Consider this lemma (my question are below):
Lemma Given three pairwise orthogonal subspaces $X$, $Y$, $Z$ of a Hilbert space $H$ that span the whole space, any vector $\nu\in H,\ ||\nu||=1$, can be written as $\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3,\ \ \ \nu_1\in X,\ \nu_2\in Y, \ \nu_3\in Z,$ with $\alpha,\beta,\gamma\geq0$ and $\alpha^2+\beta^2+\gamma^2=1$ and $\nu_1,\nu_2,\nu_3$ and unit vectors.
Proof Let $\{x_j\}$, $\{y_j\}$, $\{z_j\}$ be orthonormal bases for $X$, $Y$, $Z$. Together, they form a basis for the whole $H$. So there exist coefficients such that $ \nu=\sum_j a_jx_j + \sum_jb_jy_j+\sum_jc_jz_j. $ As $\|\nu\|=1$, $\sum_j|a_j|^2+\sum_j|b_j|^2+\sum_j|c_j|^2=1$. Let $ \alpha=(\sum_j |a_j|^2)^{1/2},\ \beta=(\sum_j |b_j|^2)^{1/2},\ \gamma=(\sum_j |c_j|^2)^{1/2},\ $ and $ \nu_1=\sum_j\frac{a_j}\alpha\,x_j,\ \nu_2=\sum_j\frac{b_j}\beta\,y_j,\ \nu_3=\sum_j\frac{c_j}\gamma\,z_j. $ Then $\nu_1,\nu_2\nu_3$ are unit vectors, $\alpha^2+\beta^2+\gamma^2=1$, and $ \nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3. $
My questions are: 1) Is this lemma true for complex Hilbert spaces ? (my guess would be "yes") 2) Is this lemma true, not for just three subspaces, but for subspace $X_1,\ldots X_n$, that are pairwise orthogonal and span the whole space ?
For who wants to know: This lemma is from here.