5.11 Gram-Schmidt

Let the input $V_1,V_2,\cdots ,V_n$ be a set of $n$ linearly independent vectors. We want to use Grame-Schmidt to obtain set of $n$ orthonormal vectors, called $v_1,v_2,\cdots ,v_n$. The notation $⟨V_1,V_2⟩$ is used to mean the inner product between any two vectors. The first vector $v_1$ is easy to find777$$\begin{array}{}\text{(1)}& v_1=\frac{V_1}{\sqrt{⟨V_1,V_1⟩}}\end{array}$$ The second $$v_2^{\prime }=V_2-v_1⟨v_1,V_2⟩$$ Where $v_2^{\prime }$ means $v_2$ but not yet normalized. Before we normalize $v_2^{\prime }$, we need to show that $⟨v_1,v_2^{\prime }⟩=0$. But $$⟨v_1,v_2^{\prime }⟩=⟨v_1,(V_2-v_1⟨v_1,V_2⟩)⟩$$ Expanding the above gives$$⟨v_1,v_2^{\prime }⟩=⟨v_1,V_2⟩-⟨v_1,v_1⟨v_1,V_2⟩⟩$$ But $⟨v_1,V_2⟩$ above is just a number. We can take it out of the second inner product term above. The above becomes$$⟨v_1,v_2^{\prime }⟩=⟨v_1,V_2⟩-⟨v_1,V_2⟩⟨v_1,v_1⟩$$ But $⟨v_1,v_1⟩=1$, since $v_1$ is normalized vector. The above becomes$$\begin{array}{rl}⟨v_1,v_2^{\prime }⟩& =⟨v_1,V_2⟩-⟨v_1,V_2⟩\\ & =0\end{array}$$

Now we normalized $v_2^{\prime }$ $$v_2=\frac{v_2^{\prime }}{\sqrt{⟨v_2^{\prime },v_2^{\prime }⟩}}$$ Now we find $v_3$$$\begin{array}{rl}{v}_3^{\prime }& =V_3-(v_1⟨v_1,V_3⟩+v_2⟨v_2,V_3⟩)\\ v_3& =\frac{v_3^{\prime }}{\sqrt{⟨v_3^{\prime },v_3^{\prime }⟩}}\end{array}$$

And so on.