# Gram-Schmidt in a linearly independent set.

We know that if $$X$$ is an inner product space with a base $$B$$, then we can construct (by Gram-Schmidt) an orthogonal base $$B'$$ for $$B$$.

But, if we only have a linearly independent subset of $$X$$, lets say it $$A=\{x_n\}_{n \in \mathbb{N}}$$ then can we construct a orthogonal set $$A'$$ from $$A$$ by doing the same ( Gram-Schmidt) process ?.

And if so, will they have the same $$Span$$?

• Yes, it will produce an orthormal basis of the subspace spanned by $X$. Jan 20 at 15:40

A proof that this works is by induction. Suppose that $$\{x_k\}_{1 \leqq k \leqq n}$$ is a linearly independent set of vectors, then (note: we will normalise the vectors at the end because this substantially simplifies things)
• for the base case clearly setting $$v_1 = x_1$$ gives us a linearly independent set $$\{v_1\}$$ whose span is the same as that of $$\{x_1\}$$ (not surprising; they are the same set).
• then assume we have a set $$S = \{v_{k}\}_{1 \leqq k \leqq n}$$ of orthogonal vectors. Then we can define $$v_{k+1}$$ as $$v_{n+1} := x_{k+1} - \sum_{1 \leqq i \leqq n} \frac{\left(x_{n+1}, v_{i}\right)}{||v_i||^2} v_i$$ then we can consider for arbitrary $$j \ne n+1$$ \begin{align} (v_{n+1}, v_{j}) &= \left(x_{k+1} - \sum_{1 \leqq i \leqq n} \frac{\left(x_{n+1}, v_{i}\right)}{||v_i||^2} v_i, v_j \right) \\ &= (x_{k + 1}, v_j) - \sum_{1 \leqq i \leqq n} \left[ \left( \frac{\left(x_{n+1}, v_{i}\right)}{||v_i||^2} \right) (v_i, v_j) \right] \\ &= (x_{k+1}, v_j) - \frac{(x_{n+1}, v_j)}{(v_j, v_j)} (v_j, v_j) \\ &= (x_{k+1}, v_j) - (x_{n+1}, v_j) \\ &= 0 \end{align} Note that we applied the inductive hypothesis in establishing that $$\sum_{1 \leqq i \leqq n} \left[ \left( \frac{\left(x_{n+1}, v_{i}\right)}{||v_i||^2} \right) (v_i, v_j) \right] = \frac{(x_{n+1}, v_j)}{(v_j, v_j)} (v_j, v_j)$$ as by the inductive hypothesis for all $$1 \leqq k \leqq n$$ as $$S$$ is an orthongonal set $$(v_k, v_j) = 0 \iff k = j$$ which means we can discard all the terms except when $$k = j$$. Now we have shown that adding the new $$v_{k+1}$$ to our set means that it still consists of orthogonal vectors. Therefore it is linearly independent, and thus it has dimension $$n+1$$, but it is also a subset of $$\{x_{k}\}_{1 \leqq k \leqq n}$$ so it must be that the span of the two sets is equal.
• can this work the set $A$ is infinite but countable ? Jan 20 at 15:49
Suppose Gram-Schmidt works on a basis of an inner product space $$(X, I)$$, and we have a linearly independent set $$A$$. Then $$A$$ is a basis for its span $$S$$. Now restrict our inner product to $$S$$ to get $$I|_S$$. We then apply Gram-Schmidt to the basis $$A$$ of the inner product space $$(S, I|_S)$$ to get an orthogonal basis $$A'$$. But of course $$A'$$ is orthogonal in $$(X, I)$$, spans $$S$$, and is linearly independent in $$X$$; we are done.