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Let $V$ be a finitely generated real inner product space with $\dim V=n$ and inner product $\langle\cdot,\cdot\rangle:V\times V\to\mathbb{R}$.

Let $U,W\subseteq V$ be sub-spaces of $V$. Is it true that we have $$\text{proj}_{U+W}=\text{proj}_U+\text{proj}_W-\text{proj}_{U\cap W}$$ (where $\text{proj}_X:V\to V$ is the orthogonal projection onto the sub-space $X$)? After some attempts and some web searches, I was unable to find any information regarding this.

I want to say this is true. Let $\mathbf{u}=\{u_1,\dots,u_m\}$ be an orthonormal basis of $U$ and $\mathbf{w}=\{w_1,\dots,w_k\}$ be an orthonormal basis of $W$. Either $U$ and $W$ intersect trivially or they don't. If they do intersect trivially, then we are done, since $\mathbf{u}\cup\mathbf{w}=\{u_1,\dots,u_m,w_1,\dots,w_k\}$ is a basis for $U+W$ and we can orthonormalize it using Grahm-Schmidt to get an orthonormal basis $\{u_1,\dots,u_m,w_1',\dots,w_k'\}$ for $U+W$. In this case its clear that $$\text{proj}_{U+W}=\text{proj}_U+\text{proj}_W$$ since $\text{proj}_{U\cap W}$ is identically the zero operator on $V$.

Having some trouble in the more general case where the sub-spaces don't intersect trivially, mainly with notation and how to go about finding an orthogonal/orthonormal basis of $U\cap W$.

Maybe there is a better way to approach this problem or some better notation and someone may be able to offer some insight.

Edit: As of Jyrki Lahtonen's counter example, the claim is false. Still open to new answers/perspectives and possibly some conditions which would allow for this to happen, if there are any.

C Squared
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    You are on the right track but not quite. Start with a basis of $U \cap W$ and expan this basis to bases of $U$ and $W.$ Notice that the basis of ${0}$ is empty. – William M. Jul 07 '21 at 00:26
  • @WillM. so start with an orthonormal basis ${v_1,\dots,v_r}$ of $U\cap W$ (assume non-empty for now). Extend to orthonormal bases ${v_1,\dots,v_r,u_1,\dots,u_m}$ of $U$ and ${v_1,\dots,v_r,w_1,\dots,w_k}$ of $W$. Then ${u_1,\dots,u_m,w_1,\dots,w_k}$ forms a basis of $U+W$. The basis for $U+W$ may not be orthonormal though, so run it through Gram-Schmidt? – C Squared Jul 07 '21 at 00:36
  • The basis of $U + W$ is orthonormal. – William M. Jul 07 '21 at 00:42
  • @WillM. oh? i guess im not seeing why. if it is then the conclusion is immediate. For example, how is it guaranteed that $u_1$ perpendicular to $w_1$ ? – C Squared Jul 07 '21 at 00:44
  • You are right, it need not be. Construct a basis of$U \cap W$ then extend it orthonormally to a basis of $U$ and then to $U+W.$ – William M. Jul 07 '21 at 01:00
  • alright. i’ll try working that way. thanks, starting with the basis of $U\cap W$ is a bit easier – C Squared Jul 07 '21 at 01:02
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    I just did a rough sketch in my notebook but my last comment seems to be right. To grow the bases one at a time. – William M. Jul 07 '21 at 01:15

2 Answers2

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The claim is false as stated. Proffering a counterexample.

Let $V=\Bbb{R}^2$ with the usual inner product. Let $U=\{(x,0)\mid x\in\Bbb{R}\}$ be the $x$-axis. Let $w=\{(t,t)\mid t\in\Bbb{R}\}$ be the line with slope one. Then $U+W=V$ and $U\cap W=\{0\}$.

Consider the vector $\vec{x}=(3,1)$. Its projection to $U$ is $p_U(\vec{x})=(3,0)$. Its projection to $W$ is $p_W(\vec{x})=(2,2)$. Of course, its projection to $U\cap W$ is the zero vector and its projection to $V$ is itself.

