Intuition behind A = $Q \Lambda Q^T$ Normally, we can diagonalize a matrix A by $A = S \Lambda S^{-1}$.
When the matrix is symmetric, we have that $A = Q \Lambda Q^{-1} = Q \Lambda Q^{T}$.
The part I'm failing to understand is: why can we just make the eigenvector matrix $S$ a set of orthonormal vectors?
Looking at the relationship $Ax = \lambda x$, if I changed the magnitude of $x$, wouldn't I then require a different $\lambda$ to satisfy the equation? Wouldn't that also have to change $\Lambda$ in the diagonalization of $A$ where $A$ is symmetric?
 A: 
The part I'm failing to understand is: why can we just make the eigenvector matrix $S$ a set of orthonormal vectors?

If $A$ is symmetric and $x,y$ are eigenvectors with eigenvalues $\lambda_x,\lambda_y$ then
$$\lambda_y \langle x,y\rangle= \langle x,Ay \rangle=\langle Ax,y \rangle=\lambda_x \langle x,y\rangle.$$ If $\lambda_x\ne \lambda_y$ it must be $\langle x,y\rangle=0.$

Looking at the relationship $Ax = \lambda x$, if I changed the magnitude of $x$, wouldn't I then require a different $\lambda$ to satisfy the equation? Wouldn't that also have to change $\Lambda$ in the diagonalization of $A$ where $A$ is symmetric?

If $y=kx$ and $Ax=\lambda x$ then $Ay=\lambda y.$ Note that $Ay=A(kx)=kAx=k\lambda x=\lambda k x=\lambda y.$
A: $x$ appears on both sides of the equation $Ax=\lambda x$, hence changing the "magnitude" of $x$ just results in another eigenvector for eigenvalue $\lambda $.  That is,  if $x$ is an eigenvector for eigenvalue $\lambda $, then so is $\alpha x$ for any scalar $\alpha \neq0$.
All symmetric matrices are diagonalizable. Note that for a symmetric matrix,  since eigenvectors for distinct eigenvalues are orthogonal,  it is possible to get an orthonormal basis of eigenvectors. 
