# Solving biased eigendecomposition problem, $\mathbf{AV_B}=\mathbf{V_B\Lambda_B} + \mathbf{Y_B}$

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 $$\mathbf{A}$$.

• 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. – obareey Jan 29 at 17:27
• 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. – Firman Jan 29 at 18:03