Suppose $\{u_1, u_2, u_3,\ldots\}$ form an orthonormal system in an inner product space $V$. Let $U_n = \operatorname{span}\{u_1, u_2, \ldots, u_n\}$, and define the operator $\Pi_n: V\to U_n$ by $$\Pi_n(f) = \sum_{i=1}^n\langle f, u_i\rangle u_i, \quad f\in V.$$ Prove that $\Pi_n$ is surjective.

I know that to prove surjectivity, I need to show that for any $u\in U_n$ that there exists a function $f\in V$ such that $\Pi_n(f) = u$. Here, this would mean I take $$u = \alpha_1 u_1 +\cdots + \alpha_n u_n,$$ which now I'm slightly confused what to do next. Likely I need to use the fact that $\{u_1, u_2,\ldots, u_n\}$ is an orthonormal basis, which here means that $U_n$ is $n-$dimensional subspace of $V$, and $$\langle u_i, u_j\rangle = \begin{cases}1, & i=j\\0, &i\ne j\end{cases}.$$ It's fairly obvious that I can hit any $u_i\in U_n$ by just choosing $u_i\in V$: $$\Pi_n(u_j) = \sum_{i=1}^n\langle u_j, u_i\rangle u_i = 0 + 0 + \cdots + 1u_j + \cdots + 0 = u_j$$ but how can I guarantee I can hit $u = \alpha_1 u_1+\cdots+\alpha_n u_n$ for arbitrary $\alpha_i\in\mathbb R$? Or do the $\alpha_i$ come from somewhere else?

  • $\begingroup$ $\Pi_n$ is a linear map, so after you have shown you can hit all those $u_i$, you are done, since you've shown that the dimension of the image is equal to the dimension of $U_n$. If you want to hit an arbitrary linear combination, just use linearity in the inputs: $\Pi_n(2 u_2 - 3 u_5)$ for example. $\endgroup$ – Joppy May 3 '18 at 2:13

As @Joppy stated in the comment above, you almost come to the end.

For $u=\alpha_1 u_1+\dotsc+\alpha_n u_n=\sum_{j=1}^n \alpha_j u_j$, just calculate $\Pi_n(u)$ as follows: $$ \begin{align*} \Pi_n(u) &\equiv \sum_{i=1}^n \langle u,u_i\rangle u_i = \sum_{i=1}^n \left\langle \sum_{j=1}^n \alpha_j u_j, u_i\right\rangle u_i \\ &= \sum_{i=1}^n \left( \sum_{j=1}^n \alpha_j \langle u_j,u_i\rangle \right) u_i = \sum_{i=1}^n \alpha_i u_i = u \end{align*} $$


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