I didn't find a pre-existing question relating to what I'm going to ask, so I apologize if this is a duplicate question for one I hadn't found:
Why is the property that eigenvalues corresponding to distinct eigenvectors of a real symmetric matrix are orthogonal useful if the eigenvectors were bundled into a matrix?
I know why it's true that they are orthongonal, and I know that there are exactly $n$ not necessarily distinct eigenvalues of a real $n\times n$ symmetric matrix. I'm not sure why the corresponding eigenvectors "bundled into a matrix" is useful. You'd have a matrix whose columns are orthogonal to each other, but beyond that what does that accomplish?
The answer may be simple and I'm just missing something. Any help is greatly appreciated.
Best Answer
Let $S\in M_n(\mathbb R)$ be the symmetric matrix under consideration.
Assume in a preliminary step, that the eigenvalues of $S$ are pairwise distinct. Picking a unit eigenvector for each eigenvalue gives us an orthogonal, even an orthonormal system in $\mathbb R^n$ of size $n$, hence an orthonormal basis.
Bundling of the chosen eigenvectors as column vectors yields an $n\times n$ matrix, let's call it $O$, and using the transpose and the identity matrix, the orthonormality can be expressed as $\,O^T\!O=\mathbb 1_n\,\!$. Which (since the dimension is finite) is equivalent to $\,OO^T=\mathbb 1_n\,$ or $\,O^{\,T}=O^{\,-1}$.
Thus $O$ is an orthogonal matrix.
Recall that orthogonal matrices (preserving orthogonality and norms) are precisely those which transform any orthonormal basis into an(other) orthonormal basis.
By definition of $O$ we have $$SO\,=\,OD\;\iff\; S\,=\,OD\,O^T$$ with $D$ denoting a diagonal matrix containing the eigenvalues in the appropriate order. So $S$ is diagonalisable, and one may say "diagonalisable with respect to an orthonormal basis".
This accomplishes a notable and most useful characteristic of symmetric matrices.
And it's valid in full generality, i.e., after raising the initial assumption of distinct eigenvalues, because in every eigenspace, independently of each other, one can choose an orthonormal basis of that subspace, and proceed in the same way.