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Some matrices can be expressed as polynomials of each other.

  • For example inverses: if $AB=I$ then $A$ can be expressed as some polynomial $p(B)$ and also it exists such polynomial $r(A)$ that $B=r(A)$.
    (standard procedure of deriving such polynomials is based on Cayley-Hamilton theorem).

Not always we have situation that if for example some matrix can be expressed as others polynomial then the reverse statement is true.

  • For example $A^2=I$ doesn't mean that it exists such polynomial $p(I)$ that $p(I)=A$
    (a polynomial of identity matrix must be a scalar matrix but $A$ in the case $A^2=I$ doesn't necessarily be a scalar matrix - the same can be said if instead of $I$ we would have on RHS above equation a scalar matrix).

From these considerations the question follows:

  • what are exactly conditions that allow for a pair of matrices $A,B$ to be expressed as polynomials of each other?

One could say that they must have exactly the same eigenvectors but I'm not sure whether it is equivalent to the condition mentioned above or that conditions can be expressed in other, more universal way..

Let entries of considered matrices be real.

Widawensen
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1 Answers1

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A partial answer is the following:

If $A,B$ are diagonalizable matrices, and each eigenvalue of each matrix has multiplicity 1, then $$ A=p(B), \, B=q(A) \iff AB=BA $$

Exodd
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  • How do you know $AB=BA $ --> $ A=p(B), , B=q(A)$? My counterexample with $I$ shows that there is no such implication for every pair of commuting matrices. – Widawensen May 12 '20 at 13:20
  • $AB=BA$ implies the matrices are symultaneously diagonalizable, so you can work with diagonal matrices. If all the entries of diagonal matrices are distinct, you can find an interpolation polynomial that bring them in any value you want – Exodd May 12 '20 at 13:22
  • @Widawensen In fact in the hypothesis, every eigenvalue has multiplicity 1, while $I$ has eigenvalues with multiplicity $n$ – Exodd May 12 '20 at 13:25
  • I see, the answer is valid for matrices with n distinct eigenvalues. Probably we can say also in this case that the sets of eigenvectors for both matrices are the same (up to the scale). – Widawensen May 12 '20 at 13:28
  • Maybe commuting, diagonalizable matrices with distinct eigenvalues are the only possibility? What do you think ? – Widawensen May 12 '20 at 13:47
  • @Widawensen No, take $B=A$ for any $A$ – Exodd May 12 '20 at 17:51
  • Good example. More generally we could take $B=kA+sI$. – Widawensen May 13 '20 at 15:00