In Linear Algebra, Gilbert Strang, $4$th edition the theorem $5$S (section $5.6$) is the following –
Every real symmetric A can be diagonalized by an orthogonal matrix Q.
Every Hermitian matrix can be diagonalized by a unitary U.
In one of the remarks to this theorem it is stated that –
A is the limit of symmetric matrices with distinct eigenvalues. As the
limit approaches, the eigenvectors stay perpendicular.
What does this limit of symmetric matrices denote?
Best Answer
Let's say we have a sequence of $m \times n$ matrices $A_k$. What does it mean to approach a limit of $A$? It means that $\|A_k - A\| \to 0$ in the real numbers. That is, for all $\varepsilon > 0$, there exists some $N$ such that $$k \ge N \implies \|A_k - A\| < \varepsilon.$$ Here, $\| \cdot \|$ is a norm on the vector space of $m \times n$ matrices. Which norm specifically? It turns out not to matter, as the vector space is finite-dimensional.
One such norm could be $$\|A\| = \sup_{\substack{1 \le i \le m \\ 1 \le j \le n}} |(A)_{ij}|,$$ and using this norm we can easily show:
That is, $A_k \to A$ if and only if the entries of $A_k$ converge to the entries of $A$. This is probably the easiest way to think of it.
Now, as it turns out, multiplication by a matrix is a continuous operation (matrix multiplication is just a series of addition and multiplication of matrix entries), so we see that, if $A_k \to A$, then $A_k B \to AB$.
How does this help us? If $A$ is symmetric, then $A$ can be decomposed like so: $$A = U^\top D U,$$ where $D$ is a diagonal matrix with real entries (specifically, the eigenvalues), and $U$ is an orthogonal matrix. Indeed, any matrix of the form $U^\top D U$ is also symmetric, and the eigenvalues of $U^\top D U$ are the entries of $D$.
Now, if $A$ does not have distinct eigenvalues, then $D$ does not have distinct diagonal entries. But, what we can do is form a sequence of diagonal matrices $D$ with distinct diagonal entries, but whose entries converge to the entries of $D$. For example, suppose your eigenvalues are $\lambda_1, \lambda_1, \lambda_2$, where $\lambda_1 \neq \lambda_2$. That is, your diagonal matrix could look like $$D = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2\end{pmatrix}.$$ You could then consider: $$D_k = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_1 + \frac{1}{k} & 0 \\ 0 & 0 & \lambda_2\end{pmatrix},$$ and we'd have $D_k \to D$. Further, unless $\lambda_2 - \lambda_1 = \frac{1}{k}$ for some $k$, each $D_k$ would have distinct eigenvalues, and hence the same is true of $U^\top D_k U \to U^\top D U$.
Note the trickiness there of avoiding other eigenvalues. There's also more trickiness when a repeated eigenvalue is repeated more than once (obviously you can't just do $\lambda_1 + \frac{1}{k}$ for each extra $\lambda_1$, as this will no longer involve distinct eigenvalues). This is fiddly, but not difficult to solve. I won't prove it, but hopefully you're convinced that it can be done, just by choosing other null-convergent sequences other than $\frac{1}{k}$.
I hope that helps!