Let $A\in M_n$. Can dimension of subspace $L(I,A,A^2,\ldots,A^k,\ldots)$ of $M_n$ can be bigger than $n$?

Using Cayley Hamilton theorem that every matrices $A^n$ can expressed as linear combination of $(A^{n-1},A^{n-2},\ldots,I)$ and $p(A)=0$,

if say opposite,and you say that it have more than n elements so it must exist some $A^{n+1}$ that you can expressed as linear combination, so that mean that we have $0=a_{n+1}A^{n+1}+a_nA^n + \cdots +a_0I$

since $p(A)=0$, $0=a_{n+1}A^{n+1}+p(A)$, that mean that $a_{n+1}A^{n+1}=0$, so it opposite what we think so, we only can have n elements, is this ok?

  • 2
    $\begingroup$ Can you break up your big paragraph into smaller pieces to make it more readable please? $\endgroup$ – max_zorn Sep 8 '18 at 7:09

An arbitrary element of $\operatorname{span}\{I, A, A^2, \ldots\}$ can be written as $f(A)$, where $f$ is a polynomial (of any degree).

Cayley-Hamilton states that $\chi(A) = 0$, where $\chi$ is the characteristic polynomial.

Dividing $f$ with $\chi$, we obtain unique polynomials $q,r$ such that $f(x) = \chi(x)q(x) + r(x)$ and $\deg r < \deg \chi = n$.

Plugging in $A$ we get $$f(A) = \chi(A)q(A) + r(A) = r(A)$$

Therefore, an arbitrary element of $\operatorname{span}\{I, A, A^2, \ldots\}$ can be written as a linear combination of only $\{I, A, A^2, \ldots, A^{n-1}\}$.

Hence $\operatorname{span}\{I, A, A^2, \ldots\} = \operatorname{span}\{I, A, A^2, \ldots, A^{n-1}\}$, and the latter clearly has dimension $\le n$.


If I understand what you wrote correctly, you are assuming that $0 = a_{n+1}A^{n+1} + p(A)$ where $p$ is the same polynomial as the one in Cayley Hamilton. I don't think this follows from what you have written.

Suppose we have the equality $A^n = \sum_{i=0}^{n-1} a_i A^i$ from Cayley-Hamilton. Then observe $$A^{n+1} = A^n A^1 = \sum_{i=0}^{n-1} a_i A^{i+1} = \sum_{i=1}^{n-1}a_{i-1}A^i + a_{n-1} A^n = \sum_{i=1}^{n-1}a_{i-1}A^i + a_{n-1} \sum_{i=0}^{n-1} a_i A^{i}= \sum_{i=0}^{n-1}b_i A^i $$

so $A^{n+1}$ can be expressed as a linear combination of $I,A,\dots,A^{n-1}$. Its not hard to see how this would work for any higher power.

(Note on terminology though, the span is a vector space, and assuming you're working over an infinite field, the span has infinite cardinality. Its basis can have at most $n$ elements.)


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.