Given a symmetric real matrix $A$ there is an orthogonal transformation that brings $A$ to a block diagonal form, where $a \in \mathbb{R}$:
$$ RAR^{T} = \left[ {\begin{array}{cc} a & 0 \\ 0 & M \end{array} } \right] $$ How could one see that such $R$ exists? Given this $R$ one could progressively diagonalise $A$, obtaining finally that a symmetric matrix can be diagonalised by an orthogonal matrix.
I am using prior knowledge that this is an approach to eigenvalue decomposition to formulate my answer.
Let $$R = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix},$$
where $q_1$ is an eigenvector of the matrix $A$.
Then $$RAR^T = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix}^T A \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix} = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix}^T \begin{bmatrix} \lambda_1q_1 \ \ q_1\tilde{R} \end{bmatrix} = \begin{bmatrix} \lambda_1 ||q_1||^2 \ \ q_1^T\tilde{R} \\ \tilde{R}^Tq_1 \ \ \tilde{R}^T \tilde{R} \end{bmatrix}$$
Now we need $q_1$ to be in the null space of $\tilde{R}^T$. So let $\tilde{R} = A-\lambda_1I$ and you have your answer.