Finding the square of an eigenvalue in a Generalized eigenvalue problem

Question

I'm looking at a generalized eigenvalue problem that is associated with a FEM structure:

$$\mathbf{A}\mathbf{v}-\lambda \mathbf{B}\mathbf{v}=\mathbf{0}$$

Where $\mathbf{A}$ is a stiffness matrix, $\mathbf{B}$ a special matrix, $\mathbf{v}$ an eigenvector and $\lambda$ an eigenvalue.

By looking at numerical results I noticed that the eigenvectors that I'm interested in are those with eigenvalues closest to zero (positive and negative eigenvalues are possible). Therefore, I was wondering if I can calculate a slightly different eigenvalue problem such that the resulting eigenvectors are the same and the eigenvalues are squared. This would make my eigenvalues all positive real, which is closer to how the classical structural eigenvalues look like (with $\mathbf{B}$ being the mass matrix) and also I'm hoping that it would be easier to solve numerically for large sparse systems.

I was able to generate such a system by working on the first equation: $$\left(\mathbf{B}^{-1}\mathbf{A}\right)\mathbf{v}=\lambda\mathbf{v} \quad \Rightarrow \quad \left(\mathbf{B}^{-1}\mathbf{A}\right)^2\mathbf{v}=\lambda^2\mathbf{v} \quad \Rightarrow \quad \left(\mathbf{A}\mathbf{B}^{-1}\mathbf{A}\right)\mathbf{v}=\lambda^2\mathbf{B}\mathbf{v}$$

However this system still requires inverting $\mathbf{B}$.

Are there some additional steps that I could take to further the derivation to a point where the eigen-system would be made of matrix multiplications and additions on $\mathbf{A}$ and $\mathbf{B}$? My naive first approach was to check if using $\mathbf{A}^2$ and $\mathbf{B}^2$ would work.

Note that $\mathbf{A}$ is real positive definite and $\mathbf{B}$ is real symmetric.

Answer 1

What if you tried taking the squared norm of your expression?

$$ (Av-\lambda Bv)^T(Av-\lambda Bv) = v^T(A^TA-\lambda(A^TB+B^TA)+\lambda^2B^TB)v=0 \,\,\,\,\,\,\,\,\,(1) $$

If $A$ and $B$ are known, you can reduce it to a polynomial expression on $v$: $$ v^T(A^TA)v -v^T(A^TB+B^TA)v\lambda+v^T(B^TB)v\lambda^2=0\,\,\,\,\,\,\,\,\,(2) $$

Note that if you only want a value that is really close to zero, you could take $\lambda^2\approx 0$, thus: $$ v^T(A^TA)v -v^T(A^TB+B^TA)v\lambda=0\,\,\,\,\,\,\,\,\,(3) $$ Thus: $$ \text{Det}\left(A^TA -\lambda (A^TB+B^TA) \right) =0 \,\,\,\,\,\,\,\,\,(4) $$ This should give you a good starting point if trying to solve the earlier equation numerically.