Taking the Inner Product with the Zero Vector

Question

Is it possible for the inner product of any vector with the zero vector $ \mathbf{0} $ to be nonzero? Or must it always be zero? I'm struggling to find a counterexample. That is, is the following statement correct?

$$ \langle \mathbf{v}, \mathbf{0}\rangle = 0 \space \forall \space v \in V$$

What if you consider this $0\mathbf{u} = \mathbf{0}$, and use linearity of inner product to conclude $\langle \mathbf{v}, \mathbf{0} \rangle = 0$. — vTx, Oct 15 '14 at 07:46

score 11 · Accepted Answer · answered Oct 15 '14 at 07:39

11

Note that an inner product is conjugate-linear in the second component (Details at http://en.wikipedia.org/wiki/Inner_product_space#Elementary_properties).

Thus, since 2 is its own conjugate, we have $2<v,0>=<v,2\cdot 0>=<v,0>$, hence $<v,0>=0$. So yes, it's true.

answered Oct 15 '14 at 07:39

Alan

16,582

Might be a beginner question, but how do you know that 2⋅0 equals 0? I mean, given that 0 in this context is the zero vector, not the number zero. – the-bass Mar 14 '24 at 10:07

score 8 · Answer 2 · edited Oct 09 '20 at 11:53

8

Think of it simply like this.

$\langle \mathbf{v} , \mathbf0 \rangle = \langle \mathbf{v}, \mathbf{0}*\mathbf{0} \rangle$

Here let one of the zeroes be the scalar, then the RHS will look like

$\langle \mathbf{v} , \mathbf0 \rangle = \langle \mathbf{v} , \mathbf{0}*\mathbf{0} \rangle = 0\cdot\langle \mathbf{v} , \mathbf{0} \rangle$

which is $0$.

edited Oct 09 '20 at 11:53

user1729

31,015

answered Oct 09 '20 at 11:18

Krishan

178

Taking the Inner Product with the Zero Vector

2 Answers2