Eigenspace (with Examples)

What is Eigenspace?

Eigenspace is the span of a set of eigenvectors. These vectors correspond to one eigenvalue. So, an eigenspace always maps to a fixed eigenvalue. It is also a subspace of the original vector space. Finding it is equivalent to calculating eigenvectors.

The basis of an eigenspace is the set of linearly independent eigenvectors for the corresponding eigenvalue. The cardinality of this set (number of elements in it) is the dimension of the eigenspace.

For each eigenvalue, there is an eigenspace. Interesting cases arise as eigenvalues may be distinct or repeated. Let us see all three possibilities, with examples in ℝ ²:

1. Distinct Eigenvalue – Eigenspace is a Line
2. Repeated Eigenvalue
   • Eigenspace is a Line
   • Eigenspace is ℝ ²

Eigenspace for Distinct Eigenvalues

Our two dimensional real matrix is $A=\left(\begin{array}{cc}1& 2\\ 3& 0\end{array}\right)$. It has two real eigenvalues 3 and −2. Eigenspace of each eigenvalue is shown below.

Eigenspace for λ = 3

The eigenvector corresponding to λ = 3 is (1, 1)^T. In the following image you can see:

• Unit Eigenvector v in red color ($v=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}{)}^T$ )
• Transformed vector Av in blue
• Eigenspace as a line (in sky-blue)

Eigenspace for λ = − 2

The eigenvector is $(\frac{-2}{3},1{)}^T$. The image shows unit eigenvector ( − 0.56, 0.83)^T. In this case also eigenspace is a line.

Eigenspace for a Repeated Eigenvalue

Case 1: Repeated Eigenvalue – Eigenspace is a Line

For this example we use the matrix $A=\left(\begin{array}{cc}2& 0\\ 1& 2\end{array}\right)$. It has a repeated eigenvalue = 2. The eigenspace is a line.

Case 2: Repeated Eigenvalue – Eigenspace is ℝ ²

In this example our matrix is $A=\left(\begin{array}{cc}3& 0\\ 0& 3\end{array}\right)$. It has a repeated eigenvalue = 3. In this case every vector in ℝ ² is an eigenvector. Therefore entire ℝ ² is eigenspace for λ = 3.

Eigenspace is a subspace

Let us say S is the set of all eigenvectors for a fixed λ. To show that S is a subspace, we have to prove the following:

1. If vectors v, w belong to S, v + w also belongs to S.
2. If vector v is in S, αv is also in S (for some scalar α).

We borrow the following from the original vector space:

• Operations like addition and scalar multiplication
• Scalars (its field)

Proof of 1

Suppose vectors v, w ∈ S.

$A(v+w)$ = Av + Aw

= λv + λw = $\lambda (v+w)$

(v + w) is an eigenvector with eigenvalue λ.

Therefore (v + w) ∈ S

Proof of 2

For a scalar α ∈ 𝔽 and a vector v ∈ S, we show that αv ∈ S.

Here 𝔽 is the field of the original vector space.

$A(\alpha v)$ = αAv = αλv = $\lambda (\alpha v)$

αv is an eigenvector with eigenvalue λ.

Therefore αv ∈ S

Related

Eigenvalues and Eigenvectors (Examples & Questions) ➤