I know the derivation is probably online in a lot of places; but perhaps I'm wording my question wrong, I cannot find an example online that satisfies what I am looking for.
My question is given a $n \ x \ n$ symmetric matrix, what would be the generalization to isolate $\lambda_i$ from $A\mathbf{x} = \lambda_i\mathbf{x}$ to $\lambda_i$ = ...
Most examples I've seen either used Spectral theorem or the values of matrix and eigenvectors are already given, is there an intuitive explanation to this question?
Take the equation and multiply by $\mathbf{x}^T$ (normalized), then $\mathbf{x}^TA\mathbf{x} = \lambda_i \mathbf{x}^T\mathbf{x} = \lambda_i$
Note Scalar-vector multiplication is commutative
\begin{equation} \left[ \begin{matrix} x_1 & x_2 \end{matrix} \right] \lambda_i \left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right] = \left[ \begin{matrix} \lambda_i x_1 & \lambda_i x_2 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right] = \lambda_i \left[ \begin{matrix} x_1 & x_2 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right] \end{equation}
General idea
Suppose $A$ has an orthonormal basis of eigenvectors $\mathbf{x}_1,\mathbf{x}_2, \dots, \mathbf{x}_n$ corresponding to distinct eigenvalues $\lambda_1>\lambda_2>\cdots>\lambda_n >0$.
Every unit vector $\mathbf{x}$ has an expansion $\mathbf{x} = \sum \alpha_i \mathbf{x}_i$, where $\sum \alpha_i^2 = 1$.
Then $A\mathbf{x} = \sum \alpha_i A\mathbf{x}_i = \sum \alpha_i \lambda_i\mathbf{x}_i$ and $\mathbf{x}^TA\mathbf{x}= \sum \alpha_i^2 \lambda_i$
The maximum value of this expression occurs when $\alpha_1=1$ and all other $\alpha_i$ are 0. If we restrict to the complement of $\mathbf{x}_1$ (i.e. requiring $\alpha_1=0$), then the max occurs for $\alpha_2 = 1$, etc.