Let $H$ be a Hilbert space. Suppose I have a basis for $H$ called $\{h_j\}$. Define $$H_n := \text{span}\{h_1,...,h_n\}.$$ Suppose now I am given an orthonormal basis for $H$ called $\{v_j\}$.

My question is, is it possible to reorder the basis $v_j$ so that

1) For all $i=1,...,n$, the element $v_i \in H_n$

2) The set $\{v_j\}_{j=1}^N$ is a basis of $H_n$

My thoughts: 1) possible since $H_n$ is subspace of $H$ so there must be a basis function in there. It's dimension is $n$ so there must be $n$ basis functions

2) Follows from above.

Am I right?

  • $\begingroup$ No. A simple counterexample: In $\Bbb R^2$, take $h_1=(1,0)$, $h_2=(0,1)$, $v_1=(1/\sqrt 2)(1,1)$, $v_2=(1/\sqrt2)(-1,1)$, and consider $H_1$. $\endgroup$ Jun 26 '13 at 18:47
  • $\begingroup$ THanks. How about if I am allowed to choose $n$ large enough so that the first condition holds? Everything is infinite-dimensional in my case. I guess the second condition need not hold.. $\endgroup$ Jun 26 '13 at 18:54
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    $\begingroup$ @michael_faber Michael your question reminds me that in numerical pde, the methods are drastically different when one choose the a global polynomial basis (Bernstein), locally supported basis (finite element), or Fourier basis to approximate the solution. $\endgroup$
    – Shuhao Cao
    Jun 26 '13 at 18:59

Take $L^2([-\pi,\pi])$, first basis set $$ \{1,x,x^2,\ldots,x^n,\ldots\}\quad \text{for } n\in \mathbb{N}\cup \{0\}. $$ and the orthonormal basis set $$ \left\{\frac{1}{\sqrt{2\pi}} e^{imx}\right\}\quad \text{for } m\in \mathbb{Z}. $$

The Fourier expansion of an $n$-degree polynomial can not be represented by $n$ terms from the sine and cosine basis. Neither does a sine or cosine's Taylor expansion contain just a finite degree polynomial.


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