When will there exist a basis of eigenvectors?

Question

From what I have found, "there exists a basis of eigenvectors if and only if the algebraic and multiplicity of each eigenvalue is the same." However, I cannot relate the multiplicity of eigenvalues to the basis of eigenvectors. How can I prove that this is true?

Answer 1

Let $\lambda_1,\ldots,\lambda_k$ be your eigenvalues and let $n$ be the dimension of the space. Be $GM$ and $AM$ we abbreviate "geometric" and "algebraic multiplicity", respectively. Moreover, let $A$ be the linear map and $V$ the space it acts on.

If algebraic and geometric multiplicities coincide for each eigenvalue, then for each eigenspace $\ker(A-\lambda_j I)$ choose a basis $B_j$. Then $\sum_{j=1}^k|B_j| = \sum_{j=1}^k GM(\lambda_j)= \sum_{j=1}^k AM(\lambda_j) = n$. So, $\bigcup_{j=1}^k B_j$ is a basis of $V$ (consisting of eigenvectors of $A$).

Conversely, let $B = \bigcup_{j=1}^k B_j$ be a basis of eigenvectors for $V$, where $B_j\subset\ker(A-\lambda_jI)$. We have that $|B_j|\le GM(\lambda_j)$ for each $j$ and hence (as $GM(\lambda_j)\le AM(\lambda_j)$ for each $j$) $$ n = \sum_{j=1}^k|B_j|\le\sum_{j=1}^k GM(\lambda_j)\le\sum_{j=1}^k AM(\lambda_j) = n. $$ This implies $GM(\lambda_j) =AM(\lambda_j)$ for each $j$.

Answer 2

If there exists a basis of eigenvectors, then the operator is diagonlizable in some eigenbasis. Now for any eigenvalue $\lambda$, if the eigenspace $E_{\lambda}$ is $n$-dimensional, then there will be exactly $n$-$\lambda$'s on the diagonal matrix, hence the characteristic polynomial has $\lambda$ as a root with multiplicity $n$. This shows that the geometric and algebraic multiplicity are the same for each eigenvalue.

Conversely if the geometric and algebraic multiplicities are the same for each eigenvalu, then you can find a basis for the eigenspace for each $\lambda$. When you put all the bases together, the new list of vectors form a linearly independent set of vectors (because eigenvectors corresponding to distinct eigenvalues are linearly independent). Then this new list of vectors has same number of elements as the sum of algebraic multiplicity, which is equal to the dimension of the vector space. So this is actually a basis for the vector space that contains only eigenvectors, i.e. it is an eigenbasis.