Proof:

Let $A$ be an $n \times n$ symmetric matrix.

(a) $A$ has $n$ real eigenvalues, counting multiplicities:
Since $A$ is symmetric, the characteristic polynomial of $A$ has real coefficients. By the Fundamental Theorem of Algebra, the eigenvalues of $A$ must all be real because symmetric matrices are self-adjoint, and the eigenvalues of self-adjoint matrices are real.

(b) The dimension of the eigenspace for each eigenvalue $\lambda$ equals the multiplicity of $\lambda$ as a root of the characteristic equation:
Let $\lambda$ be an eigenvalue of $A$. Since $A$ is symmetric, it is diagonalizable. Therefore, the geometric multiplicity (dimension of the eigenspace) of $\lambda$ equals its algebraic multiplicity (multiplicity as a root of the characteristic equation).

(c) The eigenspaces are mutually orthogonal:
Let $\lambda_1$ and $\lambda_2$ be distinct eigenvalues of $A$, with corresponding eigenspaces $E_1$ and $E_2$. Let $\vec{u} \in E_1$ and $\vec{v} \in E_2$. Since $A$ is symmetric: $$(A \vec{u}) \cdot \vec{v} = (A \vec{v}) \cdot \vec{u}.$$ Substituting $A \vec{u} = \lambda_1 \vec{u}$ and $A \vec{v} = \lambda_2 \vec{v}$, this becomes: $$\lambda_1 (\vec{u} \cdot \vec{v}) = \lambda_2 (\vec{u} \cdot \vec{v}).$$ Since $\lambda_1 \neq \lambda_2$, it follows that $\vec{u} \cdot \vec{v} = 0$, proving that $E_1$ and $E_2$ are orthogonal.

(d) $A$ is orthogonally diagonalizable:
Since $A$ is symmetric, the eigenvalues are real, the eigenspaces are mutually orthogonal, and the geometric multiplicity equals the algebraic multiplicity for all eigenvalues. Therefore, there exists an orthogonal matrix $P$ (columns are orthonormal eigenvectors of $A$) such that: $$P^TAP = \Lambda,$$ where $\Lambda$ is a diagonal matrix of eigenvalues. Thus, $A$ is orthogonally diagonalizable.

Proof:

Let $A$ be a symmetric $n \times n$ matrix. Assume $\vec{u}$ and $\vec{v}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1$ and $\lambda_2$, respectively. We aim to prove that $\vec{u}$ and $\vec{v}$ are orthogonal.

Since $\vec{u}$ is an eigenvector of $A$ with eigenvalue $\lambda_1$, we have: $$A \vec{u} = \lambda_1 \vec{u}.$$ Similarly, since $\vec{v}$ is an eigenvector of $A$ with eigenvalue $\lambda_2$, we have: $$A \vec{v} = \lambda_2 \vec{v}.$$

Taking the dot product of $A \vec{u}$ with $\vec{v}$, we get: $$(A \vec{u}) \cdot \vec{v} = (\lambda_1 \vec{u}) \cdot \vec{v}.$$ Expanding, this becomes: $$\lambda_1 (\vec{u} \cdot \vec{v}).$$

Now take the dot product of $\vec{u}$ with $A \vec{v}$: $$\vec{u} \cdot (A \vec{v}) = \vec{u} \cdot (\lambda_2 \vec{v}).$$ Expanding, this becomes: $$\lambda_2 (\vec{u} \cdot \vec{v}).$$

Since $A$ is symmetric, we know that: $$(A \vec{u}) \cdot \vec{v} = \vec{u} \cdot (A \vec{v}).$$ Substituting the earlier results, we have: $$\lambda_1 (\vec{u} \cdot \vec{v}) = \lambda_2 (\vec{u} \cdot \vec{v}).$$

Rearranging: $$(\lambda_1 - \lambda_2) (\vec{u} \cdot \vec{v}) = 0.$$

Since $\lambda_1 \neq \lambda_2$ by assumption, it follows that: $$\vec{u} \cdot \vec{v} = 0.$$

Thus, $\vec{u}$ and $\vec{v}$ are orthogonal.

Proof:

Let $A$ be an $n \times n$ matrix.

(If direction):
Assume $A$ is symmetric. By the spectral theorem, all eigenvalues of $A$ are real, and the eigenspaces corresponding to distinct eigenvalues are mutually orthogonal. Furthermore, the geometric multiplicity of each eigenvalue equals its algebraic multiplicity, allowing $A$ to be diagonalized. Since the eigenspaces are mutually orthogonal, it is possible to construct an orthogonal matrix $P$ such that: $$P^T A P = \Lambda,$$ where $\Lambda$ is a diagonal matrix containing the eigenvalues of $A$. Thus, $A$ is orthogonally diagonalizable.

(Only if direction):
Assume $A$ is orthogonally diagonalizable. Then there exists an orthogonal matrix $P$ such that: $$P^T A P = \Lambda,$$ where $\Lambda$ is a diagonal matrix. Rewriting this, we have: $$A = P \Lambda P^T.$$ Since $P$ is orthogonal, $P^T = P^{-1}$, so: $$A = P \Lambda P^{-1}.$$ The product $P \Lambda P^T$ is symmetric because $P^T$ is the transpose of $P$ and $\Lambda$ is diagonal (hence symmetric). Thus, $A$ is symmetric.