 1 \\ 2 & 3 - 2 \end{bmatrix}

= \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}

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

\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

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

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

On solving both the eq (i) and eq (ii), we get x = −y/2

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 = −1/2.

Therefore, the eigenvector for λ₁ = 1 is x₁ = \begin{bmatrix} -1/2 \\ 1 \end{bmatrix}

Now, In the similar way, you can find for λ₂ = 5

(A − λ₂I) = \begin{bmatrix} 4 - 5 & 1 \\ 2 & 3 - 5 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}

Now, (A - λ₂I) x₂ = 0

⇒ \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

-x + y = 0 . . . (iii)

2x + 2y = 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

\begin{bmatrix} 1 \\ 1 \end{bmatrix} as , you get y = 1 and x = 1

Therefore, the eigenvector λ₂ = 5, x₂ = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

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

basis of λ₁ = \begin{bmatrix} -1/2 \\ 1 \end{bmatrix}

basis of λ₂ = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

Therefore, the basis of the eigenspaces for matrix A \begin{bmatrix} -1/2 \\ 1 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \end{bmatrix}

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

Solution:

Step 1 : Find eigenvalues of matrix A.

det(A−λI) = 0

As, (A - λI) = \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} = 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 λ₁ = 1 and λ₂ = 3

For λ₁ =1

(A - λI) = \begin{bmatrix} 2 - 3 & 1 \\ 1 & 2 - 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}

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

⇒ \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

-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 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

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

(A − λ₂I) = \begin{bmatrix} 2 - 3 & 1 \\ 1 & 2 - 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}

Now, (A - λ₂I) x₂ = 0

⇒ \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

-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

\begin{bmatrix} 1 \\ 1 \end{bmatrix} as , you get x = 1 and y = -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 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

Basis of λ₂ = \begin{bmatrix} 1 \\ -1 \end{bmatrix}

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

Example 3: Find basis of eigenspaces for matrix A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}

Solution:

Step 1 : Find eigenvalues of matrix A.

det(A−λI) = 0

(A - λI) = \begin{bmatrix} 0 - \lambda & -1 \\ 1 & 0 - \lambda \end{bmatrix}

⇒ (A - λI) = \begin{bmatrix} -\lambda & -1 \\ 1 & -\lambda \end{bmatrix}

Thus, det (A - λI) = λ ² + 1 = 0

⇒ λ² + 1 = 0

⇒ (λ + i) (λ − i) = 0

So, we get Eigenvalues are λ = ± i.

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

For λ₁ = i

(A - λI) = \begin{bmatrix} 0 - i & -1 - 0 \\ 1 - 0 & 0 - (i) \end{bmatrix} = \begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix}

Now, you solve to get eigenvector x₁, (A - λ₁I)x₁ = 0

⇒ \begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

−ix - y = 0 . . . (i)

x - iy = 0 . . . (ii)

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

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 = i.

Therefore, the eigenvector for λ₁ = i is x₁ = \begin{bmatrix} i \\ 1 \end{bmatrix}

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

(A - λ₂I) = \begin{bmatrix} 0 - (-i) & -1 - 0 \\ 1 - 0 & 0 - (-i) \end{bmatrix} = \begin{bmatrix} i & -1 \\ 1 & i \end{bmatrix}

⇒ (A - λ₂I) x₂ = 0

⇒ \begin{bmatrix} i & -1 \\ 1 & i \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

ix - y = 0 . . . (iii)

y + ix = 0 . . . (iv)

On solving both the eq (iii) and eq (iv), we get x = y/i

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/i.

Therefore, the eigenvector for λ₂ = -i is x₂ = \begin{bmatrix} 1/i \\ 1 \end{bmatrix}

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

basis of λ₁ = \begin{bmatrix} i \\ 1 \end{bmatrix}

basis of λ₂ = \begin{bmatrix} 1/i \\ 1 \end{bmatrix}

Therefore, the basis of the eigenspaces for matrix A \begin{bmatrix} i \\ 1 \end{bmatrix} , \begin{bmatrix} 1/i \\ 1 \end{bmatrix}

Practice Questions on How to find Basis for Eigenspaces

Problem 1: Find the basis for the eigenspaces of the following matrices:

• A = \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix}
• B = \begin{bmatrix} 6 & 2 \\ 2 & 6 \end{bmatrix}
• C = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}
• D = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}
• E = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}

Conclusion

In conclusion, understanding eigenspaces and their basis are important in linear algebra. By finding the eigenvectors and corresponding eigenvalues of a square matrix, we can identify the directions in which the matrix scales vectors. The basis of an eigenspace consists of linearly independent eigenvectors associated with a specific eigenvalue. These eigenvectors form a foundation for expressing any vector within the eigenspace as a combination of them.

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