1
$\begingroup$

Does someone know if in the problem the projection of x onto U is defined like that :

$x_u = \displaystyle \frac{\langle x,u\rangle}{u. u}$ $u$

Problem:

Let $U,V\subset\mathbb C^n$ be two subspaces, such that $\mathbb C^n = U+V$ and further assume $U\cap V = \{0\}$.

Show that every $x\in\mathbb C^n$ can be written as $x=x_u+x_v$ with $x_u\in U$ and $x_v\in V$ and that this decomposition is unique.

$\endgroup$
  • 2
    $\begingroup$ That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem. $\endgroup$ – Michael Burr Jan 5 at 11:06
1
$\begingroup$

If $x\in \mathbb{C}^n$, then since $\mathbb{C}^n = U+V$ there must exist $x_u\in U$ and $x_v\in v$ such that $$x=x_u+x_v$$ Suppose now that there exist alternative $y_u,y_v$ with $$x=y_u+y_v$$ Then $$x-x=0=(x_u-y_u)+(x_v-y_v)$$ Hence $$y_u-x_u=x_v-y_v$$ Since $U$ and $V$ are subspaces, $y_u-x_u\in U$ and $x_v-y_v\in V$. Thus these vectors are in $U\cap V=\{0\}$, hence $y_u-x_u=x_v-y_v=0$, and hence the decomposition is unique.

Clearly we didn't need projection here. Indeed this proof works for infinite dimensional vector spaces as well, as well as vector spaces over an arbitrary field (not necessarily real or complex numbers, or even characteristic $0$), and an inner product is not required.

$\endgroup$
  • $\begingroup$ Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ? $\endgroup$ – Kai Jan 5 at 16:12
  • $\begingroup$ @Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection. $\endgroup$ – Matt Samuel Jan 5 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.