Generalized spectral radius and matrix norm

I am reading a paper that discusses the design of an approximate matrix in the context of numerical methods for PDEs. However, I do not understand the following step, which I believe should be independent of all the context:

We are trying to design a matrix $$A$$ such that its norm (when seen as an operator $$A:V\rightarrow W$$, where V and W are ordinary column vector spaces) is equal to that of another matrix $$B$$ that belongs to the same space. The inner product to be considered in $$V$$ is defined with the help of a positive-definite matrix $$M$$ as

$$(v, w) = v^T M^{-1} w \tag{1}$$

where $$v,w \in V$$ and where its corresponding norm is denoted as $$\left | \cdot \right |_{M^{-1}}$$. The norm of any operator $$C:V\rightarrow W$$ is then expressed as

$$\left \| C \right \| = \sup_{\left | x \right |_{M^{-1}} = 1}{x^T C x} \label{norm} \tag{2}$$

Closely paraphrasing the paper, a practical way to impose $$\left \| A \right \| = \left \| B \right \|$$ is to compute the spectrum of matrices $$A$$ and $$B$$, with respect to matrix $$M^{-1}$$ and impose that at least the spectral radius be the same. According to the explanations given in the same paper, the referred spectral radius (of an operator $$C$$) is to be taken as the largest of the absolute values of the solutions for $$\lambda$$ of the generalized eigenvalue problem $$CU = \lambda M^{-1} U$$.

Could someone help me break down the logic behind this equivalence? That is, why does the equality of the $$M^{-1}$$-spectral radii of $$A$$ and $$B$$ imply $$\left \| A \right \| = \left \| B \right \|$$?

EDIT: I have realized that the matrices that the paper is applying this equivalence to are symmetric, so I am not sure (although I suspect so) if the equivalence is intended to be true only in this case, since it is not explicitly mentioned in the abstract discussion. For this case, I have found an explanation which I will post as an answer.

• If they have the same spectral radius, you can apply unitary operators to give you the same eigenvalues, ergo the same system. The set of unitary operators on a matrix forms an equivalence class on the set of matrices of the same size. Feb 1, 2021 at 16:53
• Thank you, Cuhurazatee. Do you mean that the result of applying unitary operations will leave the eigenvalues and the norm unchanged? OK, this is true (see, e.g., core.ac.uk/download/pdf/82001682.pdf) however, how can I apply this fact to derive the condition being discussed? Feb 4, 2021 at 17:58

I have found a way to explain this equivalence for the case of both $$A$$ and $$B$$ symmetric, based on this material.
$$\left \| C \right \| = \sup_{x \neq 0}{R(x)} \tag{2}$$
where the generalized Rayleigh coefficient, $$R(x)$$, is defined as $$R(x) := \frac{x^T C x}{x^T M^{-1} x}$$ The maximum of $$R$$ is attained at a stationary point of R (sum is assumed over repeated indices): $$\frac{\partial }{\partial x_m} \left ( \frac{x_i C_{ij} x_j}{x_k M^{-1}_{kl} x_l} \right ) = 0 \Rightarrow \left ( \delta_{im} C_{ij} x_j + x_i C_{ij} \delta_{jm}\right ) x_k M^{-1}_{kl} x_l - x_i C_{ij} x_j \left ( \delta_{km} M^{-1}_{kl} x_l - x_k M^{-1}_{kl} \delta_{lm}\right ) \\ = 2C_{jm} x_k M^{-1}_{kl} x_l - 2 x_i C_{ij} x_j M^{-1}_{ml} x_l = 0 \Rightarrow (x^T M^{-1} x) C x = (x^T C x) M^{-1}x$$ where $$\delta_{ij}$$ is the Kronecker delta. Note that we have used the symmetry of $$A$$ and $$B$$. That is, $$C x = R(x) M^{-1}x$$ So that the maximum is attained at an eigenvalue of the generalized problem. In particular, the maximum eigenvalue will yield the greatest value of $$R$$. Thus, imposing that the (generalized) spectral radius of the two matrices coincides will force their norms to be equal too.