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I see the Cauchy-Schwarz Inequality written as follows

$$|\langle u,v\rangle| \leq \lVert u\rVert \cdot\lVert v\rVert.$$

Why the is the absolute value of $\langle u,v\rangle$ specified? Surely it is apparent if the right hand side is greater than or equal to, for example, $5$, then it will be greater than or equal to $-5$?

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The absolute value gives a stronger result, then. – Dylan Moreland Apr 19 '12 at 14:44
up vote 5 down vote accepted

I assumed that we work in a real inner product space, otherwise of course we have to put the modulus.

The inequality $\langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$ is also true, but doesn't give any information if $\langle u,v\rangle\leq 0$, since in this case it's true, and just the trivial fact that a non-negative number is greater than a non-positive one. What is not trivial is that $\lVert u\rVert\lVert v\rVert$ is greater than the absolute value. But in fact the assertions $$\forall u,v \quad \langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$$ and $$\forall u,v\quad |\langle u,v\rangle|\leq \lVert u\rVert\lVert v\rVert$$ are equivalent. Indeed, the second implies the first, and consider successively $u$ and $-u$ in the first to get the second one.

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What if the inner product is a complex number which can happen if $u$ and $v$ are vectors of complex numbers?

For real vectors, the Cauchy-Schwarz Inequality is better written as $$-||u||\cdot ||v|| \leq \langle u, v \rangle \leq ||u||\cdot ||v||$$ where, if $||v|| > 0$, then equality holds in the right inequality if $u = \lambda v$ with $\lambda \geq 0$ and in the left inequality if $u = \lambda v$ with $\lambda \leq 0$.

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