TITLE: Question related to Gram schmidt process.
QUESTION [0 upvotes]: We have a set of linearly independent vector $u_0,u_1,\cdots,u_{n-1}$ and $A$ is a positive definite matrix.Gram schmidt process is given as;
$$v_0=u_0$$$$v_k=u_k-\sum_{i=0}^{k-1}\frac{v_i^TAu_k}{v_i^TAv_i}v_i\,,k=1,2,\cdots n-1.$$We have to prove that $v_k^TAv_i=0$ for $i=0,1,2,\cdots k-1.$
MY TRY:We know that $A$ is positive definite.so,$<u,v>=u^TAv$,then $v_k^TAv_i=<v_k,vi>$ ,where $v_k$ is orthogonal to  $v_i$ for $i=0,1,2,\cdots k-1.$,hence $$v_k^TAv_i=0\text{ for}\, i=0,1,2,\cdots k-1.$$.Is it correct $?$.Thank you

REPLY [1 votes]: Every symmetric positive definite $n\times n$ Matrix $A$ over a field $\mathbb{F}$ allows you to define a new scalar product, via
$$
\langle x, y \rangle_A := x^T A y. \tag{$\ast$}
$$
Since $A$ is symmetric we have that
$$
\langle x, y \rangle_A = x^T A y = (x^T A y)^T = y^T A^T x = y^T A x = \langle y, x \rangle_A.
$$
This is a necessary condition for $(\ast)$ to define a scalar product.
Note that for $A = E_n$, where $E_n$ is the identity matrix you get the usual scalar product, i.e.
$$
\langle x, y \rangle_{E_n} = x^T E_n y = x^T y = \langle x, y \rangle.
$$
With this notion your Gram-Schmidt process looks like this:
You start with a set of linearly independent vectors $u_0,u_1,\cdots,u_{n-1}$ and a symmetric positive definite matrix $A$. Then,
$$v_0 :=u_0$$
and 
$$
v_k
:=u_k-\sum_{i=0}^{k-1}\frac{v_i^TAu_k}{v_i^TAv_i} v_i 
= u_k - \sum_{i = 0}^{k - 1} \frac{\langle v_i, u_k \rangle_A}{\langle v_i, v_i \rangle_A} v_i
$$
for all $k=1,2,\cdots n-1$.
The rest of the proof goes as for the usual Gram-Schmidt process:
We prove the statement via induction over $k$.
Let $k = 1$. Then
$$
\langle v_1, v_0 \rangle_A = \langle u_1, v_0 \rangle_A - \frac{\langle v_0, u_1 \rangle_A}{\langle v_0, v_0 \rangle_A} \langle v_0, v_0 \rangle_A = \langle u_1, u_0 \rangle_A - \langle u_0, u_1 \rangle_A = 0.
$$
Now assume the statement to hold for all natural numbers strictly less than $k$.
Taking the scalar product with $v_j$ for $j =  1, \dots, k-1$ gives
$$
\langle v_k, v_j \rangle_A = \langle u_k, v_i \rangle_A - \sum_{i = 0}^{k - 1} \frac{\langle v_i, u_k \rangle_A}{\langle v_i, v_i \rangle_A} \langle v_i, v_j \rangle_A = \langle u_k, v_i \rangle_A - \langle v_i, u_k \rangle_A = 0.
$$
Note that in the last equality we used the symmetry of our scalar product.
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