How to find the orthogonal projection of the given vector on the given subspace $W$ of the inner product space $V$. $V=P(R)$ with the inner product $\langle f(x),g(x) \rangle$=$\int_0^1 f(t)g(t )dt$, $h(x)=4+3x-2x^2$ and $W=P_1(R)={\{1,x}\}$.
I don't know how to  do this the question. All I know is that it has something do with the Gram-Schmidt process.
Please provide full details ,if possible as I don't know much about about orthogonalization projection. 
 A: The inner product structure of your vector space $V$ is $$\langle f | g 
\rangle = \int_0^1 f(x)g(x) dx$$
To project a vector
$$h(x) = 4 + 3x - 2x^2$$
on the subspace $W$ of the vector space $V$, you just add the projections of $h$ on each of the basis vectors of the subspace.
In this case, since 
$$W = P_1 = \left\{ 1, x \right\}$$
and the vector we wish to project is $h$, we need to find
$$
w = 1 \times\langle h |1\rangle + x \times\langle h|x\rangle 
$$ 
Where $w$ is the projection of $h$ in $W$
Let's now compute $w$
$$
w = 1 \times\langle h |1\rangle + x \times\langle h|x\rangle \\
= 1 \times \int_0^1h \cdot 1dx + x \times \int_0^1 h \cdot x dx \\
= \int_0^1 (4 + 3x - 2x^2)dx + x\int_0^1 (4 + 3x - 2x^2)x dx \\
= \int_0^1 (4 + 3x - 2x^2)dx + x\int_0^1 (4x + 3x^2 - 2x^3) dx \\
= 4x + \frac{3x^2}{2} - \frac{2x^3}{3} \bigg \vert_0^1 +
x \left( \frac{4x^2}{2} + \frac{3x^3}{3} - \frac{2x^4}{4} \bigg \vert_0^1 \right) \\
= \left(4 + \frac{3}{2} - \frac{2}{3} \right) + x \left(\frac{4}{2} - \frac{3}{3} - \frac{2}{4}\right) \\
= \frac{12 + 9 - 4}{6} + x \left(2 - 1 - \frac{1}{2} \right )\\
= \frac{17}{6} + \frac{x}{2}
$$ 
Hence, the projection of $h$ on $W$, or
$$
w = \langle h | W \rangle = \frac{17}{6} + \frac{x}{2}
$$
A: The orthogonal projection of $h$ onto a subspace $W$ is the unique $w\in W$ such that $(h-w)\perp W$. It's the same orthogonal projection you learned in Calculus. So, you want to find scalars $\alpha,\beta$ such that $w(x)=\alpha +\beta x$ satisfies
$$
          \int_{0}^{1}(4+3x-2x^2-\alpha -\beta x)dx = 0 \\
          \int_{0}^{1}(4+3x-2x^2-\alpha-\beta x)xdx = 0.
$$
That's two equations in two unknowns $\alpha$,$\beta$:
$$
           4+\frac{3}{2}-\frac{2}{3}-\alpha-\frac{1}{2}\beta = 0 \\
           \frac{4}{2}+\frac{3}{3}-\frac{2}{4}-\frac{\alpha}{2}-\frac{\beta}{3} = 0.
$$
You could apply Gram-Schdmit to $\{ 1,x \}$ in order to find an orthonormal basis $\{ w_1,w_2 \}$ of $W$; then the orthogonal projection is
$$
           w=(h,w_1)w_1+(h,w_2)w_2
$$
because $h-(h,w_1)w_1-(h,w_2)w_2$ is orthogonal to $w_1$ and to $w_2$. Doesn't seem worth the effort in this case where the equivalent system is a 2x2 matrix problem.
A: Here’s a method that’s admittedly overkill for this simple problem.  
First, express everything as coordinates relative to the standard basis $\{1,x,x^2\}$. The inner product can then be expressed as $\langle v,w\rangle = w^TGv$, where $G$ is the Gram matrix with elements $$G_{ij}=\langle e_i,e_j\rangle=\int_0^1x^{i+j-2}\,dx = \frac1{i+j-1}.$$ Abusing the notation somewhat, let $W$ be the matrix that has the basis vectors of the subspace $W$ as its columns. Then the orthogonal projection onto this subspace is given by $$\pi_W = W(W^TGW)^{-1}W^TG.$$ This expression for the projection onto $W$ is valid for any basis for $W$, not just for an orthonormal one.  
In this case, we compute $$\pi_W = \pmatrix{1&0&-\frac16\\0&1&1\\0&0&0}$$ and so $\pi_w h = \pi_W(4,3,-2)^T = \left(\frac{13}3,1,0\right)^T = x+\frac{13}3$.
