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I'm trying to check whether or not this set is linearly independent for all $n$, where $A$ is $n \times n$ and $A, A^2, \dots, A^{n^2}$ are distinct matrices and $I_n$ is the identity matrix.

Clearly, if we take $n = 2$, and $A = 3 I_n$ then the set $\{I, A, A^2, A^3, A^4 \}$ is not linearly independent. Is this good enough?

Rodrigo de Azevedo
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Matheus Andrade
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  • I recommend you read on minimal polynomial. The way it is constructed has a clear relation with your question. – user2345678 Jan 23 '18 at 14:24
  • I do not understand why this question was not put on hold. Especially when we read the OP's work: "Clearly, if we take $n = 2$, and $A = 3 I_n$ then the set ${I, A, A^2, A^3, A^4 }$ is not linearly independent. Is this good enough?". Note that no one has explained how absurd his reasoning is. In short, this question is even weaker than showing that "find the $A\in M_2$ s.t. $A^2=0$" or "find $A,B$ s.t. $AB-BA=I$" or "if $A,B$ are square matrices s.t. $AB=I$, then show that $BA=I$" and many more... –  Jan 24 '18 at 11:14
  • cf. also https://math.stackexchange.com/questions/1427469/if-a-in-kn-times-n-is-i-n-a-a2-a3-dots-an2-1-lin?rq=1 where we can find a very strange answer. (to be taken as a second or third degree joke). –  Jan 24 '18 at 11:14
  • @loupblanc How is my original reasoning absurd? – Matheus Andrade Jan 24 '18 at 11:18
  • OK, I do the job. The required answer is "for every $A$, the considered set is linearly dependent". When you choose $A=3I$, you show that "there is $A$ s.t. the considered set is linearly dependent". Doing that, you do not find any contradiction but you do not prove anything. Finally the answer to your question "Is this good enough? " is "not at all". –  Jan 24 '18 at 11:28
  • @loupblanc Isn't considering $A = 3I$ just offering a counter example? It is my understanding that when one makes a claim that "this property holds for every...", it's enough to find a counter example, so that there is at least one case where the property does not hold and so the claim is false. Where did that fail here? The claim I want to prove false is "for every $A$, the set is linearly independent", as I previously stated, so finding a counter example is enough. – Matheus Andrade Jan 24 '18 at 11:32
  • I do agree with you that if the problem wanted me to "prove whether or not this set is linearly independent for every $A$", you'd be absolutely correct. The question was one of those true or false exercises though, where the claim was "for every $A$, this set is linearly independent." – Matheus Andrade Jan 24 '18 at 11:36
  • You wrote "I'm trying to check whether or not this set is linearly independent...". You had the wrong intuition. That's exactly what I blame you. If you want to do mathematics, then you have to learn to guess what is true. –  Jan 24 '18 at 11:40
  • You previously stated my reasoning was absurd. I have now explained my reasoning, and you still have not clarified your original statement further. Now you blame me for "not having intuition", when in fact, I had a counter example to disprove the claim from the beginning. As one often learns when advancing their studies, intuition is a very tricky thing in mathematics, so saying that "to do mathematics, you have to learn to guess what is true" is at best naive. To do mathematics, one has to learn how to prove what is true, not guess it. – Matheus Andrade Jan 24 '18 at 11:44
  • OK I spent my time. The discussion is closed. –  Jan 24 '18 at 11:48

3 Answers3

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The vector space of $n \times n$ -matrices has dimension $n^2$, the set $\{I_n, A, A^2, \cdots, A^{n^2} \}$ has $n^2+1$ elements.

Conclusion ?

Fred
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By the way, you can show something (much) stronger:

Given any $n \times n$ matrix $A$, the set $\{I_n, A, A^2, \dots, A^n\}$ is linearly dependent.

For proof, use the Cayley-Hamilton theorem.

Pavel Čoupek
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Consider $V$ a vector space over a field $\mathbb{K}$. Then, if you take any matrix $M \in \mathcal{M}_n(\mathbb{K})$ you can compute its characteristic polynomial which you know has degree $n$. Then, notice this:

Let $m_A(t)$ be the minimal polynomial of $A$. Then, if you call $k=deg(m_A(t))$, $k$ is the dimension of the vector space generated by the powers of the matrix (i.e. Span$(I,A,A^2,\dots,A^{k-1})$.

Conclude by saying that $deg(m_A(t)) \leq deg(p_A(t)) = n$ where $p_A(t)$ is the characteristic polynomial.

Alberto Andrenucci
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