Algebraic multiplicity is equal to the geometric multiplicity for all eigenvalues of
Proof: Let , an eigenvalue
Then, geometric multiplicity of alg. multiplicity of
HW:
, where is , is , is , and is .
Then,
Proof cont.
Suppose the eigenspace associated to has dimension .
Let be a bass of this eigenspace.
Now, complete to a bass of by considering a basis of . Say the basis is . Note that .
Define
and consider
Since
Coming back to the proof:
Since , alg. mult. of
Proof:
Case 1:
be an eigenbasis of associated to .
Define,
Consider
Which is a diagonal matrix, so this case is proven.
Case 2:
Since is diagonalizable, there exists an invertible such that is a diagonal matrix
Therefore, are eigenvectors associated with eigenvalue .
Thus, is an eigenbasis of associated to
Case 3
Alg. mult. = geo. mult. for all eigenvalues
sum of geo. mult. = sum of all dim of eigenspaces = There exists an eigenbasis
Case 4:
There exists an eigenbasis. This implies that the sum of dimension of eigenspace is equal to the sum of geo. mult. which is less that the sum of alg. mult.
Therefore, geo. mult. is equal to the alg. mult. for each eigenvalue.
Corollary: has -distinct eigenvalues, then is diagonalizable.
Corollary: Let be a polynomial.
If is diagonalizable, then so is .
Proof: There exists non-singular such that is
Substituting this into our polynomial
Therefore,
So, is similar to the diagonal matrix
Hence, it is also diagonalizable
Spectral theorem:
Recall: is called symmetric if is called Hermitian if
Let denote the dot product.
, .
Therefore,
Proposition: Let is such that . Then,
All eigenvalues of are real numbers
Proof: Let be an eigenvector of with an associated eigenvector
Now,
Therefore,
Therefore, , meaning that is real.
Proposition: Let and be eigenvectors of associated with with with . Then,
Proof:
Definition: A matrix is said to have simple spectrum if all its eigenvalues are different.
Lemma: Any symmetric matrix can be approximated by matrices with simple spectrum.