Is the following Proof Correct? For the facility of the reader the formal statement of the Gram-Schimdt Procedure is provided below.

Proposition. If $\{w_1,w_2,\dots, w_n\}$ is an orthogonal set of vectors, then the vectors $v_1,v_2,\dots, v_n$ derived from the Gram-Schmidt process satisfy $v_i=w_i,\forall i = 1,2,\dots,n$.

Proof. Define $I_n = \{1,2,\dots,n\}$ and $S_k' = \{v_1,v_2,\dots,v_k\},\forall k\in I_n$. We proceed by recourse to Induction. For $i=1$ it is evident by the Gram-Schmidt procedure that $v_1 = w_1$. Now assume for an arbitrary $k\in I_n\backslash\{n\}$ that the set $S_k'$ satisfies the proposition that $v_i=w_i,\forall i=1,2,\dots,k$.

Then by using the the inductive hypothesis in tandem with the procedure to determine $v_{k+1}$ yields $$v_{k+1} = w_{k+1}-\sum_{i=1}^{k+1-1}\frac{\langle w_{k+1},v_i\rangle}{||v_i||^2}v_i = w_{k+1}-\sum_{i=1}^{k}\frac{\langle w_{k+1},w_i\rangle}{||w_i||^2}w_i = w_{k+1}$$ where the last equality follow from our original hypothesis that $\{w_1,w_2,\dots, w_n\}$ is orthogonal completing the induction.


Theorem. Let $V$ be an inner product space and $S = \{w_1,w_2,\dots.w_n\}$ be a linearly independent subset of $V$. Define $S' = \{v_1,v_2,\dots,v_n\}$, where $v_1 = w_1$ and $$v_k = w_k-\sum_{j=1}^{k-1}\frac{\langle w_k,v_j\rangle}{||v_j||^2}v_j\text{ for }2\leq k\leq n.$$ The $S'$ is an orthogonal set of nonzero vectors such that $\operatorname{span}(S') = \operatorname{span}(S)$.


Yes, other than the fact that there's a typo. What you meant to write was:\begin{align}v_{k+1}&= w_{k+1}-\sum_{i=1}^{k+1-1}\frac{\langle w_{k+1},v_i\rangle}{||v_i||^2}v_i\\&=w_{k+1}-\sum_{i=1}^{k}\frac{\langle w_{k+1},w_i\rangle}{||w_i||^2}w_i\\&=w_{k+1}.\end{align}

  • $\begingroup$ Thankyou Sir. Your counsel is appreciated as always. $\endgroup$ – Atif Farooq Jun 11 '18 at 17:15
  • $\begingroup$ @AtifFarooq I'm glad I could help. $\endgroup$ – José Carlos Santos Jun 11 '18 at 17:16

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