Final answer:
The inner product of any vector and the null vector in an inner product space is zero due to the linearity property (distributivity) of inner products which allows us to multiply by zero, annihilating the inner product.
Step-by-step explanation:
The question asks us to use the properties of inner products to prove that \( \langle\mathbf{v}, \mathbf{0}\rangle=0 \) in any inner product space \( V \) for any \( \mathbf{v} \) in \( V \). To do so, we need to understand the three main properties of inner products, which are conjugate symmetry, linearity in the first argument, and positive-definiteness.
The second property, linearity in the first argument (also known as distributivity), tells us that:
\( \langle\mathbf{u}+\mathbf{w}, \mathbf{v}\rangle=\langle\mathbf{u}, \mathbf{v}\rangle + \langle\mathbf{w}, \mathbf{v}\rangle \)
And also
\( \langle c\mathbf{u}, \mathbf{v}\rangle = c\langle\mathbf{u}, \mathbf{v}\rangle \)
Where \( \mathbf{u} \) and \( \mathbf{w} \) are vectors in \( V \), \( \mathbf{v} \) is another vector in \( V \), and \( c \) is a scalar.
We can use these properties to show that:
\( \langle\mathbf{v}, \mathbf{0}\rangle = \langle\mathbf{v}, 0\cdot \mathbf{v}\rangle = 0\cdot \langle\mathbf{v}, \mathbf{v}\rangle = 0 \)
This equality holds because multiplying by zero annuls the inner product, as dictated by the linearity property.
The complete question is: Using only the 3 properties of (real or complex) inner products, prove that in any inner product space \( V \), for any \( \mathbf{v} \) in \( V \), \[ \langle\mathbf{v}, \mathbf{0}\rangle=0 \] is: