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I am trying to prove that for an $n\times n$ matrix $A$ there exists a non-zero polynomial $p(x)$ such that $p(A) = 0$.

Is my proof valid?

Proof:

Suppose $\deg(p(x)) = n$. Since the coefficients of $p$ can be whatever I want them to be $p(A)$ is simply a linear combination of the set: $$B_n = \{I,A,A^2,\dots,A^n\}$$

Now consider:

$$\lambda_0I + \lambda_1A + \dots+\lambda_nA^n = 0$$

If $\lambda_i = 0 \space \forall i \in \{0,1,\dots,n\}$ then $B_n$ is a linearly independent set. Now I consider a polynomial of degree $n+1$. So $B_{n+1} = \{I,A,A^2,\dots,A^n,A^{n+1}\}$. I do the same check and if we get the same result I continue the process until $\deg(p(x)) = n^2 +1$.

Since $\dim(M_{n\times n}(F)) = n^2 \implies B_{n^2+1}$ is linearly dependent $\iff (\lambda_0I + \lambda_1A +\dots+\lambda_{n^2+1}A^{n^2+1} = 0 \implies$ for some $i$, $\lambda_i \neq 0 $.)

Using this I just construct $p(x)$ with these coefficients, and we then have $p(A) = 0$.

Is there a better way? Or is my method okay?

Kenta S
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Governor
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    Your proof is fine. You only need the last two paragraphs really. See also https://math.stackexchange.com/questions/1310609/avoiding-the-cayley-hamilton-theorem – lhf Jan 06 '21 at 13:21
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    You cannot start with "Suppose $\deg p=n$" when attmepting to prove that $p$ exists in the first place. And even then, you have no reason to suppose that the degree of $p$ equals the dimension $n$ of the matrix. – Hagen von Eitzen Jan 06 '21 at 13:22
  • @lhf cheers man – Governor Jan 06 '21 at 13:22
  • @HagenvonEitzen I should start from the last step right? – Governor Jan 06 '21 at 13:24
  • Also: "If $\lambda_i=0$ ... then .... linearly independent" - No. "If this implies that all $\lambda_i=0$, then ..."! – Hagen von Eitzen Jan 06 '21 at 13:24
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    @Govind75 I think you are overcomplicating things. Basically, the set ${A^k : 0 \leq k \leq n^2}$ is a set in $Hom(\mathbb R^n,\mathbb R^n)$ of size $n^2+1$, more than the dimension of that space, so it is linearly dependent. Then just writing down the linear combination equaling zero as a polynomial in $A$ does the job (nobody cares which coefficients are zero, in the end the polynomial simplifies itself to a conventional one). – Sarvesh Ravichandran Iyer Jan 06 '21 at 13:26
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    @Govind75 Yeah, just not that $n^2+1$ powers of $A$ (e.g., $I,A, A^2, \ldots A^{n^2}$) must be linearly dependent and the coefficients in a non-trivial dependency give you the coefficients of your desired polynomial $p$ – Hagen von Eitzen Jan 06 '21 at 13:27

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