But $\vec{x}\neq p_U(\vec{x})+p_W(\vec{x}).$

Jyrki Lahtonen
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  • oh wow. nice counter example. wonder where my proof went wrong – C Squared Jul 07 '21 at 03:38
  • one more follow up question if you don't mind. do you think that one can write $\text{proj}_{U+W}$ in terms of $\text{proj}_U$ and $\text{proj}_W$? possibly there are some conditions that allow for this to happen? – C Squared Jul 07 '21 at 04:06
  • @CSquared Your recipe (and proof) works if we have direct sum decompositions $U=U'+ (U\cap W)$, $W=W'+(U\cap W)$ such that $U'\perp (U\cap W)$, $W'\perp (U\cap W)$ and $U'\perp W'$. I didn't check, but it "feels" like this is the case when the two orthogonal projections commute: $p_U\circ p_W=p_W\circ p_U$. When that composition would also be equal to $p_{U\cap W}$. Sorry, I don't have time to think about this right now. May be later. – Jyrki Lahtonen Jul 07 '21 at 05:29
  • no worries. and thank you so much again, this was really helpful – C Squared Jul 07 '21 at 05:30
  • May be start here? @CSquared – Jyrki Lahtonen Jul 07 '21 at 05:30
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Expanding on WillM.'s comments:

Choose an orthonormal basis $\mathbf{int}_{U,W}=\{v_1,\dots,v_r\}$ of $U\cap W$ (assuming they intersect non-trivially).

Extend this to an orthonormal basis $\mathbf{u}=\{v_1,\dots,v_r,u_1,\dots,u_m\}$ of $U$.

Further extend this to an orthornomal basis $\mathbf{sum}_{U,W}=\{v_1,\dots,v_r,u_1,\dots,u_m,w_1,\dots,w_l\}$ of $U+W$.

Note that $\mathbf{w}=\{v_1,\dots,v_r,w_1,\dots,w_l\}$ is an orthonormal basis of $W$.

Then we have \begin{align*} \text{proj}_{U}(x)+\text{proj}_{W}(x)-\text{proj}_{U\cap V}(x) &=\left(\sum_{i=1}^r\langle x,v_i\rangle v_i+\sum_{j=1}^m\langle x,u_j\rangle u_j\right)+\left(\sum_{i=1}^{r}\langle x,v_i\rangle v_i+\sum_{k=1}^{l}\langle x,w_k\rangle w_k\right)-\sum_{i=1}^r\langle x,v_i\rangle v_i\\ &=\sum_{i=1}^r\langle x,v_i\rangle v_i+\sum_{j=1}^m\langle x,u_j\rangle u_j+\sum_{k=1}^{l}\langle x,w_k\rangle w_k\\ &=\text{proj}_{U+W}(x) \end{align*} which completes the proof.

C Squared
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  • Why are $w_i$s orthogonal to $u_j$s? – Jyrki Lahtonen Jul 07 '21 at 03:39
  • @JyrkiLahtonen I had planned to extended them in such a manner to be orthonormal to each other. – C Squared Jul 07 '21 at 03:41
  • I think a problem is that you cannot select the $w_i$s from $U^\perp\cap W$ as that intersection may have uncomfortably low dimension. You can extend an orthonormal basis of $U$ to an orthonormal basis of $U+W$, but there is no way to guarantee that basis would contain an orthonormal basis of $W$. – Jyrki Lahtonen Jul 07 '21 at 03:43
  • hmm. in my head i had thought of extending the bases using gram-schmidt or something similar so i didn't have to choose them from $U^{\perp}\cap W$, but that is the space where they would naturally lie in anyways. – C Squared Jul 07 '21 at 03:55
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    The problem with that is that in Gram-Schmidt you may need to add something from $U$ to vectors from $W$, and end up outside of $W$. – Jyrki Lahtonen Jul 07 '21 at 03:57
  • @JyrkiLahtonen this is even false when the subspaces intersect trivially, as you have shown. guess i'm misunderstanding a couple of things. thanks for the counter example – C Squared Jul 07 '21 at 04:00
  • Yeah, testing the claim with 1-dimensional subspaces already reveals a problem :-) – Jyrki Lahtonen Jul 07 '21 at 04:04
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    If $U \cap W ={0}$ but $U$ and $W$ are not orthogonal to each other (i.e. $\langle u, w\rangle =0$ for all $u\in U, w\in W$), you will not be able to find orthogonal bases of these two spaces such that the union of the two bases is still orthogonal. – Eike Schulte Jul 07 '21 at 04:58
  • If $U \cap W$ is not trivial, you can consider the orthogonal complements of $U\cap W$ in $U$ and $W$. Again, these complements need to be orthogonal if you want to find compatible orthogonal bases for $U\cap W$, $U$, $W$ and $U+W$. – Eike Schulte Jul 07 '21 at 05:06
  • Very interesting that my intuition was somewhat wrong. I did a rough sketch and I don't see an immediate flaw. You can still orthonormalise and the space spanned is the samem, but hey, you already gave a counterexample. So I am sure if I start digging what went wrong, I will find it. Thanks – William M. Jul 07 '21 at 13:38