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I came across the following in a linear algebra book:

We know already that if $U$ is a subspace of a vector space $V$, then there are, usually, many other subspaces $W$ in $V$ such that $U \oplus W = V$. There is no natural way of choosing one from among the wealth of complements of $U$.

In other words, there exist at least $2$ subspaces $W_1$ and $W_2$, both disjoint from $U$, such that $U+W_1=U+W_2=V$. If $\{v_1, \ldots, v_m\}$ is a basis of $U$, that can be extended to a basis $\{v_1, \ldots, v_m, v_{m+1}, \ldots, v_n\}$ of $V$. Won't the elements $\{v_{m+1}, \ldots, v_n\}$ form the basis of the direct sum complement? And no matter how the original basis of $U$ was extended, won't the new vectors in the extended set span the same vector space regardless? In such a case, how can we get different direct-sum complement subspaces $W_1$ and $W_2$?

(I know there's a flaw in my thinking above. Would be great if anyone could point them out. How would I prove that the direct sum complement is non-unique?)

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The error lies in assuming that the only possibility for the supplementary subspace is $\langle v_{m+1},\ldots,v_n\rangle$. It isn't. If $V=\mathbb{R}^2$, $U=\langle(1,0)\rangle$ and if you complete the basis, which vector will you take? You will choose $(0,1)$ probably. And indeed $\langle(0,1)\rangle$ is a supplementary subspace. However, $\langle(1,1)\rangle(=\langle(1,0)+(0,1)\rangle)$ will also work, for instance.

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  • $\begingroup$ I see. So if $U$ is a subspace, it's possible for it to have disjoint complementary subspaces. I guess that's the counter-intuitive part. $\endgroup$ Jun 5, 2018 at 13:43
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    $\begingroup$ @ShirishKulhari They are not disjoint; the null vecor belongs to both of them. And they may well have more vectors in common (although not in the situation that I described). $\endgroup$ Jun 5, 2018 at 13:45
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Consider a vector space of dimension $2$ and $(e_1,e_2)$ a basis, $(e_1,e_1+e_2)$ is also a basis of the space. $W_1=Vect(e_2)$ and $W_2=Vect(e_1+e_2)$ are different supplementary spaces of $vect(e_1)$.

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  • $\begingroup$ More generally, if $V = \mathbb R^2$, we have $V = U \oplus W$ for any two distinct one-dimensional subspaces $U$ and $W$. $\endgroup$ Jun 5, 2018 at 13:38
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I would like to add a few points here to give a bigger picture. Given a subspace $W\subset V$, there always exists a complement to $W$ in $V$. This is not unique. A simple example can illustrate this.

Take $\mathbb R^2$, and a subspace $W=span\{(1,0)\}$. Clearly $W'=span\{(0,1)\}$ is a complement, i.e. $W \oplus W'=\mathbb R^2$. Now consider $W''=span\{(1,1)\}$. This is also a complement. Now, it should be intuitively obvious that you can generate a whole bunch of such complementary subspaces (imagine it geometrically). Further in your argument, the extension of basis of the subspace to the entire space exists, but is not unique in any sense.

Extra. So, is there any canonical complementary space? Well, yes. Given an inner product $g$, define the subspace $W^g = \{v\in V:g(v,w)=0, \forall w\in W\}$. It will turn out that $W\oplus W^g =V$.

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