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:
- $\innerprod x {0_V} = \overline 0 = 0$
$\blacksquare$