# Finding an orthogonal basis for an inner product space $\mathbf{P}_2$

Look at the inner product in $\mathbf{P}_2$ given by $\langle p,q \rangle = p(0)q(0) + p(\frac{1}{2})q(\frac{1}{2}) + p(1)q(1)$.

Find an orthogonal basis $\mathbf{C}$ for $\mathbf{P}_2$.

Now, I know I can use the standard basis for $\mathbf{P}_2$, i.e. $B=\{1,t,t^2 \}$ and apply Gram-Schmidt. So: $\mathbf{v}_1 = \mathbf{x}_1 = 1$

$\mathbf{v}_2 = \mathbf{x}_2 - proj_{\mathbf{v}_1} \mathbf{x}_2 = \mathbf{x}_2 - \frac{\langle \mathbf{x}_2, \mathbf{v}_1 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle}\mathbf{v}_1$ and so on.

My question is incredibly trivial: I just can't figure out how to calculate the inner products in the equation above using the one in the problem.

From a solution of the problem I know that $\langle \mathbf{v}_1, \mathbf{v}_1 \rangle = 1$ and $\langle \mathbf{x}_2, \mathbf{v}_1 \rangle = \langle t,1\rangle= 3/2$. Could someone please show me the calculations, or how to insert these into the equation in the problem?

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$\langle \mathbf{x}_2, \mathbf{v}_1\rangle = \mathbf{x}_2(0) \; \mathbf{v}_1(0) + \mathbf{x}_2(1/2) \;\mathbf{v}_1(1/2) + \mathbf{x}_2(1) \; \mathbf{v}_1(1)= 0 \cdot 1 + \frac{1}{2} \cdot 1 + 1 \cdot 1 = \frac{3}{2}$.