Inner product of vector sub space and its orthogonal complement in Hilbert space

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So im attempting a problem but there is 1 thing I don't understand.

In a Hilbert space of an infinite series $\vec{a}_{i=0}^{\infty}$ we have a sub vector space V for which the following holds: ${a}_{2k+1} = {a}_{2k}$.

Then, in the solutions, the author constructs a basis for the hilbert space $\vec{e}_{i=0}^{\infty}$ which is fine, but then for the inner product, which is defined as $ \langle a | b \rangle $, they write they write the element $a$ as $$\langle {\vec{e}_{2k+1} + \vec{e}_{2k}|b \rangle}= {b}_{2k+1} + {b}_{2k} = 0$$

It equals $0$ because we are constructing the orthogonal complement to V. $\vec{b}$ is thought of as a vector belonging to the orthogonal complement.

What I don't understand is how we go from the 'condition' of $\vec{e}_{2k+1} = \vec{e}_{2k}$ to, instead of having them equal one another, we write them as a sum.

My first thought is that it is written as such due to bilinearity? I.e.,

$$\langle {\vec{e}_{2k+1} + \vec{e}_{2k}|b \rangle}= \langle{\vec{e}_{2k+1} | b \rangle} + \langle{\vec{e}_{2k} | b \rangle} = {b}_{2k+1} + {b}_{2k} = 0 $$

If so, then I understand the motivation. But how come it is $\vec{e}_{2k+1} + \vec{e}_{2k}$ and not $\vec{e}_{2k+1} - \vec{e}_{2k} = 0$. And why does it specifically yield ${b}_{2k+1} + {b}_{2k}$ ?

$\vec{e}_{i}$ is an orthonormal basis defined as $\vec{e}_{i}= (\delta_{i,j})_{j}$