Uniqueness of Eigenspaces

Question

If you consider the $3$ by $3$ matrix $$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0& 0\\ 1 & 0 & 1\end{bmatrix}$$you can find the eigenvalues to be $2,0,0$, so it is degenerate. The eigenvalue $2$ corresponds the the eigenvector $\{1,0,1\}$, while the eigenvector $0$ gives a certain degree of freedom; for example the remaining eigenvectors could be $\{-1,0,1 \}$ and $\{0,1,0 \}$ or they could be $\{-1,-1,1 \}$ and $\{-1,2,1 \}$. Does this mean the eigenspace of $A$ is not unique, and if so why? If not, or if I am "not even wrong", how do you manage having the choice of different eigenvectors?

Answer 1

Eigenspaces are vector spaces. Note that the eigenvector $\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ is not the only eigenvector for eigenvalue 2. Any nonzero scalar multiple will also be an eigenvector. For example $\begin{bmatrix}-2 \\ 0 \\ -2\end{bmatrix}$

For the eigenspace of 0, when you give two eigenvectors typically what you are saying is you have a basis for the eigenspace. There will be infinitely many choices of basis.

Answer 2

If $A$ is any matrix, then its eigenspaces are vector spaces. This means that they have at least one basis (a set of vectors that are linearly independent and together generate the eigenspace). However in general there is not a unique choice of basis (for if $B$ is a basis, then surely $2B$, the set containing the same vectors as in $B$ but multiplied by $2$, is also a basis), and this is a perfectly fine fact that holds for all vector spaces. Without this fact, it would not be possible to diagonalize matrices, since diagonalization is essentially changing the given matrix by moving to a different basis from a given one.

There are several ways to obtain different bases from a given one (such as multiplying each vector by an appropriate nondegenerate matrix); the most notable of them is the Gram-Schmidt process, which however makes sense only in the presence of a bilinear form over the space.