From what I know the Lanczos algorithm takes a symmetric matrix $A$ and a vector $v$ of norm $1$ as inputs, then generates tridiagonal matrix $J$ and orthonormal matrix $V$ such that $J = V^TAV$. I read the paper A Survey of Matrix Inverse Eigenvalue Problems By Boley and Golub (1986) that makes me confused.
In this paper having constructed a bordered diagonal matrix $B$ (a matrix whose graph is a star graph) using values $\lambda=\{\lambda_1, \lambda_2, \cdots, \lambda_n\}$ and $\mu=\{\mu_1, \mu_2, \cdots, \mu_{n-1}\}$ eigen data such that $\sigma(B)=\lambda$ and $\sigma(B_1)=\mu$ where $B_1$ is the matrix $B$ after removing its first row and column and
$$
B = \begin{bmatrix}
a1&b^T\\b&M
\end{bmatrix}
=\begin{bmatrix}
a1&b_1&b_2& \dots&b_{n-1}\\
b_1&\mu_1&0&\cdots&0\\
b_2&0&\mu_2&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\\
b_{n-1}&0&0&\cdots&\mu_{n-1}
\end{bmatrix}
$$
it constructs tridiagonal matrix $J$ such that $\sigma(J)=\lambda$ and $\sigma(J_1) = \mu$ in the following way:
Using the orthogonal tranformation (Householder) we have
$$
\begin{bmatrix}
1&0\\0&Q_1^T
\end{bmatrix}
\begin{bmatrix}
a_1&b_1e_1^T\\b_1e_1&J_1
\end{bmatrix}
\begin{bmatrix}
1&0\\0&Q_1
\end{bmatrix} = \begin{bmatrix}
a_1&b^T\\b&M
\end{bmatrix}
$$
It is clear the above transformation keeps the eigenvalues of $B$ and $B_1$. On carrying out the multiplication, we find
$$
Q_1^TJ_1Q_1 = M, Q_1b_1e_1 = b
$$
The first equation shows that the eigenvectors of $J_1$ are the columns of $Q_1$ and the second equation shows that, apart from the factor $b_1$, the vector $b$ is the vector of first components of the eigenvectors of $J_1$:
$$
b = b_1\{q_{11}, q_{12}, \cdots, q_{1, n-1}\}
$$
This is what I can't understand
Thus, apart from the factor $b_1$, $b$ is the vector $v$ needed for the construction of $J_1$ from the Lanczos algorithm.
It is clear the method first builds a bordered diagonal matrix in some way, then uses the $b$ as the starting vector. If the Lanczos algorithm works by any starting vector with norm $1$, then why it tries to build this using the $b$ vector?
Best Answer
The starting vectors in most of the eigenvalue algorithms are initialized with a random vector with entries $X \sim N(0,1)$ If you look in Trefethen. You should note that they randomly initialize most of the algorithms in numerical linear algebra.
It says "arbitrary" but what it means it is random. Some code follows after an explanation. The Lanzcos algorithm is in a suite of algorithms called the Krylov methods. The Krylov methods are a specific type of iterative eigenvalue algorithm as opposed to direct methods. The Krylov methods are listed below.
To explain how Lanczos works I am going to explain how Arnoldi works. The basis of both of these eigenvalue methods is first to use a similarity transform reducing to Hessenberg form. That is the following
$$A = QHQ^{*} $$
or $$AQ = QH $$
Instead, we compute $Q_{n} $ letting it be a $m \times n$ matrix whose first $n$ columns are of $Q$
Additionally, we have
In the book it calls it Hessenberg matrix however my professor said it is "hessenbergish". We end up with
$$ A Q_{n} = Q_{n+1} \bar{H_{n+1}} $$
The idea is simple that we are actually building Krylov subspaces
$$ \mathbf{K_{n}} = \langle b, Ab, \cdots, A^{n-1}b \rangle = \langle q_{1} ,q_{2} , \cdots q_{n} \rangle $$
We form these throught the QR decomp
$$ K_{n} = Q_{n}R_{n}$$
Now the we had $H_{n+1}$ to remove this row we note the following. $ Q_{n}^{*}Q_{n+1}$ is the $n \times (n+1)$ identity then $Q_{n}^{*}Q_{n+1}\bar{H_{n}}$ is $ n \times n$ Hessenberg matrix obtained by removing the last row.
$$ H_{n} = Q_{n}^{*}AQ_{n} $$
Now if you see we are iteratively building our projections through these Krylov subspaces.
The Lanczos iteration is simply when the matrix is symmetric.
if you note if v is not specified it is these lines
It is constructing an orthogonal eigenvector. $q_{1}$ and each one after that from it.