# Sets forming orthonormal basis

So the question is:

Which of the following sets of vectors form an orthonormal basis for $\mathbb{R}^2$
$(a) \{(1,0)^T, (0,1)^T\}$
$(b) \{(\frac{3}{4},\frac{4}{5})^T,(\frac{5}{13},\frac{12}{13})^T\}$

I know that a is a basis and b is not. The thing is, I just don't know why. I know the definition of orthonormal is if two vectors are perpendicular and of unit length. But I don't understand how to prove or even find for that matter, the orthonormal basis of a space. Any ideas? Thanks

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## 2 Answers

Two vectors $x,y \in \mathbb{R}^n$ are orthogonal if their dot product equals zero; that is $$x \cdot y = x_1 y_1 + \ldots + x_n y_n = 0,$$ where $x = (x_1, \ldots, x_n)$, $y = (y_1, \ldots, y_n)$. A vector $x$ is of unit length if $x \cdot x = 1$. A set is called orthonormal if every vector has unit length and any two different vectors are orthogonal. We also know the following:

Theorem. If a set is orthonormal, then it is linearly independent.

So we can just compute the dot products between all the vectors in your examples, and check whether they equal 0 when dotting two different vectors, and equals 1 when dotting a vector with itself. Then, using the theorem, you'll automatically know whether it's a orthonormal basis or not.

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Yah i just figured this out at the same time, thank you for the reply. – Charlie Yabben Oct 11 '12 at 23:08
but how can you ensure that it spans the vector space?? – TheJoker Oct 11 '12 at 23:18
@TheJoker, well if you know the set if orthonormal, then it's linearly independent, so you just count how many elements there are. – Christopher A. Wong Oct 11 '12 at 23:31

Proving something to be orthonormal basis is simple. First prove the set of vector to be a basis (set is linear independent and set spans the vector space, here $\mathbb R^2$). Then check for orthogonal prperty, i.e dot product is $0$ and vectors are of unit length).
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