For a symmetric matrix, the geometric and algebraic multiplicities are equal Could anyone tell me how to prove the following proposition?
Let $\lambda$ be an eigenvalue of a symmetric matrix. Then, its geometric multiplicity equals its algebraic multiplicity.
 A: 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. 
A: Let's review some basic matrix theorems first:

*

*Geometric Multiplicity $\gamma(\lambda)$ is the dimension of $ker(A-
\lambda I)$, and the number of Jordan Blocks for an eigenvalue $\lambda$; Algebraic Multiplicity $\mu(\lambda)$ is the sum of the sizes of all Jordan Blocks for 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.
A: I saw this question in Anton's Elementary Linear Algebra (Exercise 7.2.31).
At there, this question was used to prove the spectral theorem.
So I was supposed to unable to use the theorem.
Anton had given an outline of the proof.
I just give the detail.

*

*Let $B_0=\{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k\}$ be a basis for the eigenspace $E_{\lambda}$.
Extend this basis to an orthonormal basis $B=\{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k, \mathbf{u}_{k+1}, ..., \mathbf{u}_n\}$ for $\mathbb{R}^n$.


*Let $P=[\mathbf{u}_1|\mathbf{u}_2|\cdots |\mathbf{u}_k |\mathbf{u}_{k+1} |\cdots |\mathbf{u}_n]$. Then $AP=P\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}$ and $P^{-1}=P^T$ and $P^T AP=\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}$.


*Since $A$ is symmetric, we have
$$
\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}
=P^T AP=P^T A^T P=(P^T AP)^T
=\begin{bmatrix}\lambda I_k & X \\ O & Y \end{bmatrix}^T
=\begin{bmatrix}\lambda I_k & O \\ X^T & Y^T \end{bmatrix}.
$$
This follows that $X=O$.
Hence, $AP=P\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$ and $P^T AP=\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$.


*Note that $A$ is similar to $\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$. So they have the same characteristic polynomial.
The characteristic polynomial of $\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}$ is $\det{(\lambda I_k-xI_k)}\det{(Y-xI_{n-k})}$.
We will show that $\lambda$ cannot be a root of $\det{(Y-xI_{n-k})}$.
Then the algebraic multiplicity of $\lambda$ is $k$, equals to geometric multiplicity of $\lambda$.


*Suppose that $\lambda$ is a root of $\det{(Y-xI_{n-k})}$.
Let $\mathbf{y}$ be an eigenvector of $Y$ with eigenvalue $\lambda$.
Then
$$
A\left(P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\right)
=\left(P\begin{bmatrix}\lambda I_k & O \\ O & Y \end{bmatrix}P^T\right)\left(P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\right)
=\lambda\left(P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\right).
$$
That is, $P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}$ is an eigenvector of $A$ with eigenvalue $\lambda$.
Which implies that $P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}\in E_{\lambda}=\text{span}(\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k)$.


*On the other hand, note that
$$
P\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}
=[\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k, \mathbf{u}_{k+1}, ..., \mathbf{u}_n]\begin{bmatrix}0\\0\\\vdots\\0\\\mathbf{y}\end{bmatrix}
\in \text{span}(\mathbf{u}_{k+1}, \mathbf{u}_{k+2}, ..., \mathbf{u}_{n}).
$$
This is impossible because $\text{span}(\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k) \cap \text{span}(\mathbf{u}_{k+1}, \mathbf{u}_{k+2}, ..., \mathbf{u}_{n})=\{0\}$.
A: If $A$ is real symmetric then it is diagonalizable.  That is, there is some orthogonal $P$ such that $AP=PD$, where $D$ is diagonal.  The columns of $P$ are each eigenvectors, and form a basis.  Hence all geometric multiplicities equal algebraic multiplicities.
