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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,$=u^TAv$,then $v_k^TAv_i=$ ,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

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    You have a typo in your definition of the Gram-Schmidt process. It should be $\sum_{i = 0}^{k - 1} \frac{v_i^T A u_k}{v_i^T A v_i} v_i$. – 2017-02-11

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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.