Inner Product with Zero Vector

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Theorem

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.

Let $0_V$ be the zero vector of $V$.


Then:

$\innerprod {0_V} x = \innerprod x {0_V} = 0$

for all $x \in V$.


Proof

We have:

\(\ds \innerprod {0_V} x\) \(=\) \(\ds \innerprod {0_V + 0_V} x\) Definition of Zero Vector
\(\ds \) \(=\) \(\ds \innerprod {0_V} x + \innerprod {0_V} x\) linearity of inner product in first argument

so:

$\innerprod {0_V} x = 0$

From conjugate symmetry, we have:

$\innerprod x {0_V} = \overline {\innerprod {0_V} x}$

so:

$\innerprod x {0_V} = \overline 0 = 0$

$\blacksquare$