Let $$V = \left\{ f : [0,1] \to \mathbb R\ :\ f \text{ is a polynomial of degree} \leq n \right\}$$ Let $f_j(x)=x^j$ for $0 \le j \le n$ and let $A$ be the $(n+1) \times (n+1)$ matrix given by
$$a_{ij} = \int_{0}^{1} f_i(x) \, f_j(x) \, \mathrm d x$$
Then which of the following is/are true?
$1.$ $\ \dim V=n$.
$2.$ $\ \dim V>n$.
$3.$ $\ A$ is non-negative definite, i.e., for all $v \in \Bbb R^n$, $\left < Av,v \right > \ge 0$.
$4.$ $\ \det A>0$.
Clearly $\{1,x,x^2,\cdots, x^n \}$ is a basis for $V$. This shows that $\dim V=n+1>n$. Hence $(1)$ is incorrect but $(2)$ is a correct option. For determining whether or not $(3)$ and $(4)$ are correct options I first find the matrix $A$ which is
$$A=\begin{bmatrix} 1 & {\frac {1} {2}} & {\cdots} & {\frac {1} {n+1}} \\ {\frac {1} {2}} & {\frac {1} {3}} & {\cdots} & {\frac {1} {n+2}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {\frac {1} {n+1}} & {\frac {1} {n+2}} & {\cdots} & {\frac {1} {2n+1}} \end{bmatrix}.$$
Here I got stuck. How do I find $\det A$? Also after evaluating $ \langle Av,v \rangle$ I can't figure out whether or not $\langle Av,v \rangle \ge 0$ for all $v \in \Bbb R^n$. Please help me in this regard. Thank you very much.
Let $v=(v_0,\ldots,v_n)^T$ and $f=\sum_{i=0}^n v_if_i.$ Then $$ \langle Av,v \rangle = \sum_{i=0}^n\sum_{j=0}^n v_iv_j\int_0^1f_i(x)f_j(x)dx \\ = \int_0^1\left(\sum_{i=0}^nv_if_i(x)\right)\left(\sum_{j=0}^nv_jf_j(x)\right)dx =\int_0^1 \left(f(x)\right)^2dx $$ Therefore $\langle Av,v\rangle > 0$ if $v\neq 0.$ The matrix is positive definite. (3) and (4) are correct.