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?

Solution 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.

Solution 2:

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.