One can see from the Cayley-Hamilton Theorem that for a $n\times n$ matrix, we can write any power of the matrix as a linear combination of lesser powers and the identity matrix, say if $A\neq cI_n$, $c\in \Bbb{C}$ is a given matrix, it can be written as a linear combination of $I_n,A^{-1},A,A^2,\cdots,A^{n-1}$. Is this representation unique? If so, is it under special cases? As an example let's consider this:

Let $A\neq cI_n$ , $c\in \Bbb{C}$ be a $3\times 3$ matrix over $\Bbb{C}$ and $A^3=k_3A^2+k_2A+k_1I_3=m_3A^2+m_2A+m_1I_3$ for $k_i,m_i\in \Bbb{C}$. Does it hold that $k_i=m_i \forall i=1,2,3?$

If we take a formalistic approach I guess we can say that the equivalence above consitutes a system of $3$ equations with $6$ variables ($k_i,m_i$).

That is, we evaluate the two linear combinations and by matrix equivalence we demand that all entries are equal: $a_{ij}=b_{ij}=c_{ij}$. But in general, unless the matrix $A$ has a "special" form, that will have infinite solutions.

Is this-overly simplistic-approach correct? Should I try to find counter examples perhaps or the question has some trivial answer that I overlook?


2 Answers 2


The situation is as follows: let $d$ be the degree of the minimal polynomial of $A$. Then the matrices $\{I,A,A^2,\dots,A^{d-1}\}$ are linearly independent and form a basis for the span of the powers of $A$.

In particular, this means that $A$ will have the uniqueness property you describe if and only if the minimal and characteristic polynomials coincide. That is, the following are equivalent:

  1. The matrices $\{I,A,\dots, A^{n-1}\}$ are linearly independent
  2. $\Bbb R^n$ is $A$-cyclic
  3. $A$ is similar to a companion matrix
  4. $A$ is non-derogatory
  5. The minimal polynomial of $A$ is the same as its characteristic polynomial
  6. The Jordan form of $A$ has one Jordan block for each eigenvalue
  7. All eigenspaces of $A$ have dimension at most $1$
  8. A matrix $B$ satisfies $AB = BA$ if and only if there is a polynomial $f(x)$ for which $B = f(A)$.

So, for example: any $A$ with $n$ distinct eigenvalues will have this property, and any $A$ which consists of a single Jordan block will have this property. Any diagonalizable $A$ with a repeated eigenvalue will not have this property.

  • 1
    $\begingroup$ Thank you very much for your thorough and informative answer! $\endgroup$ May 16, 2016 at 16:45

Take $A=2I$ the identity matrix, $A^3=8I =2A^2$ so the coefficients are not unique.

Take $A$ nilpotent and not zero $A^3=0$, you have $A^3=0A^2=0A$.

  • $\begingroup$ Thanks yet again! I will make sure that no multiples of $I_n$ are considered valid-otherwise we have trivial counterexamples. $\endgroup$ May 16, 2016 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.