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?
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- 5$\begingroup$ The eigenspace is unique, but it has (just like any non-trivial linear space) several bases. Note that for the eigenvalue $2$ you also have some freedom: any $(x,0,x)$ with non-zero $x$ would do. $\endgroup$Arnaud Mortier– Arnaud Mortier2018-05-11 20:17:40 +00:00Commented May 11, 2018 at 20:17
- 3$\begingroup$ Moreover, be careful to distinguish the notations for the set $\{-1,0,1\}$ and for the vector $(-1,0,1)$. $\endgroup$paf– paf2018-05-11 20:20:54 +00:00Commented May 11, 2018 at 20:20
- $\begingroup$ @ArnaudMortier Wow we answered very similarly giving you a +1 :) $\endgroup$N8tron– N8tron2018-05-11 20:23:45 +00:00Commented May 11, 2018 at 20:23
- $\begingroup$ Is the ability of an eigenspace to be unique but capable of changing bases dependent on the non-uniqueness of the eigenvectors which generate the eigenspace? $\endgroup$Zach– Zach2018-05-11 20:23:57 +00:00Commented May 11, 2018 at 20:23
- $\begingroup$ @ZacharyCarter The only space that has a unique basis is the trivial space. That is the space that consists of the zero vector. The basis of this space is the empty set. $\endgroup$N8tron– N8tron2018-05-11 20:26:21 +00:00Commented May 11, 2018 at 20:26
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2 Answers
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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.
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4 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.
- $\begingroup$ If the eigenvectors were unique, would that imply a single choice of basis? $\endgroup$Zach– Zach2018-05-11 20:26:11 +00:00Commented May 11, 2018 at 20:26
- $\begingroup$ @ZacharyCarter Eigenvectors will never be unique (at least as long as you don't go into finite fields) $\endgroup$N8tron– N8tron2018-05-11 20:28:11 +00:00Commented May 11, 2018 at 20:28
- $\begingroup$ I understand that; but hypothetically speaking if eigenvectors $were$ unique (in some fantasy land) would that mean there is a single basis? Asked another way, is the multiplicity of bases a result of the non-uniqueness of eigenvectors? $\endgroup$Zach– Zach2018-05-11 20:32:21 +00:00Commented May 11, 2018 at 20:32
- 2$\begingroup$ @ZacharyCarter the fantasy land does exist in vector spaces over the field with exactly 2 elements. Having a higher geometric multiplicity (dimension of the eigenspace) does result in more bases so you are correct there. Though you must be careful of which multiplicity you are referring to. The algebraic multiplicity (order of the root in the characteristic polynomial) is not always equal to the geometric multiplicity. For example take a matrix $\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ it has eigenvalue 1 with algebraic multiplicity 2 and geometric multiplicity 1. $\endgroup$N8tron– N8tron2018-05-11 20:38:42 +00:00Commented May 11, 2018 at 20:38