Projection Theorem I've been trying to apply the projection theorem to the following problem with no success. I've spent a few hours on this today, any help would be appreciated.
Let H be a finite dimensional Hibert Space and $\{{v_{1},v_{2}\}}$ be two linearly independet vecotrs in H. Let $b_{1}$,$b_{2}\in R$. Show that, among all vectors $x\in$ H, which satisfies:
$$<x,v_{1}> = b_{1}$$
$$<x,v_{2}> = b_{2}$$
the vector $x'\in$ H has the minimum norm if $x'$ satisfies:
$$x' = \alpha_{1}v_{1} + \alpha_{2}v_{2}$$
With
$$<v_{1},v_{1}>\alpha_{1} + <v_{2},v_{1}>\alpha_{2} = b_{1}$$
$$<v_{1},v_{2}>\alpha_{1} + <v_{2},v_{2}>\alpha_{2} = b_{2}$$
I only managed to get some substitutions done and defined the subspace for the projection theorem as follows:
M = $\{x\in H : x = \frac{b_{1}}{\|{v_{1}}\|}v_{1} + \frac{b_{2}}{\||{v_{2}\|}}v_{2}\}$
I really don't know how to proceed..
 A: The set
$$
       \mathcal{M}=\{ x \in H : (x,v_1)=b_1,\; (x,v_2)=b_2 \}
$$
is a translation of a subspace. Every $m\in\mathcal{M}$ can be written as
$$
                 m = \alpha_1 v_1 + \alpha_2 v_2 + y
$$
where $y \perp v_1$, $y\perp v_2$ and where $\alpha_1$, $\alpha_2$ are the unique scalars for which
$$
\begin{array}{ccc}
            (\alpha_1 v_1 + \alpha_2 v_2,v_1) & = & b_1 \\
            (\alpha_1 v_1 + \alpha_2 v_2,v_2) & = & b_2.
\end{array}   \;\;\;\; (\dagger)
$$
This is a well-posed system for the $\alpha_j$ because $\{ v_1,v_2 \}$ is an independent set of vectors. Therefore,
$$
       \|m\|^2 = \|\alpha_1 v_1 + \alpha_2 v_2 \|^2+\|y\|^2
$$
which means that the element $m\in\mathcal{M}$ of minimal norm is $\alpha_1 v_1+\alpha_2 v_2$. The system $(\dagger)$ has the matrix form
$$
      \left[\begin{array}{cc}(v_1,v_1) & (v_2,v_1) \\ (v_1,v_2) & (v_2,v_2)\end{array}\right]\left[\begin{array}{c}\alpha_1 \\ \alpha_2\end{array}\right]=\left[\begin{array}{c}b_1 \\ b_2\end{array}\right] \\
   \left[\begin{array}{c}\alpha_1 \\ \alpha_2\end{array}\right]
   = \frac{1}{(v_1,v_1)(v_2,v_2)-|(v_1,v_2)|^2}\left[\begin{array}{cc}(v_2,v_2) & -(v_2,v_1) \\ -(v_1,v_2) & (v_1,v_1)\end{array}\right]\left[\begin{array}{c}b_1 \\ b_2\end{array}\right]
$$
