Proof dot product identity

Question

From the dot product properties we can write:

$$ \langle v, u \rangle = v^{T} u = \sum_{j}^{} u_j v_j $$

I want to prove that:

$$ \langle v, u \rangle u = (u u^{T}) v $$

I tried rewriting the first expression as

$$ \langle v, u \rangle u = (v^{T} u)u = \cdots = (u u^{T}) v $$

But I don't know how to continue to reach the right hand side. Is this approach the right way?

Answer 1

The $i$-th component of $\langle v, u \rangle u$ is $$\sum_j v_ju_j u_i$$ While the $i$-th component of $(u u^t) v$ is $$\sum_{j} (uu^t)_{ij}v_j=\sum_j u_iu_jv_j$$

Answer 2

We have$\, \langle v, u \rangle u = u \langle u, v \rangle = u (u^{T} v) = (u u^{T})v \,$ where the first equality is commutativity of multiplication of a scalar times a vector. The second equality by definition of dot (or inner) product. The third equality by associativity of multiplicaton of matrices and vectors.