How can an eigenspace have more than one dimension?

Question

This is a simple question. An eigenspace is defined as the set of all the eigenvectors associated with an eigenvalue of a matrix. If $\lambda_1$ is one of the eigenvalue of matrix $A$ and $V$ is an eigenvector corresponding to the eigenvalue $\lambda_1$. No the eigenvector $V$ is not unique as all the multiples of this vector, $a*V$, are also eigenvectors corresponding to $\lambda_1$. But all these vectors have the same direction as $a*V$ ; it just means that vector $V$ is stretched or squished. So how can an eigenspace have dimension more than 1 as all the vectors corresponding to $\lambda_1$ just stay in 1 dimension and doesn't break out of it.

Answer 1

Consider the matrix $A := \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \in \mathbb{R}^{2 \times 2}$. Its has only one eigenvalue $\lambda = 2 \in \mathbb{R}$, but the eigenspace to the eigenvalue $2$ is $\mathbb{R}^2$ as $$A \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

and

$$A \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix}= 2 \begin{pmatrix} 0 \\ 1 \end{pmatrix} ~~.$$

Both $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ are eigenvectors, yet both are linearly independent.

---

Given a matrix $A \in \mathbb{R}^{n \times n}$ with eigenvalue $\lambda \in \mathbb{R}$ and an eigenvector $v \in \mathbb{R}^n$ to $\lambda$, it is true that all vectors $\mu v$ (with $\mu \in \mathbb{R}$) are also eigenvectors of $A$ to the eigenvalue $\lambda$. However, they are not all of them, as the example shows - different eigenvectors to an eigenvalue may be linearly independent.

Answer 2

Other concrete examples from elementary geometry :

1) the matrix of projection $P$ in $\mathbb{R^3}$ onto a plane, for example plane $x0y$ is such that all vectors $V$ in this plane verify $PV=1V$. The other eigenspace is generated by all vectors $V=k(0,0,1)$ which are projected onto $0$, thus verifying $PV=0V$. Therefore the eigenspace associated to $1$ is 2-dimensional, and is the whole plane $x0y$.

2) the matrix of symmetry $S$ in $\mathbb{R^3}$ with respect to plane $x0y$ is such that all vectors $V$ in this plane verify as well $SV=1V$. The other eigenspace is generated by all vectors $V=k(0,0,1)$ verifying $SV=-1V$. Therefore (again) the eigenspace associated to $1$ is 2-dimensional, and is the whole plane $x0y$.