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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.

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  • Really? nobody ? –  Aug 23 '12 at 16:50
  • THe positivity of the coefficients is rather arbitrary, in my opinion. You can just multiply some $v_j$ by $-1$. – Siminore Aug 23 '12 at 16:57
  • But isn't this a simple consequence of the Gram-Schmidt orthonormalization method? At least if $H$ is separable. – Siminore Aug 23 '12 at 16:59
  • It seems you intended to make some assumptions about some or all of these vectors being unit vectors but forgot to explicate them in introducing those vectors. Without that aspect, the lemma is trivial. I suspect that it's straightforward to prove even with those assumptions, but you should explicate them to disambiguate the question. Note that state vectors in physics are typically assumed to be unit vectors, but "any vector" carries no implication of the vector being a unit vector. – joriki Aug 23 '12 at 17:08
  • @joriki Yes, indeed, $\nu_i$ should additionally be unit vectors. For unclarity please see the original post, where this lemma was stated in the answer –  Aug 23 '12 at 17:58
  • @Simonore I'm not sure... –  Aug 23 '12 at 17:58
  • I downvoted. The question is unclear as stated, and you didn't clarify it when I pointed this out. Pointing out in a comment that it would become clear if one followed a link to a lengthy answer is not a form of clarifying the question; neither is pointing out that one out of four vectors is a unit vector when all four ($\nu,\nu_1,\nu_2,\nu_3$) are assumed to be unit vectors. Note also that it's very confusing to copy something that begins with "Edit" in bold from another post into your question where this is actually not an edit but an essential part of your question as originally stated. – joriki Aug 23 '12 at 18:27
  • @joriki Ok, given. @ Everybody else: If someone is not yet so mad at me for improperly formatting the lemma or forgetting to mention that all of the $\nu$'s are uni vectors, I would he/she would provide an answer. I hope everythings clear now as it is. –  Aug 23 '12 at 18:53
  • @user36772: a) It's not a question of being mad at you; b) I didn't criticize that you'd forgotten to mention this, but that you refused to mention it after I pointed out that you'd forgotten it. – joriki Aug 23 '12 at 18:59
  • @joriki ah, sorry for that. I was in a terrible hurry than, but didn't want to leave it hanging, so that why I just refered to the other post. –  Aug 23 '12 at 20:31
  • I've cancelled my downvote. – joriki Aug 24 '12 at 04:28

1 Answers1

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The lemma holds for any finite number of real or complex Hilbert spaces of any finite or infinite dimension. The proof is unnecessarily complicated and involves some subtleties in the case of infinite-dimensional spaces that a straightforward proof doesn't need to get into; also it doesn't cover the case where one of $\alpha$, $\beta$, $\gamma$ is zero.

Since the subspaces $X_1,\dotsc,X_n$ span the space, there are vectors $x_i\in X_i$ such that

$$ \nu=x_1+\dotso+x_n\;. $$

Take $u_i=x_i/\|x_i\|$ if $x_i\ne0$ and $u_i$ an arbitrary unit vector in $X_i$ otherwise. Then

$$ \nu=\lambda_1u_1+\dotso+\lambda_nu_n $$

with $\lambda_i=\|x_i\|$ if $x_i\ne0$ and $\lambda_i=0$ otherwise, so $\lambda_i\ge0$. Since the $u_i$ are orthonormal, $\|\nu\|=1$ implies $\lambda_1^2+\dotso+\lambda_n^2=1$.

joriki
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