Usefulness of Why Eigenvectors Corresponding to Distinct Eigenvalues of Symmetric Matrix are Orthogonal

Question

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.

Answer 1

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.

Answer 2

I'm going to give a slightly different example of why this is useful in Quantum Mechanics and how from it we can arrive at Quantum Numbers. A quick background:

1. In Q.M a complete description of the state of a physical system is given by a normalized vector $|\phi>$ in the Hilbert space appropriate to the system
2. Observable quantities (energy, position,momentum etc.) are represented by hermitian operators. Our goal is to have a hermitian operator which we can diagonalise to form an eigenbasis of the Hilbert space.

Now given an observable $Q$ represented as an operator we have that our eigenvalues $q_j$ will label our eigenvectors as $|q_j>$. Now suppose that the eigenvalues are not all distinct: then the eigenvectors will not give us a complete orthogonal basis of the Hilbert space. In this case, we use additional labels or 'quantum numbers' corresponding to a different observable $R$.

E.g:
$$ |q_j,1>,|q_j,2>\\ Q|q_j,i>=q_j|q_j,i>\\ R|q_j,i>=r_i|q_j,i> $$ And then the e-vectors can be labelled $$ |q_j,r_i> $$

Thus we can see that having non-distinct eigenvalues of an operator leads to the idea of a quantum number.

As an aside, when does this happen?

Given two operators $A,B$,that give a complete basis, we require that: $$ A|a_i,b_j>=a_i|a_i,b_j>\\ B|a_i,b_j>=b_j|a_i,b_j> $$ Now let $[A,B]$ be the commutator of two matrices then: $$ [A,B]|a_i,b_j>=(a_ib_j-b_ja_i)|a_i,b_j>=0 $$ Now as the vectors form a complete basis they are nonzero thus $[A,B]=0$.

Similarly the converse can be shown: $[A,B]=0 \implies $complete set of common eigenvectors.

Answer 3

We often want to change basis from the standard basis to a basis of eigenvectors for a particular matrix. Suppose that $\beta = (v_1, \ldots, v_n)$ is an ordered basis of eigenvectors for a matrix $A$. If $$Q = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}, $$ then $Q^{-1}$ is the change of basis matrix from the standard basis to the basis $\beta$. If $Q$ is orthogonal, then $Q^{-1} = Q^T$, and so $Q^T$ is our change of basis matrix.