Prove $\sum_{i=1}^{n}v_i^t \cdot v_i = I$ for orthonormal base Let 
$$
\{v_1,..,v_n\} 
$$
Be an orthonormal base in $R^n$ with the standard inner product. 
I need to prove that: 
$$
\sum_{i=1}^{n}v_i^t \cdot v_i = I
$$
Where $v_i$ is a row vector. 
What i tried:
I tried to look at an example but still - i dont feel its getting me somewhere. 
Lets take $R^2$
The base will be:
$$
\{v_1,v_2\}
$$
Let $v_1 = [a_1,a_2], v_2 = [b_1,b_2]$
As an orthonormal base we know that: 
$$
||v_1|| = ||v_2|| = 1
$$
And:
$$
<v_1,v_2> = 0
$$
Therefore: 
$$
||v_1||^2 = a_1^2 + a_2^2 = 1, ||v_2||^2 = b_1^2 + b_2^2 = 1
$$
$$
<v_1,v_2> = a_1b_1 + a_2b_2 = 0
$$
$$
\sum_{i = 1}^{n = 2}v_i^t \cdot v_i = 
\begin{bmatrix} a_1^2&a_2a_2 \\ a_2a_1&a_2^2\end{bmatrix} + \begin{bmatrix} b_1^2&b_2b_2 \\ b_2b_1&b_2^2\end{bmatrix}
$$
But i dont see how i get here to $I_2$? Am i even in the right direction? what am i missing? 
I would prefer a hint than a full answer - as those are my homework. 
And thanks for the help. 

Hint: Rewrite $\sum_{i=1}^n v_i^Tv_i$ as a matrix product.
