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Show that the set $\{A, A^2, A^3, A^4\}$ of $2 \times 2$ matrices is linearly dependent.

Cayley–Hamilton theorem states that every square matrix $A$ satisfies its own characteristic polynomial equation $\det(A-\lambda I) = 0$ such that $A^n+a_{n-1}A^{n-1}+\cdots+a_2A^2+a_1A+a_0I=0$.
How can I show that the given set of matrices is linearly dependent because $I$ is not in the set. Am I missing some subtlety here?

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  • $\begingroup$ Sorry, I meant linearly dependent. I'm not sure if your statement applies to linear dependence too. What I'm saying is, is there some definition or workaround that allows me to use the $I$ in the linear combination? Or even just some truth that I am not aware of. $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:13
  • $\begingroup$ No problem. Just separated out the question and your attempt for clarity. $\endgroup$ Jul 16, 2017 at 13:22

2 Answers 2

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The matrix $A$ satisfies its characteristic polynomial, so $$A^2 + a_1 A + a_0I=0.$$ Multiplying by $A$ we obtain $$A^3+a_1A^2+a_0A=0,$$ that is, already $A, \, A^2, \, A^3$ are linearly dependent.

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  • $\begingroup$ Wow, that's awesome. $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:20
  • $\begingroup$ Sorry @Francesco. Yours is an awesome answer, but I extracted some more info from the other answer. Cheers. $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:31
  • $\begingroup$ No problem, but is not the other answer saying the same things as mine? :-) $\endgroup$ Jul 16, 2017 at 13:33
  • $\begingroup$ Yes it is, so who do I give it to? Let me see if I can select your answer also. $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:36
  • $\begingroup$ Just select the answer you prefer, I was only joking $\endgroup$ Jul 16, 2017 at 13:37
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Let the characteristic equation of $A$ be: $$A^2+pA+qI=0$$ Then $$A^2=-pA-qI$$ Thus $$A^3=-pA^2-qA$$ Hence $$A^4=-pA^3-qA^2$$ That is $\{A^2,A^3,A^4\}$ is linearly dependent. As a superset of a linearly dependent set is linearly dependent, therefore $\{A,A^2,A^3,A^4\}$ is also linearly dependent.

Edit: Let $\{x,y\}$ be linearly dependent. Then there exists scalars $a,b$ such that $ax+by=0.$ Hence, $ax+by+0z=0$. Thus, $\{x,y,z\}$ is also linearly dependent.

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  • $\begingroup$ Are you saying @Sahiba, that even if you have a set of 100 elements, you only need two elements to be linearly dependent for the set to be known as linearly dependent? $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:24
  • $\begingroup$ @Bucephalus Yes I'm saying that. $\endgroup$ Jul 16, 2017 at 13:28
  • $\begingroup$ OK, that's good. $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:30
  • $\begingroup$ @Bucephalus I have added the reason why that is true in the answer. $\endgroup$ Jul 16, 2017 at 13:30
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    $\begingroup$ Yeah, that's a good explanation @Sahiba. Thanks for your help. $\endgroup$
    – Bucephalus
    Jul 16, 2017 at 13:32

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