# Prove that $a_1A^1 + a_2A^2 + … + a_5A^5 = 0$ [duplicate]

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.

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 $$c_i=0$$. $$\square$$

• 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? – WaterBro Jul 8 '19 at 0:40
• @WaterBro Alright, I proved this elementary lemma for you. Make sure to check out this question for more informations and ideas on this. – Maximilian Janisch Jul 8 '19 at 0:55
• Thank u very much! This really helps me a lot! – WaterBro Jul 8 '19 at 1:00
• @WaterBro You're welcome! I'm glad to have helped. – Maximilian Janisch Jul 8 '19 at 1:00

$$M_2(\mathbb R)$$ is a four dimensional vector space and hence any $$5$$ elements in it are linearly dependent.

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 $$A$$: $$(\det A) A - (\operatorname{tr} A) A^2 + A^3 = 0 .$$

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