# When powers of a matrix are linearly independent

There is a question in my math material I don't know how to solve. It states that $A$ is a $2\times 2$ matrix, I need to determine whether the set $\{I, A, A^2\}$ is linearly independent.

I think this set is linearly independent, but I am not really know how to prove this. Anyone would like to help me about this? Thanks.

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For $A=I$ it is certainly not independent. What do you know about $A$? –  Elmar Zander Mar 5 '13 at 17:55
I only know A is a random 2*2 matrix.So I just suppose A= {a,b; c,d} –  JavaLeave Mar 5 '13 at 17:56
What does Cayley-Hamilton say again? –  Georges Elencwajg Mar 5 '13 at 17:58
@GeorgesElencwajg it said that A^2 = tr(A)A- det(A)I. What should I know from this ? –  JavaLeave Mar 5 '13 at 18:09
@KellyAnn : That $\exists a,b,c\in\mathbb{K}, a A^2 + b A + c I = 0$ with $(a,b,c)\not= (0,0,0)$ –  xavierm02 Mar 5 '13 at 18:17

$A\in M_2(\mathbb{K})$

The characteristic polynomial of $A$ is $\chi_A=X^2 -\operatorname{tr}(A)X+ \det(A)I_2$

By Cayley-Hamilton, $\chi_A(A)=0_2$, that is $A^2 -\operatorname{tr}(A)A+ \det(A)I_2=0_2$

So $\exists (a,b,c)\in\mathbb{K}^3, aA^2+bA+cI_2=0_2$ namely $(a,b,c)=(1, -\operatorname{tr}(A),\det(A) )$ where $a=1\not= 0$

Which means that $\left\{I_2,A,A^2\right\}$ is linearly dependent.

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Considering $A=I_2$ we see that $-2I+A+A^2=0$ where $\mathbb R$ is taken as the corresponding field of scalars. Consequently, $\left\{I, A, A^2\right\}$ is not linearly independent.

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Please note the point by @xavierm02, that $\left\{I_2,A,A^2\right\}$ is linearly dependent for every $2 \times 2$ matrix $A$. –  Andreas Caranti Mar 6 '13 at 7:52
aah...forgot to recall Cayley-Hamilton. –  Sugata Adhya Mar 6 '13 at 7:54