Answer
$B'=\{((0,1),(\frac{4}{5},(1,0)\}$ is orthonormal basis for $R^2$.
Work Step by Step
Let $B=\{ v_1=(0,1),v_2=(2,5) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(0,1)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(2,5)-5(0,1)=(2,0).$$
Normalizing $w_1$and $w_2$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=(0,1)$$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{1}{2}(2,0) =(1,0).$$
Hence, $B'=\{((0,1),(\frac{4}{5},(1,0)\}$ is orthonormal basis for $R^2$.
