# Prove the uniqueness of orthogonal function without considering sign

Suppose that the group $$p_0(x), p_1(x),... p_N(x)$$ are orthonormal in the interval $$[-1,1]$$, which means with 2 arbitrary functions $$p_i(x), > p_j(x)$$, the conditions below is satisfied.

$$\int_{-1}^1p_i(x), p_j(x)dx = \begin{cases} 1,& (i = j)\\ 0,& (i \not= j) \end{cases}$$

In this case, $$p_i(x)$$ is the polynomial with degree $$i$$.

Prove the uniqueness of $$p_N(x)$$ without considering sign by the procedure as below:

1. Generally the N-degree polynomial $$f_N(x)$$ can be written as below: $$f_N(x) = \sum_{k=0}^N c_kp_k(x)$$

Demonstrate the coefficient $$c_k$$ $$(k = 0,1,2...,N)$$by $$p_0(x), > p_1(x),... p_N(x)$$ and $$f_N(x)$$ .

1. Let $$p_N(x)^*$$ is an N degree polynomial other than $$p_N(x)$$ so that $$p_0(x), p_1(x),... p_N(x)^*$$ are also orthonormal. Let $$f_N(x) > = p_N(x)^*$$, prove that $$p_N(x) = -p_N(x)^*$$

Here is what I am thinking:

First, demonstrate $$f_N(x)$$ in matrix form as below:

$$f_N(x)$$ = $$\pmatrix{p_0(x)&p_1(x)&...&p_N(x)} \pmatrix{c_0\\c_1\\...\\c_N}$$.

Then multiply both sides with $$\pmatrix{p_0(x)\\p_1(x)\\...\\p_N(x)}$$, so we have: $$\pmatrix{p_0(x)f_N(x)\\p_1(x)f_N(x)\\...\\p_N(x)f_N(x)} =\pmatrix{p_0(x)p_0(x)&p_0(x)p_1(x)&...&p_0(x)p_N(x)\\p_1(x)p_0(x)&p_1(x)p_1(x)&...&p_0(x)p_1(x)\\...........&...........&...&...........\\p_N(x)p_0(x)&p_N(x)p_1(x)&...&p_N(x)p_N(x)}\pmatrix{c_0\\c_1\\...\\c_N}$$

Integral both side from $${-1}$$ to $$1$$, and use the properties of orthonormal functions, we have:

$$\pmatrix{\int_{-1}^1p_0(x)f_N(x)dx\\\int_{-1}^1p_1(x)f_N(x)dx\\...\\\int_{-1}^1p_N(x)f_N(x)dx} =\pmatrix{1&0&...&0\\0&1&...&0\\...&...&...&...\\0&0&...&1}\pmatrix{c_0\\c_1\\...\\c_N} = \pmatrix{c_0\\c_1\\...\\c_N}$$

So number 1 is satisfied (we already demonstrated the coefficients by $$f_N(x)$$ and $$p_0(x),p_1(x),...p_N(x)$$

Then for 2, when $$f_N(x)=p_N(x)^*$$, we have:

$$\pmatrix{c_0\\c_1\\...\\c_N}= \pmatrix{\int_{-1}^1p_0(x)p_N(x)^*dx\\\int_{-1}^1p_1(x)p_N(x)^*dx\\...\\\int_{-1}^1p_N(x)p_N(x)^*dx} = \pmatrix{1\\1\\...\\\int_{-1}^1p_N(x)p_N(x)^*dx}$$

Then: $$f_N(x)$$ =$$p_N(x)^*$$=$$\pmatrix{p_0(x)&p_1(x)&...&p_N(x)}\pmatrix{1\\1\\...\\\int_{-1}^1p_N(x)p_N(x)^*dx}$$ =$$p_N(x)*\int_{-1}^1p_N(x)p_N(x)^*dx$$

So $$p_N(x)^*$$=$$p_N(x)*\int_{-1}^1p_N(x)p_N(x)^*dx$$

Then I stuck here. Can we show that $$\int_{-1}^1p_N(x)p_N(x)^*dx = -1$$ in this case so that $$p_N(x)^* = - p_N(x)$$?