geometric multiplicity= algebraic multiplicity for a symmetric matrix
Could any one tell me how to prove:
$\lambda$ be an eigen value of a symmetric matrix $A$ then how to show that the geometric multiplicity and algebraic multiplicity are equal?
Every symmetric matrix is diagonalizable (this can be proved by small perturbation argument), that is: it has a full set of orthogonal eigenvectors and is conjugate to a diagonal matrix. So, you only need to prove the statement for diagonal matrix. Symmetric matrices have no Jordan block in their spectral decomposition, that cause discrepancy in the geometric and algebraic multiplicities of eigenvalues.
Let's review some basic matrix theorems first:
- Geometric Multiplicity $\gamma(\lambda)$ is the dimension of $ker(A- \lambda I)$; Algebraic Multiplicity $\mu(\lambda)$ is the sum of the sizes of all Jordan Blocks corresponding to an eigenvalue $\lambda$; and $\gamma\leq \mu$.
- Suppose $A$ is symmetric, we can diagonalize it as $A=Q\Lambda Q^{T},\ Q^{T}=Q^{-1}$.
$\Lambda$ is a diagonal matrix with eigenvalues of matrix $A$ on the main diagonal line; $Q$ is an Orthonormal matrix with column space as span of Eigenspaces. We can always construct an Eigenspace for each $\lambda$ with size of Algebraic Multiplicity $\mu(\lambda)$.
For a specific eigenvalue $\lambda$, if Geometric Multiplicity $\gamma(\lambda)$ is equal to Algebraic Multiplicity $\mu(\lambda)$, this means the size of the largest Jordan Block should be 1 and there are $\mu=\gamma$ blocks for $\lambda$. In another words, for any eigenvaule $\lambda$ and its eigevector $v$: $$(A-\lambda I)^{\mu}v=0\iff (A-\lambda I)v=0$$
Prove:
$$(A-\lambda I)^{\mu}v=(Q\Lambda Q^{T}-\lambda I)^{\mu}v=[Q(\Lambda-\lambda I)Q^{T}]^{\mu}v=[Q(\Lambda-\lambda I)^{\mu}][Q^{T}v]=0$$Since column vectors of $Q$ are constructed as orthonormal basis, we get the equation above. Take an example of $\lambda=\lambda_{i}$ for clarification, assume Algebraic Multiplicity $\mu(\lambda_{i}) = 2$, $v = v_{i_{1}}$ as its eigenvector and construct another base vector $v_{i_2}$ subject to $v_{i_{2}} \bot v_{i_{1}}$, which span eigenspace of $\lambda_{i}$. Rewrite equation above as 2 separate parts considering orthonormality of eigenvectors:
$$ \left(
\begin{array}{cccccc}
|&\cdots&|&|&\cdots&| \\
(\lambda_{1}-\lambda)^{2}v_{1}&\cdots&(\lambda_{i}-\lambda)^{2}v_{{i}_{1}}&(\lambda_{i}-\lambda)^{2}v_{i_{2}}&\cdots&(\lambda_{n}-\lambda)^{2}v_{n} \\
|&\cdots&|&|&\cdots&|
\end{array}
\right)
\left(
\begin{array}{ccc}
0\\
\vdots\\
1\\
0\\
\vdots\\
0
\end{array}
\right) = (\lambda_{i} -\ \lambda)^{2}v=0 $$
$$hence\ for\ \mu\in \mathbb{Z}^{+},\ (A-\lambda I)^{\mu}v=Q(\Lambda-\lambda I)^{\mu}Q^{T}v=0$$
$$(A-\lambda I)v=0$$
vice versa.