Show that if $T$ is a symmetric tridiagonal matrix and an eigenvalue $\lambda$ has multiplicity $k$, then at least $k−1$ subdiagonal elements of $T$ are zero.
If we consider a submatrix $B$ that has its diagonal elements as the subdiagonal elements of the matrix $T$, then we can say that the rank of $B$ is not more than the rank of $T$. Can we use this information to prove that the diagonal elements of B has at least $k-1$ zeros?
The most natural explanation for this (in my view) lies in the fact that the eigenvectors of such a matrix solve a difference equation. More precisely, if we write $T_{nn}=b_n$, $T_{n,n+1}=T_{n+1,n}=a_n$, then $Ty=zy$ if and only if $y$ solves $$ a_n y_{n+1} + a_{n-1}y_{n-1} + b_n y_n = zy_n, \quad 1\le n\le N , $$ and satisfies the boundary conditions $y_0=y_{N+1}=0$ (and here we can set $a_0=1$).
Now clearly, if $a_n\not= 0$ for all $n$, then the solution space is at most one-dimensional, because you must start with a multiple of $y_0=0, y_1=1$, and then the subsequent $y_n$'s are delivered to you by the recursion. The dimension can only get bigger if $a_n=0$ somewhere.
Finally, fewer than $k-1$ indices with $a_n=0$ won't suffice: Let's say you had $j<k-1$ of them, then your matrix has block structure, with $j+1<k$ blocks of the type discussed above (all $a$'s within the block non-zero), so the blocks have only simple spectrum.
