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I'm taking a course on linear algebra and have been asked to prove the following theorem.

If we let $A$ be an $n \times n$ matrix over a field $\mathbb{F}$, then there exist scalars $a_1, ..., a_{n^2} \in \mathbb{F}$ which are not all zero such that: $$\sum_{i=0}^{n^2} a_i A^i = O $$ where $O$ is the zero matrix and $A^0$ denotes the identity matrix.

My problem is that I'm not sure how to approach such a proof. I have attempted to apply induction. Here I've first shown the case to be true for $n=1$, but then I'm not sure how to use the induction hypothesis to finish the proof. I'm also not sure induction is the right choice here.

Any pointers on how to proceed would be much appreciated.

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  • $\begingroup$ @ArpanBanerjee this is not the Cayley Hamilton theorem, this is merely a matter of dimensions. The theorem you referred to goes way deeper. $\endgroup$
    – Gabriel Romon
    Apr 2, 2015 at 10:50
  • $\begingroup$ @LeGrandDODOM What do you mean by a matter of dimensions? $\endgroup$
    – Arpan
    Apr 2, 2015 at 10:53
  • $\begingroup$ @ArpanBanerjee Have you read martini's answer ? $\endgroup$
    – Gabriel Romon
    Apr 2, 2015 at 10:59
  • $\begingroup$ I'm actually not very familiar with these concepts, I just remembered the theorem and this looked similar to it. $\endgroup$
    – Arpan
    Apr 2, 2015 at 11:01

1 Answer 1

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Hint. The main ingredient is to observe that the space $\def\Mat{\mathord{\rm Mat}}\Mat_n(\def\F{\mathbf F}\F)$ of $n \times n$ matrices over the field $\F$ has dimension $n^2$. Do you see why? Hence the $n^2 + 1 > n^2$ elements $A^0, \ldots, A^{n^2}\in \Mat_n(\F)$ are linearly dependent.

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    $\begingroup$ Aha, so the idea was to realize that linear dependence is necessary. I think $\text{dim}(\text{Mat}_n(\mathbb{F})) = n^2$ because it's possible to define an isomorphism between $n$ by $n$ matrices and vectors in $\mathbb{F}^{n^2}$ which have dimension $n^2$. Is there a simpler way to see this? $\endgroup$
    – TorbenB
    Apr 3, 2015 at 6:46

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