Prove that exists an inner product such that $\{1+x,x+x^2,x^2+x^3,x^3+x^4,x^4-1\}$ is orthogonal

Question

I've been asked to prove or disprove that exists an inner product in $\mathbb{R}_5[x]$ such that

$A = \{1+x,x+x^2,x^2+x^3,x^3+x^4,x^4-1\}$

is orthogonal.

How does one approach this?

Answer 1

If these vectors are linearly independent, then they can be extended to a basis*, which then determines a unique inner product for which the basis is orthonormal (the standard inner product in terms of coordinates with respect to this basis), and we will be done. (Since orthogonal families of nonzero vectors in an inner product space are always linearly independent, that condition is clearly necessary, so it is if and only if.)

So it remains to see whether these vectors are linearly independent. Since all $5$ are contained in the $5$ dimensional (sub-)space of polynomials of degree less than $5$, and moreover they all have zero evaluation at $x=-1$ (which is a nontrivial linear condition on that space), they lie in a $4$-dimensional subspace, so they must be linearly dependent. Therefore they cannot be orthogonal for any inner product.

Since finally it is the implication "linear dependence forbids being mutually orthogonal" for sets of nonzero vectors that solves this question, let me recall the argument proving that. One argument is by contradiction: if some vector $v$ equals a linear combination of the other vectors supposed to be orthogonal to $v$, then $v$ is orthogonal to itself, which is impossible for a nonzero vector for a (positive definite) inner product. Another argument is by contrapositive: if a family $(v_i)_{i\in I}$ of nonzero vectors is orthogonal for an inner product, then in any linear combination $v=\sum_{i\in I}c_iv_i$ one can recover the coefficients from $c_i=(v_i\mid v)/(v_i\mid v_i)$, and in particular $v=0$ implies $c_i=0$ for all $i$, proving linear independence of the family.

*It depends on what $\Bbb R_5[x]$ means; normally the notation means polynomials over $\Bbb R_5$, but since that does not designate any ring as far as I can tell, I'll assume it means polynomials in $\Bbb R[x]$ whose degree bears some relation to the number $5$. If the relation is "less than" then these vectors are already a the right number for a basis; if it is "less than or equal", then an extension would have been needed if they were linearly independent. These two relations seem to be the most plausible candidates for this unspecified notational convention.

Answer 2

Hint: $v_5 = v_4 - v_3 + v_2 - v_1 \implies$ $\langle v_5,v_j \rangle = \pm 1$ for $j=1,\dots4$.