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I believe that for a linear space (assume finite first, but I'd like to hear about the case for infinite dimensional space.) $$V_1 + V_2 = V_3$$ does not imply that $$V_1 = V_3 - V_2.$$ Simply consider the case that $V_1$ is spanned by $\{e_1, e_2\}$, but $V_2$ is spanned by $\{e_2, e_3\}$, They sum up to $V_3 = \mathbb{R}^3$, but $V_3 - V_1 = e_3 \neq V_2$.

Therefore, operations on linear space does not obey basic arithmetic operations, right?

Why this question?

I was working on my bonus question The proof in textbook on The Transversality Theorem. So from transversality theorem, we have $$\operatorname{im} df_x T_x(X) + T_z(Z) = T_y(Y)$$ I was trying to make this question independent from that post, so I replaced with vector spaces $V_1, V_2, V_3$.

In Andreas Blass's great answer for this question, How to show $\dim T_x(X) = \dim df_x T_x(X).$, he proved $\dim T_x(X) = \dim df_x T_x(X)$ by transversality $$\dim f_xT_x(X)\geq \dim Y-\dim Z$$ and rank-nullity that $$\dim f_xT_x(X)\leq\dim X.$$

But they I start to wonder that transversality may be enough by its own, since dimensional complementarity of $X,Z$ to $Y$ is given: $$\operatorname{im} df_x T_x(X) + T_z(Z) = T_y(Y) \Rightarrow \operatorname{im} df_x T_x(X) = T_y(Y) - T_z(Z) = T_x(X).$$

Hence $\dim T_x(X) = \dim df_x T_x(X)$.

This "shortcut" is wrong since the minus operation is not granted for vector space. So Andreas blass' original approach with inequalities is correct and no shortcut known. X-)

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1LiterTears
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    How do you find that $V_3−V_1=e_3$? –  Jul 22 '13 at 17:12
  • er... I didn't intend to assign it a particular meaning. Just a regular minus in vector space. So what $V_3 - V_1$ should be? I don't really know. Thanks @AsalBeagDubh – 1LiterTears Jul 22 '13 at 17:14
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    What regular minus? There is no standard subtraction operation for subspaces of a vector space. – Chris Eagle Jul 22 '13 at 17:16
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    What does $V_3-V_1$ mean? If it is the set ${v_3-v_1\ : v_3\in V_3, v_1\in V_1}$ then it is exactly the same as $V_3+V_1$. – Giuseppe Negro Jul 22 '13 at 17:16
  • Thanks @ChrisEagle, that actually is more or less why I asked. Because if it does, it actually confuses me. So I can not really carry around the equality for vector spaces, since minus is not defined? – 1LiterTears Jul 22 '13 at 17:18
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    What do you mean by "regular minus in a vector space"? I might use that phrase to mean subtraction of one vector from another, but you're subtracting vector subspaces, not vectors. So you might mean the set-theoretic difference between the subspaces viewed as sets, but your mention of $e_3$ suggests you have something else in mind. What is it? – Andreas Blass Jul 22 '13 at 17:19
  • @Jellyfish You cannot perform the standard arithmetic manipulations on vector subspaces, no. – Alex Becker Jul 22 '13 at 17:19
  • Hi @AndreasBlass, it is greatly delightful to hear your comment. Because the reason for this question is that I found a "shortcut" for the last proof you helped me with. And then I found the shortcut is wrong for the reason asked in this question, that minus does not exist for vector space. I'll add the story in the question statement in a minute. – 1LiterTears Jul 22 '13 at 17:25
  • And right @AndreasBlass, for the addition and hypothetical minus, I meant set-theoretic difference. – 1LiterTears Jul 22 '13 at 17:27
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    You are thinking, more or less, of the quotient vector space, or orthogonal complement. And to say that $U+V=W$ is to say that mod $V$, $U$ maps onto $W/V$. – Ted Shifrin Jul 22 '13 at 17:37
  • So under the premises of set-theoretic minus, $V_3 - V_1$ equals to the vector space spanned by ${e_3}$, right? I must get something wrong if what I said suggested I have something else in mind. @AndreasBlass – 1LiterTears Jul 22 '13 at 17:38
  • Hi @TedShifrin, if $V_1$ and $V_2$ are orthogonal complement, then the argument or addition and minus shall be correct, I think. However, I was talking about addition in the context of transversality theorem, hence I should mean set addition, I guess. Therefore, I can't just move around and do the minus. – 1LiterTears Jul 22 '13 at 17:45
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    Honestly, you need to forget about subtraction. It is useful to interpret transversality in terms of passing to the quotient space: $T_xX$ should map surjectively to $T_yY/T_yZ$ (which can be interpreted as the normal space of $Z$ at $y$). – Ted Shifrin Jul 22 '13 at 17:51
  • Hi @TedShifrin, I heard of the definition of quotient space, but I was not able to follow two of your claims, "$U+V=W$ is to say that mod $V, U$ maps onto $W/V$." and "$T_xX$ should map surjectively to $T_yY/T_yZ$", though I got some vague intuition on the latter one. – 1LiterTears Jul 22 '13 at 18:07
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    Referring to your (currently) third-from-last comment: No, the set-theoretic difference of two vector subspaces will never be a vector subspace. I agree with Ted Shifrin that you should forget about "subtracting" vector spaces and learn about quotient spaces. – Andreas Blass Jul 22 '13 at 19:19
  • Got it, because the origin is subtracted? Thanks a lot @AndreasBlass! – 1LiterTears Jul 22 '13 at 19:23
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    Maybe one way to define a "subtraction" (scare quotes, because it is not a subtraction in the ordinary sense) could be $V_1-V_2:=V_1\cap V_2^\bot$ where $V_2^\bot$ is the orthogonal subspace to $V_2$, that is, the set of vectors orthogonal to all vectors of $V_2$. Of course that's only defined in inner product spaces. But then, assuming the standard inner product, indeed for your $V_1$ and $V_2$, $V_1-V_2$ is spanned by ${e_3}$. – celtschk Jul 23 '13 at 19:54
  • Great point, @celtschk, that is exactly how I am trying to think of it. But I was more or less confused by the notion of quotient space by many expert discussion above.. – 1LiterTears Jul 23 '13 at 20:05

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To have "arithmetical operations" you need to define the symbol "$-$" to mean the additive inverse of "$+$".

For this to be possible, you need your addition to satisfy cancellation, i.e. $x+z=y+z$ should imply $x=y$. This is not the case with you "addition": using your own example, $$ V_1+V_2=\mathbb R^3+V_2, $$ but you cannot conclude that $V_1=\mathbb R^3$.

The notation $V_1-V_2$ is used from time to time to mean the set $\{v_1-v_2:\ v_1\in V_1,\ v_2\in V_2\}$ but, as was already mentioned in the comments, for subspaces it is the same as $V_1+V_2$. The point is that this idea is meant to be used with subsets and not with subspaces.

The set difference, as you also mention, could also be denoted with the symbol "$-$", but in this case the "difference" will usually not even be a subspace.

Martin Argerami
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