While studying differential geometry I have formulated a result in linear algebra which I haven't been able to prove:

Let $V$ be a real vector space with an inner product, and let $\{e_1,\dots,e_k\}\subset V$ be an orthonormal basis. Let $W\subset V$ be a subspace of dimension $n\geq 1$. There exist $i_1,\dots,i_n$ such that $P_S|_W:W\to S$ is an isomorphism, where $S=\operatorname{span}(e_{i_1},\dots,e_{i_n})$ and $P_S:V\to S$ is the orthogonal projection onto $S$.

Here's (what I think is) the most fruitful approach I've tried.

I have geometrically visualized that "almost always" any $(i_1,\dots,i_n)$ will do, except when some $e_i\in W$. So, I assert that we can pick $(i_1,\dots,i_n)$ such that $W$ is not perpendicular to $\operatorname{span}(e_{i_j})$ for all $j=1,\dots,n$.

In that case, I try to prove that $P_S|_W$ is surjective. Let $v=\sum_j \langle v,e_{i_j}\rangle e_{i_j}\in S$. Define $w_{i_j}=P_W(e_{i_j})$ and $w=\sum_j \langle v,e_{i_j}\rangle w_{i_j}$. A computation gives $$P_S(w)=\sum_k \langle v,e_{i_k}\rangle P_S P_W(e_{i_k})$$

If I could prove that $P_S P_W(e_{i_k})\in \operatorname{span}(e_{i_k})$ (which I geometrically believe, although I have a feeling it might fail in higher-than-imaginable dimensions), then I could, redefining $w_{i_j}$ by a scalar multiple, get that $P_SP_W(e_{i_k})=e_{i_k}$ thus obtaining $P_S(w)=v$ and finishing the proof. But I haven't been able to obtain a useful expression of $P_SP_W(e_{i_k})$.

I'd appreciate any proof, but I'd especially appreciate any comments on this approach (I've been at it for some hours now).

  • $\begingroup$ You must mean isomorphism of just vector spaces; there is in general no way the restriction of the projection can be made to respect the inner product. It would seem the inner product is altogether irrelevant for this question. $\endgroup$ – Marc van Leeuwen Apr 16 '13 at 20:18
  • $\begingroup$ @MarcvanLeeuwen: yes, I mean isomorphism of plain vector spaces. I don't think the inner product is irrelevant, how do you make sense of the orthogonal projection if it isn't there? $\endgroup$ – Bruno Stonek Apr 16 '13 at 20:37
  • $\begingroup$ You are right. But you can replace orthogonal projection by the projection parallel to the span of the remaining standard basis vectors as I did in my answer, and then the inner product is no longer needed. $\endgroup$ – Marc van Leeuwen Apr 16 '13 at 20:40

Take a basis of $W$ and express its vectors in the basis $e_1,\ldots,e_k$, forming a $k\times n$ matrix with those columns (remember $n\leq k$). Since you had a basis this matrix has (full) rank $n$. So you can select $n$ linearly independent rows from the matrix, which means a subset of $n$ coordinates of those $n$ basis vectors that are linearly independent. Then the projection on the subspace spanned by the corresponding $n$ vectors among $e_1,\ldots,e_k$, parallel to the space spanned by the remaining ones, is an isomorphism of vector spaces (the projection matrix is formed by just those selected $n$ rows, and has nonzero determinant).

  • $\begingroup$ I think you mean that the matrix has rank $n$, as $n\leq k$, so you select $n$ linearly independent rows... $\endgroup$ – Bruno Stonek Apr 16 '13 at 21:08
  • $\begingroup$ @BrunoStonek: Yes, thank you. I get profoundly confused when the dimension of the whole space is not called $n$... $\endgroup$ – Marc van Leeuwen Apr 16 '13 at 21:20
  • $\begingroup$ I can certainly empathize with that :) But in this case $n$ will be the dimension of a manifold, and $k$ the dimension of the ambient space it sits in. $\endgroup$ – Bruno Stonek Apr 16 '13 at 21:23
  • $\begingroup$ Ok, I finished grasping your answer, I like it, it's slick :) Thank you. $\endgroup$ – Bruno Stonek Apr 16 '13 at 21:35

Induct on the dimension of $V$: The statement is clear when $n = 1$. First consider the projection $PW$ onto the span of $e_2,\ldots,e_n$. By induction, there are some $\{e_i\}_{i \in I}$ such that the projection $P'$ of $PW$ onto the span of the $e_i$ for $i \in I$ is an isomorphism. ($I \subseteq \{2,3,\ldots,n\}$). The kernel of $P$ when restricted to $W$ is either zero-dimensional or one-dimensional. In the first case we are done: $P'P$ is an isomorphism of $W$ onto the span of the $\{e_i\}$ for $i \in I$. In the second case, we must have $e_1 \in W$, and so we simply need to add $e_1$ to the list of vectors.


