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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 $Span$?

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    $\begingroup$ Yes, it will produce an orthormal basis of the subspace spanned by $X$. $\endgroup$
    – StiftungWarentest
    Jan 20 at 15:40

2 Answers 2

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Yes. Note that the Gram-Schmidt process requires only a linearly independent set (and all bases are linearly independent).

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.
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  • $\begingroup$ can this work the set $A$ is infinite but countable ? $\endgroup$
    – lebong66
    Jan 20 at 15:49
  • $\begingroup$ Yes this works for countably infinite sets - we can just carry out Gram-Schmidt indefinitely. $\endgroup$
    – Teymour
    Jan 20 at 16:16
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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.

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