Basis of a closed subspace in a separable Hilbert space

Question

Suppose $H$ is a separable Hilbert space and $M$ is a closed subspace of $H$. Suppose $\{u_{j}\}_{j}$ is a countable basis of $H$. Is $\{u_{j}\}_{j}\cap M$ always a basis of $M$? If not, is there any counterexample?

Thanks very much.

Answer 1

Let $H=L^2[-\pi,\pi]$, which has an orthonormal basis $\{ u_n=\frac{1}{\sqrt{2\pi}}e^{inx} \}_{n=-\infty}^{\infty}$. Let $M$ be the subspace with finite basis $\{ 1,x,x^2,\cdots,x^{N}\}$. Then $M$ is closed. However, $M\cap \{ u_n \} = \{ 1 \}$ is not a basis of $M$ for $N \ge 1$.