Given a matrix $\mathbf{A}\in\mathbb{R}^{N\times N}$ with $\mathbf{\Lambda_B}\in\mathbb{R}^{B\times B}$ as the diagonal matrix containing $B\ (B\ll N)$ eigenvalues of matrix $\mathbf{A}$. Is there any iterative algorithm that could obtain $\mathbf{V_B}\in\mathbb{R}^{N\times B}$ that satisfies the following equation, $$ \mathbf{AV_B} =\mathbf{V_B \Lambda_B} + \mathbf{Y_B}, $$ given the bias matrix $\mathbf{Y_B}\in \mathbb{R}^{N\times B}$?

The reason I'm looking for an iterative algorithm is that the matrix $\mathbf{A}$ cannot be stored explicitly, but we can calculate the matrix-vector multiplication, $\mathbf{Av}$.

Edit (additional note): The each column of $\mathbf{Y_B}$ is guaranteed to be perpendicular to the corresponding eigenvector of $\mathbf{A}$.

  • $\begingroup$ Note that solving this for $B=1$ is enough, since this can be decomposed into $B$ distinct equations. So, you only need to solve for $(A-\lambda_1 I)v_1=y_1$. Since $\lambda_1$ is an eigenvalue of $A$, this matrix is singular, so an exact solution might not exist. However you can find "closest" solutions with pseudo-inverse for example. Unfortunately, I don't know any way to calculate pseudo-inverse only using matrix-vector multiplication. $\endgroup$ – obareey Jan 29 at 17:27
  • $\begingroup$ Thanks for the comments. I added an additional note in the post saying that each column in $\mathbf{Y_B}$ is perpendicular to the corresponding eigenvectors. So the value of $\mathbf{v_i}$ that satisfies $(\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v_i} = \mathbf{y_i}$ can actually be found. They just might not be unique. $\endgroup$ – Firman Jan 29 at 18:03

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