How to quickly check if vectors are an orthonormal basis of a vector space? Let's say we got $3$ vectors given and we need to check if they are an orthonormal basis of some vector space. How would that be done quickly? I have read on several sites on the Internet and here is my summary, correct me if I'm wrong please:


*

*All vectors need to be linearly independent.

*Vectors are perpendicular aka orthogonal to each other (if $3$ vectors given, I have to do it as pairs of $2$, right?).

*Each vector has length $1$.

 A: What you write down is correct, and that is exactly the definition of orthonormal bases.
If in the Euclidean spaces, you know we can check the linear dependence by its determinant, which is the only thing making the matter easier.
A: Assume the vectors are in 3-dimensional Euclidean space and have these coordinates: $$(x_1,y_1,z_1), \ \ (x_2,y_2,z_2), \ \ (x_3,y_3,z_3).$$
To verify orthogonality, check the following:
$$x_1\cdot x_2+y_1\cdot y_2+z_1\cdot z_2=0,$$
$$x_1\cdot x_3+y_1\cdot y_3+z_1\cdot z_3=0,$$
$$x_2\cdot x_3+y_2\cdot y_3+z_2\cdot z_3=0.$$
Orthonormal means that, in addition to orthogonality, each vector has length 1. We can check the following to ensure each vector has unit length:
$$x_1^2+y_1^2+z_1^2=1,$$
$$x_2^2+y_2^2+z_2^2=1,$$
$$x_3^2+y_3^2+z_3^2=1.$$
A: We speak about Orthogonality in a vector space when it has an inner product( which is a generalization of dot product) defined over it.
Read about Inner Product Spaces here -> https://en.wikipedia.org/wiki/Inner_product_space
Say we have an Innerproduct space $X$= (X, < , > )  where the dimension of X as a vector space is  3 , then we say given  vectors $x$ ,$y$ and $z$ form an $orthonormal$ $basis$ if  < x , y > = 0 , < x , z > = 0 and < y , z > = 0 also < x , x >=1 , < y , y > = 1 and < z , z > = 1.
Actually orthonormal vectors are anyway linearly independent so you need not check for linear independence of vectors seperately.
A: If you are using a computing environment where matrix operations are fast, you can check that 
$$A^T \cdot A = I$$
where $A$ is a matrix of your basis of column-vectors vectors: $(i_1|i_2|i_3)$.
Note that according to matrix multiplication semantics, each element in the result matrix corresponds to the dot-product of a pair of basis vectors. Hence it exactly matches the definition of orthonormality: the dot-product $<i_j,i_k>$ is 1 on the diagonal (when $j = k$) and 0 elsewhere (when $j \ne k$).
A: That is correct.

  
*
  
*All vectors need to be linearly independent
  

This is by definition the case for any basis: the vectors have to be linearly independent and span the vector space. An orthonormal basis is more specific indeed, the vectors are then:


*

*all orthogonal to each other: "ortho";

*all of unit length: "normal".


Note that any basis can be turned into an orthonormal basis by applying the Gram-Schmidt process. 

A few remarks (after comments):


*

*the vector space needs to be equipped with an inner product to talk about orthogonality of vectors (you're then working in a so called inner product space);

*if all vectors are mutually orthogonal, then they are definitely linearly independent (so you wouldn't have to check this separately, if you check orthogonality).

