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Question: Let $A\in M_2(\mathbb R)$, Prove that $a_1A^1 + a_2A^2 + ... + a_5A^5 = 0$ for some $a_i\in\mathbb R$ which are not all zero.

First I figured out the point of this proof is to show $A^1, A^2... A^5$ are linearly dependent.

While since $A^5$ is a 2x2 matrix, there must be a relationship between $A^1$ to $A^5$ such that $A^5$ can be represented by a combination of $A^1,A^2,A^3,A^4$. I can compute that but I think there should be a smarter way to get the dependent we want. Thx.

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The space of $2\times 2$ matrices is 4-dimensional. Since $A^1,\dots,A^5$ are five matrices, they must be linearly dependent.


EDIT (as requested in the comments, see this question for many complete proofs of the Lemma):

Lemma. Let $(\Bbb V, +, \cdot)$ be an $n$-dimensional vector space ($n\in\Bbb N$) with basis $e_1,\dots,e_n$. Then for $m>n$, any vectors $v_1,\dots,v_m\in\Bbb V$ are lineraly dependent.

Proof. We can write for some constants $a_{11}, \dots, a_{mn}$:
\begin{gather} v_1 = a_{11} e_1 + a_{12} e_2 + \dots + a_{1n} e_n,\\ \vdots \\ v_m = a_{m1} e_1 + a_{m2} e_2 + \dots + a_{mn} e_n. \end{gather}

Suppose that $\sum_{i=1}^m c_i v_i = 0$. Then $\sum_{i=1}^m c_i v_i = \sum_{i=1}^n e_i \left(\sum_{j=1}^m c_j a_{ji}\right)=0$ for all $i=1,\dots,n$. It follows from basic theory of linear equation systems that all $c_i=0$. $\square$

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  • $\begingroup$ Perfect! While I thought the lemma that states "when $m\gt n$, any m vectors in a n-dimensional vector space must be linearly dependent" needs a further proof for our class. Could you please give a simple proof for this? $\endgroup$ – WaterBro Jul 8 '19 at 0:40
  • $\begingroup$ @WaterBro Alright, I proved this elementary lemma for you. Make sure to check out this question for more informations and ideas on this. $\endgroup$ – Maximilian Janisch Jul 8 '19 at 0:55
  • $\begingroup$ Thank u very much! This really helps me a lot! $\endgroup$ – WaterBro Jul 8 '19 at 1:00
  • $\begingroup$ @WaterBro You're welcome! I'm glad to have helped. $\endgroup$ – Maximilian Janisch Jul 8 '19 at 1:00
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$M_2(\mathbb R)$ is a four dimensional vector space and hence any $5$ elements in it are linearly dependent.

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As others have pointed out in good answers, the problem statement---in particular, the use of five coefficients---suggests that the intended solution is just to consider the dimension of the real vector space $M_2(\Bbb R)$.

Here's an alternative method that lets us sharpen the result: The Cayley-Hamilton Theorem says that any $n \times n$ matrix $A$ satisfies its own characteristic polynomial, $$p_A(t) = \det (t I - A) ,$$ which in particular has degree $n$: $$p_A(A) = 0 .$$ In particular, if $A$ is $2 \times 2$, the left-hand side is a polynomial of degree $2$ in $A$. So, if we multiply both sides by $A$, we get $$A p_A(A) = 0 ,$$ the left-hand side of which is a (monic) cubic polynomial in $A$ with no constant term, which proves the claim and shows that we can furthermore impose $a_3 = 1, a_4 = a_5 = 0$.

As usual we can write the characteristic polynomial of a $2 \times 2$ matrix in terms of its trace and determinant, and substituting gives a simple formula for explicit choices of the remaining coefficients ($a_1, a_2$) as functions of $A$: $$(\det A) A - (\operatorname{tr} A) A^2 + A^3 = 0 .$$

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  • $\begingroup$ +1 - I like this answer because it takes advantage of the fact that we are dealing with powers of the matrix $A$ $\endgroup$ – Maximilian Janisch Jul 9 '19 at 7:36

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