Why can this inner product condition be expressed as this sum?

Question

I am studying a simple example of the projection theorem for Hilbert spaces.

Within a Hilbert space $H$ there is a subspace $U$ spanned by the vectors $u = (u_1, ..., u_k)^T$. So $U$ is just all the linear combinations $a^Tu$ with $a = (a_1, ..., a_k) \in \mathbb{R}^k$. The book says that, for an arbitrary $h \in H$, the projection of $h$ onto $U$ is the unique element $a_0^Tu$ that satisfies:

$\langle h - a_0^Tu, a^Tu \rangle = 0$ for all $a = (a_1, ..., a_k) \in \mathbb{R}^k$

I understand this to mean that if all the elements of $u$ are orthogonal to $h$, then the entire $U$ must also be orthogonal to $h$. But then it goes on to say that an equivalent condition is:

$\sum_{j=1}^k a_j\langle h - a_0^Tu, u_j \rangle = 0$ for all $a_j, j = 1,...,k$

My question is, isn't it possible for some of the $\langle h - a_0^Tu, u_j \rangle$ to be positive and some negative, so that they cancel each other out and we end up with the sum being equal to zero, even though none of the $u_j$ are orthogonal to $h - a_0^Tu$?

Answer 1

The inner product is bilinear...

$$ \langle h-a_0^Tu, a^T u \rangle = \langle h-a_0^Tu, \sum_{i=1}^k a_i u_i \rangle = \sum_{i=1}^k a_i \langle h-a_0^Tu, u_i \rangle $$

Answer 2

$\sum\limits_{j=1}^{k}a_jb_j=0$ for all choices of $a_1,a_2,...,a_k$ iff $b_j=0$ for all $j$. This can be proved by choosing (for any given $j$), $a_i=0$ for $ i\neq j$ and $a_j=1$.