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Suppose we have $m$ vectors of $n$ coordinates, then $A$ is an $n\times m$ matrix. If $m > n$, then it is a theorem that the vectors must be linearly dependent.

Prove this theorem.

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What is giving you trouble in this proof? – Clayton Jan 26 at 18:57
Do you know the dimension of the vector space of all vectors with $n$ coordinates? It is $n$. So any set of linearly independent vectors contains at most $n$ vectors. – julien Jan 26 at 18:57
How are you trying to prove the linear dependence? – Sigur Jan 26 at 18:58
I was actually trying to formalize it, I guess I'm not used to rigorous proofs yet...it sounded too obvious to actually need a proof. Hence my difficulty. – Sawyier Jan 26 at 19:07
@Sawyier I included a PDF that gives you a proof in my answer. If you follow along, step by step, you will gain insight. – Rustyn Yazdanpour Jan 26 at 19:13

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enter image description here

See Link or the image, also remember to include your thoughts/perspectives on the problem.

This PDF is brought to you in part by:
Dan Singer, Associate Professor of the Department of Mathematics and Statistics in Minnesota State University, Mankato

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Rustyn, @DoctorBatmanGod, while it's not encouraged to use "thank you" and "regards" in posts, it is certainly not discouraged as in other sites across the SE network. – Asaf Karagila Jan 26 at 19:22
@AsafKaragila Thanks, I will retain my original "letter style" – Rustyn Yazdanpour Jan 26 at 19:27
@AsafKaragila the faq says otherwise, but you would know better than I, with 66k reputation. – DoctorBatmanGod Jan 26 at 19:40
@DoctorBatmanGod Thanks, in that case, deleted. – Rustyn Yazdanpour Jan 26 at 19:41
@DoctorBatmanGod: The FAQ is relatively uniform across the network. But in many discussions on the meta site it came up and it was usually a consensus that signatures are not to be removed, unless it's part of some vastly larger overhaul to the post. And even then, it's fine if they stay. – Asaf Karagila Jan 26 at 19:43
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