What is the difference between orthogonal and orthonormal in terms of vectors and vector space? I am beginner to linear algebra. I want to know detailed explanation of what is the difference between these two and geometrically how these two are interpreted?
 A: You can think of orthogonality as vectors being perpendicular in a general vector space. And for orthonormality what we ask is that the vectors should be of length one. So vectors being orthogonal puts a restriction on the angle between the vectors whereas vectors being orthonormal puts restriction on both the angle between them as well as the length of those vectors. These properties are captured by the inner product on the vector space which occurs in the definition.
For example, in $\mathbb{R}^2$ the vectors $(0,2)$ and $ (1,0)$ are orthogonal but not orthonormal because $(0,2)$ has length $2.$
A: Two vectors are orthogonal if their inner product is zero.  In other words $\langle u,v\rangle =0$.  They are orthonormal if they are orthogonal, and additionally each vector has norm $1$.  In other words $\langle u,v \rangle =0$ and $\langle u,u\rangle = \langle v,v\rangle =1$.
Example
For vectors in $\mathbb{R}^3$ let
$$
u \;\; =\;\; \left[ \begin{array}{c}
1\\
2\\
0\\
\end{array} \right ] \hspace{2pc} v \;\; =\;\; \left [ \begin{array}{c}
0\\
0\\
3\\
\end{array} \right ].
$$
The vectors $u$ and $v$ are orthogonal since
$$
\langle u, v\rangle \;\; =\;\; 1\cdot 0 + 2\cdot 0 + 0\cdot 3 \;\; =\;\; 0
$$
but they are not orthonormal since $||u|| = \sqrt{\langle u,u\rangle } = \sqrt{1 + 4} = \sqrt{5}$ and $||v|| = \sqrt{\langle v,v\rangle } = \sqrt{3^2} = 3$.  If we define new vectors $\hat{u} = \frac{u}{||u||}$ and $\hat{v} = \frac{v}{||v||}$ then $\hat{u}$ and $\hat{v}$ are orthonormal since they each now have norm $1$, and orthogonality is preserved since $\langle \hat{u}, \hat{v}\rangle = \frac{\langle u,v\rangle }{||u||\cdot ||v||} = 0$.
