The vectors $\begin{bmatrix}a \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix}0 \\ b \\ 0\end{bmatrix}$, and $\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}$ are orthogonal under the dot product.

The vectors $1$, $x$, and $x^2 - 1/3$ are orthogonal under the standard inner product for polynomials, $\int_{-1}^1 p(x)q(x)\ dx$.

While both collections meet their respective definitions of orthogonality, there seems to be some sense in which the former is more orthogonal than the latter.

No linear combination of $1$ and $x$ produces $x^2 - 1/3$, but the combination $-1/3$ of $1$ and $1$ of $x$, yielding $x - 1/3$, appears to make some progress toward the task. This isn't to claim that $x^2 - 1/3$ and $x - 1/3$ evaluate to similar values, but merely that if it were possible to reach $x^2 - 1/3$ by taking linear combinations of $1$ and $x$, this attempt would be a better start than, say, taking $-500$ of $1$ and $\pi$ of $x$. of By contrast, in the case of $\begin{bmatrix}a \\ 0 \\ 0\end{bmatrix}$, $\begin{bmatrix}0 \\ b \\ 0\end{bmatrix}$, and $\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}$, it feels like no progress can be made no matter what combination is chosen.

Is there some sense in which the former, $\mathbb{R}^3$ vectors are orthogonal to a stronger degree than the latter, polynomial vectors?

  • $\begingroup$ How is this a progress? Could you quantify that? Speaking in terms of vector spaces, there is no difference between the two cases you gave. $\endgroup$ – Dirk Mar 21 at 15:47
  • $\begingroup$ I suppose it's in part the fact that polynomial degrees are ordered. Fourth and fifth degree polynomials are clearly more intrinsically similar than second and fifth degree polynomials. Additionally, the fact that we can construct at least one term of $x^2 - 1/3$ with the polynomial $1$ seems to constitute progress, compared to the $R^3$ case. $\endgroup$ – user10478 Mar 21 at 16:10
  • $\begingroup$ I suspect that whatever difference you’re sensing is a result of the relationship between the inner product and the basis that you’re using. If you take the three polynomials as your basis instead of $\{1,x,x^2\}$, then the situation looks exactly like the first one. $\endgroup$ – amd Mar 21 at 18:28
  • $\begingroup$ BTW, I might quibble about your characterization of that inner product as “the standard” one. I see the integral taken from $0$ to $1$ instead at least as often. $\endgroup$ – amd Mar 21 at 18:29

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