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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Cuhrazatee
    Feb 1, 2021 at 16:53
  • $\begingroup$ 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? $\endgroup$ Feb 4, 2021 at 17:58

1 Answer 1


I have found a way to explain this equivalence for the case of both $A$ and $B$ symmetric, based on this material.

Note that the norm above can be expressed as

$$ \left \| C \right \| = \sup_{x \neq 0}{R(x)} \tag{2} $$

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.


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