Well, for surjectiveness, we would need that $e_{i_k}\in P_S(W)$, because then any of their linear combination is also present. I tried to prove it, but, after rearranging the indices so that $S={\rm span}(e_1,..,e_n)$, it leads to ensuring that certain elements like $(1,0,0,..,\alpha_{n+1},\alpha_{n+2},..)$ are in $W$...

However, trying from the other end: we have $\ker P_S=S^\perp$, and we need $P_S|_W$ be injective, and $\dim S=\dim W$. Injectivity means $\ker P_S|_W=W\cap S^\perp=\{0\}$.

So, we can choose $i_1$ so that $e_{i_1}\not\perp W$, this way, with $S_1:= {\rm span}(e_{i_1})$, we have $(W\cap S_1^\perp)\subsetneq W$.
Then, choose $i_2$ so that, with $S_2:={\rm span}(e_{i_1},e_{i_2})$, we have $$W\cap S_2^\perp\,\subsetneq\, W\cap S_1^\perp\,,$$ and so on...

  • $\begingroup$ Thank you for your answer! While riding back home on the bus today I found a similar argument, I believe, which I think is easier to justify by reductio ad absurdum. Would you care to look at it? Perhaps you can help me with the combinatorial argument :) $\endgroup$ – Bruno Stonek Apr 16 '13 at 1:59
  • $\begingroup$ I think I don't understand your argument. Why can you choose $i_1$ such that $e_{i_1}\perp W$? The same example you gave on your comment to my answer is a counterexample. $\endgroup$ – Bruno Stonek Apr 16 '13 at 12:32
  • $\begingroup$ No, $e_{i_1}$ is not perpendicular to $W$. (If all $e_i$ are perpendicular, then $W=\{0\}$.) So, you can start the procedure. You still need the next steps.. $\endgroup$ – Berci Apr 17 '13 at 9:31
  • $\begingroup$ Oh, I see you wrote \not\perp but somehow I see it rendered as $\perp$. I'll reread your answer then. $\endgroup$ – Bruno Stonek Apr 17 '13 at 11:17

Update: this answer is wrong. See the comments below.

Ok, so I was riding the bus back home, a little obsessed by this problem, and I think I finally got around it:

We can suppose $n<k$, the theorem is obvious if $n=k$.

Now observe that $\operatorname{ker}(P_S|_W)=S^\perp \cap W$. So, we want to find indices $t_1,\dots,t_{k-n}$ such that $W\cap \operatorname{span}(e_{t_1},\dots,e_{t_{k-n}})=\{0\}$.

Assume by contradiction that for every choice of indices $t_1,\dots,t_{k-n}$ we have that $W\cap \operatorname{span}(e_{t_1},\dots,e_{t_{k-n}})\neq\{0\}$. Therefore $W$ contains some $e_i$ of each of these $\binom{k}{k-n}$ subspaces. A combinatorial argument (using $n<k$) shows that $W$ must have at least $k-(k-n)+1=n+1$ different $e_i$, which is absurd since $\dim W=n$.

I think the argument is nice, but I would appreciate how to really prove the "combinatorial argument" (I'm really bad at these).

  • $\begingroup$ Well, $W\cap{\rm span}(e_{t_1},..,e_{t_{k-n}})\ne\{0\}$ doesn't imply that effectively some $e_{t_i}}$ is in $W$, think about for example $W=\,(x-y=0)$, this intersects ${\rm span}(e_1,e_2)$, but doesn't contain neither $e_1$ nor $e_2$... The argument is nice, but I can't see right now how to continue.. $\endgroup$ – Berci Apr 16 '13 at 7:38
  • $\begingroup$ @Berci: shame on me, of course, you're right. $\endgroup$ – Bruno Stonek Apr 16 '13 at 12:07

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