# Is orthonormality equivalent to orthogonality and normalization in a normed inner product space?

Given the concepts of orthogonality and unit vector in a normed inner space defined as:

Two vectors $$x,y$$ are orthogonal if $$\langle x, y \rangle = 0$$

A vector $$x$$ is a unit vector if $$\|x\| = 1$$.

Then, is the following definition of orthonormality

A set of vectors $$\{v_1, v_2, \dots, v_n\}$$ is orthonormal if $$\langle v_i, v_j \rangle = \delta_{ij}$$, where $$\delta_{ij}$$ is the Kronecker Delta

equivalent to

A set of vectors $$\{v_1, v_2, \dots, v_n\}$$ is orthonormal if $$\forall(v_i \ne v_j), v_i$$ and $$v_j$$ are orthogonal and all $$v_i$$ are unit vectors

The Wikipedia page on orthonormality seems to state both these definitions, but they don't seem equivalent to me. Is the inner product of a unit vector with itself necessarily 1?

For any inner product space with inner product $$(\cdot, \cdot)$$, we define the norm on that space by $$\|x\| = \sqrt{(x,x)}$$. For more info, see here. With the norm so defined, if $$\|x\|= 1$$, then $$(x,x) = \|x\|^2 =1$$, and so the definitions are equivalent.
Morally, yes, they are equivalent, but technically, no. Not unless you specify what the norm is. Usually, in an inner product space the norm is defined naturally by $$\langle v,v\rangle^{1/2}$$, but that's not the only way. I could define the norm as $$2\langle v,v\rangle^{1/2}$$, in which case $$\langle v_i, v_j \rangle = \delta_{ij}$$ would not imply that the $$v_i$$'s have unit norm (they would in fact have norm 2).
Of course usually one defines the norm as $$\|v\|=\langle v,v \rangle^{1/2}$$, so the two are equivalent.