Let $V$ be the vector space of polynomials in the variable $t$ of degree at most $2$ over $\mathbb{R}$. An inner product on $V$ is defined by $<f,g>=\int_{0}^{1} f(t)g(t)dt$ for ${f,g} \in V$. Let $W=span\{1-t^{2}, 1+t^{2}\}$ and ${W^{⊥}}$ be the orthogonal complement of $W$ in $V$. Which of the following conditions is satisfied for all $h \in{W^{⊥}}$ ?

  1. $h$ is an even function.
  2. $h$ is an odd function.
  3. $h(t)=0$ has a real solution.
  4. $h(0)=0$.

Orthogonal complement $W^{⊥}= \{x\in V | <x,y>=0 \, y\in W\}$. Please give some hint.

  • $\begingroup$ The downvote (wasn't me) is probably because you didn't tell us what you tried or found. You should add that if possible $\endgroup$
    – Aldoggen
    Nov 12, 2019 at 15:02
  • $\begingroup$ I know the orthogonal complement definition .But how I approach to this..... $\endgroup$
    – Sachin
    Nov 12, 2019 at 15:09
  • $\begingroup$ Try taking an arbitrary $h\in V$ and taking the inner product with the basis vectors of $W$. What does that tell you? $\endgroup$
    – Aldoggen
    Nov 12, 2019 at 15:12
  • 1
    $\begingroup$ Then for finding $h(t)$ these should be hold $ \int_{0}^{1} (1-t^{2})h(t)dt=0, \int_{0}^{1} (1+t^{2})h(t)dt=0$ $\endgroup$
    – Sachin
    Nov 12, 2019 at 15:19
  • $\begingroup$ If you get stuck again, since no one answered yet you can probably expand your question with what you've found and where you're stuck. If you solve the problem, you can make an answer yourself. $\endgroup$
    – Aldoggen
    Nov 12, 2019 at 15:24

1 Answer 1


You should consider an $h\in V$, and take the inner product with the basis vectors of $W$. The inner products $$\int_0^1(1\pm t^2)(at^2+bt+c)\, dt$$ should evaluate to zero. They evaluate to $$\frac{8a}{15}+\frac{3b}{4}+\frac{3c}{2} = 0$$ and $$\frac{2a}{15}+\frac{b}{4}+\frac{c}{2} = 0$$ for the plus and minus respectively. Subtracting three times the second one from the first yields $2a/15=0$. Subtracting four times the second one from the first yields $-b/4-c/2=0$. We get $W^\perp=\mathrm{span}\{-2t+1\}$. That means only the third condition is always satisfied.

  • $\begingroup$ I think that my calculations are off. The method still works, but the conclusions might be wrong. $\endgroup$
    – Aldoggen
    Nov 14, 2019 at 13:07

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