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We know that in general, if $X$ is a vector space and $Y, Z, W$ are subspaces such that $X=Y\oplus Z$ and $X=Y\oplus W$ then it may not be true that $Z=W$.

But if we are given further that $Z\subseteq W$, then this should be true. A proof I can think of goes like this: Let $w\in W$. As $X=Y\oplus Z$, we get $w = y+z$ for some $y \in Y$ and $z \in Z \subseteq W$. But also $X=Y\oplus W$ and $w=0+w$, so the uniqueness of the decomposition forces $y=0,z=w$, whence $w=z \in Z$. So $W \subseteq Z$, and equality follows.

Somehow, I feel that there should perhaps be a cleaner and more straightforward way to see that $W=Z$, but I haven't been able to work one out.

My question is: Is there any other way this could be seen more directly?

Edit: Perhaps I should have added that these spaces could be infinite-dimensional.

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Your argument is sound, but it can be streamlined a bit.

What you need to show is that $W\subseteq Z$. So take $w\in W$; since $X=Y\oplus Z$, you can uniquely write $$ w=y+z $$ with $y\in Y$ and $z\in Z$; however $z\in W$, so $y=w-z\in W$ and therefore $y=0$.

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We can use dimension if $Z$ is of finite dimension, we have $dim(Z)=dim(X)-dim(Y)=dim(W)$ and $Z\subset W$ so $$ Z=W $$

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  • $\begingroup$ What if they are infinite dimensional? $\endgroup$ – Math Maniac Feb 23 '16 at 23:15
  • $\begingroup$ you can see the egreg answer :) $\endgroup$ – Hamza Feb 23 '16 at 23:18
  • $\begingroup$ You should state the finite-dimension restriction in your answer. $\endgroup$ – Rory Daulton Feb 23 '16 at 23:36

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