# Cauchy-Schwarz Inequality and Linear Dependence

Given a real inner product space with the standard Euclidean inner product, I read that equality in the Cauchy-Schwarz Inequality implies linear dependence. Does strict inequality, $|\left<u,v\right>|<||u||*||v||$, imply linear independence?

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## 3 Answers

Actually, $| \langle u,v \rangle | = \|u\| \|v\|$ if and only if $u=\lambda v$ for some scalar $\lambda$. You read the "hard" part of the "iff", since equality is trivial when $u$ and $v$ are linearly dependent.

Hence a strict inequality is a necessary and sufficient condition for the linear independence of $u$ and $v$.

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Yes. If $u$ and $v$ are linearly dependent, then we can assume that there exists a non-zero scalar $k$ such that $u = kv$. Then $|<u,v>|=|k|*||v||^2=||u||*||v||$.

So, linearly dependent implies equality.

Therefore, strict inequality implies linearly independent.

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Sure. If $u,v$ are linearly dependent, that means $u=cv$ for some scalar $c$, and then you can show that you get equality (or $u=0$, but that's easily taken care of).

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