# Gram-Schmidt process vectors

• Apr 2nd 2008, 02:10 PM
Gram-Schmidt process vectors
Prove that if $\displaystyle \{ w_{1}, w_{2}, ... , w_{n} \}$ is an orthogonal set of nonzero vectors, then the vectors $\displaystyle v_{1}, v_{2}, . . . , v_{n}$ derived from the Gram-Schmidt process satisfy $\displaystyle v_{i} = w_{i} \ \ \ \forall i$

my proof so far:

I intend to use induction.

Now, $\displaystyle v_{1} = w_{1}$ is travial.

Suppose that n = k is true, then I have $\displaystyle v_{k} = w{k-1} - \sum ^{k-2}_{j=1} \frac {<w_{k-1},v_{j}>}{ || v_{j} || ^2 } v_{j} = w_{k}$

Now, how would I use that information as well as the fact that the vecters are orthogonal to get k+1 is true?

thanks
• Apr 3rd 2008, 12:08 AM
Opalg
Quote:

Prove that if $\displaystyle \{ w_{1}, w_{2}, ... , w_{n} \}$ is an orthogonal set of nonzero vectors, then the vectors $\displaystyle v_{1}, v_{2}, . . . , v_{n}$ derived from the Gram-Schmidt process satisfy $\displaystyle v_{i} = w_{i} \ \ \ \forall i$
Now, $\displaystyle v_{1} = w_{1}$ is travial.
Suppose that n = k is true, then I have $\displaystyle v_{k} = w{k-1} - \sum ^{k-2}_{j=1} \frac {<w_{k-1},v_{j}>}{ || v_{j} || ^2 } v_{j} = w_{k}$
This is easy if you use strong induction (in other words, assume that the inductive hypothesis holds for all n≤k, not just for n=k). The formula for v_{k+1} is $\displaystyle v_{k+1} = w_{k+1} - \sum ^{k}_{j=1} \frac {<w_{k+1},v_{j}>}{ || v_{j} || ^2 } v_{j}$. But if $\displaystyle v_j = w_j$ for 1≤j≤k then each term in that sum will vanish, because the w's are orthogonal to each other.