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I have the quadratic equation $$ \hat{f}^2+\hat{B}\hat{f}+\hat{C}=0 $$ for the matrix $\hat{f}$, where $\hat{B}$, $\hat{C}$ are some known matrices. Do any general methods of its solution exist?

To be more specific, in my case $\hat{B}$ is the infinite diagonal matrix with the elements $B_{nm}=\delta_{nm}(i\varepsilon+b[n+\frac12])$, $i$ is the imaginary unit, and $\hat{C}$ is proportional to the unit matrix: $C_{nm}=\delta_{nm}c$. Should the matrix $\hat{f}$ be only diagonal in this case?

Update I have found the related questions Is there a unique solution for this quadratic matrix equation? and Solve $AX^2 + BX + C = 0$ about general solutions of quadratic matrix equations, so now my question is simpler: if $\hat{B}$ is a diagonal matrix and $\hat{C}$ is proportional to the unit matrix, so is it true that $\hat{f}$ should be also diagonal matrix, or there exist some nondiagonal solutions?

Alexey Sokolik
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  • In general you have symbolically $2f= -B\pm\sqrt{B^2-4C}$. The problem is that, unlike, for example, the function $f (x) = e ^ x$ with which $f (M) = e ^ M$ always makes sense, the same does not happen with the function $f (x) =\sqrt{B^2-4C}$ because not always $B ^ 2-4C$ is the square of a matrix (this is called symbolic calculation in algebras where it is said that the numerical function $ f$ operates in the algebra (All analytic function operates in a certain algebra of harmonic analysis and the reciprocal of this theorem is a famous difficult result of Katznelson). – Ataulfo May 07 '18 at 15:31
  • @Piquito , that you write is absolutely false. 1. The relation $(2f+B)^2=B^2-4C$ is true almost only when $BC=CB$. 2. Generically $B^2-4C$ is invertible and has distinct eigenvalues; thus it admits $2^n$ complex square roots. Conclusion. If $BC=CB$, then generically, there are $2^n$ solutions. 3. It's false if $BC\not= CB$; in this case, there are generically $\binom{2n}{n}$ complex solutions. –  May 07 '18 at 21:35
  • Read better your Wikipedia. – Ataulfo May 09 '18 at 01:37

1 Answers1

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$\newcommand{\trace}[0]{\text{trace}}$Consider this: $$ X = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} -2 & 0\\ 0 & -2 \end{bmatrix}, C = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}. $$ Then $$ X^{2} = \begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix}, $$ and $$ X^{2} + B X + C = 0. $$

More generally, if $X$ is any $2 \times 2$ matrix, then $X$ is a root of its characteristic polynomial, so that $$ X^{2} + B X + C = 0, $$ where $B$ and $C$ are the scalar matrices $$ B = \begin{bmatrix} - \trace(X) & 0\\ 0 & -\trace(X) \end{bmatrix}, \qquad C = \begin{bmatrix} \det(X) & 0\\ 0 & \det(X) \end{bmatrix}. $$

Andreas Caranti
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  • Thank you! But what about my specific case, where the matrices are generally of infinite size (although can be truncated to a finite size for a practical treatment), and $B$ is not proportional to the unit matrix, in contrast to your example? – Alexey Sokolik May 07 '18 at 16:19