Let us consider the real symmetric tridiagonal matrix $T=(t_{kl})$ in $M_n(\mathbb{R})$ with
$$t_{1,2}=t_{2,3}=\cdots=t_{n-1,n}=1$$
How can we compute the eigenvectors of $T$?
Eigenvectors of the symmetric tridiagonal matrices whose entries above the diagonal are all the same
$\begingroup$
$\endgroup$
2
-
$\begingroup$ I don't see how we can do that without knowing what's on the diagonal ... $\endgroup$– Michael EngelhardtCommented Feb 27, 2023 at 19:05
-
1$\begingroup$ To complement @MichaelEngelhardt : it isn't possible to compute the eigenvectors of the system when the diagonal is not constant, as it would be is a finite difference approximation of $u^{\prime\prime}(x) +q(x)u(x)=\lambda u(x)$, $u(0)=u(1)=0$, which do not have closed form solutions. In the constant diagonal case, it is explicit. $\endgroup$– usernameCommented Feb 27, 2023 at 19:24
Add a comment
|