Timeline for Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional
Current License: CC BY-SA 3.0
29 events
| when toggle format | what | by | license | comment | |
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| S Oct 10, 2017 at 13:59 | history | suggested | Mizar | CC BY-SA 3.0 |
added field to which the dimensions refer, other minor corrections
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| Oct 10, 2017 at 13:17 | review | Suggested edits | |||
| S Oct 10, 2017 at 13:59 | |||||
| Mar 17, 2016 at 21:46 | comment | added | Duchamp Gérard H. E. | Nice proof, btw +1. | |
| S Mar 17, 2016 at 20:11 | history | suggested | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Made explicit a hidden assumption "(in this first step I am assuming $card(k)<dim(E)$)" + [hols]--->[holds]
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| Mar 17, 2016 at 19:44 | review | Suggested edits | |||
| S Mar 17, 2016 at 20:11 | |||||
| Mar 17, 2016 at 19:26 | comment | added | Andrea Ferretti | @DuchampGérardH.E. In the first step I am assuming dim E > card(k) | |
| Mar 17, 2016 at 8:02 | comment | added | Duchamp Gérard H. E. | @AndreaFerretti It seems to me that the cardinality of $V\cong k^{(E)}$ (the bracket means "finitely supported") is that of $k$ (if $E$ is infinite and $card(E)<card(k)$), no ? | |
| Nov 17, 2015 at 16:44 | comment | added | Exterior | @KConrad which part of this argument cannot be generalized to projective, or at least free modules over a noncommutative ring? | |
| Feb 15, 2015 at 22:19 | comment | added | KConrad | @LeonAvery, your error is that you are not constructing $W$ correctly. The example you give, where $f$ is the real part function on $\mathbf C$, is not an element of $W$. | |
| Feb 15, 2015 at 22:16 | comment | added | KConrad | @LeonAvery, you are using $h = \mathbf R$, $k = \mathbf C$, and $V = \mathbf C$. The space $W$ is defined to be the finitely supported functions from a $k$-basis of $V$ to $h$. Here $V$ has a basis of size one, say $E = \{z_0\}$ with $z_0 \not= 0$. Then $W$ is all functions $E \rightarrow \mathbf R$, which is canonically $\mathbf R$ as a real vector space by identifying such a function with its value at $z_0$. Then $W^*$ (real dual space) has dim. 1 over $\mathbf R$ and any $f \in W^*$ extends to $V^*$ (cpx. dual space) by $z \mapsto (z/z_0)f(z_0) = (f(z_0)/z_0)z$. That's $\mathbf C$-linear. | |
| Feb 15, 2015 at 18:35 | comment | added | Leon Avery | @KConrad: I'm sure I've got this wrong, but it seems to me there is still a hole in the argument. $f_i$ is a homomorphism from $W$ to $h$, but its extension to $V$ and $k$ may not be a homomorphism. For instance, $\mathbb{C}$ is a 2D vector space over $\mathbb{R}$, and $f:a+bi\mapsto a$ is a perfectly fine linear functional on this vector space. However, it doesn't extend to $\mathbb{C}$, since, for instance, $i=i\cdot f(1)\ne f(i\cdot 1)=0$. In fact, the extension of $\mathbb{C}$ over $\mathbb{R}$ to $\mathbb{C}$ over $\mathbb{C}$ is 1D. Where am I going wrong? | |
| Oct 17, 2014 at 15:13 | comment | added | user71815 | @ShayBenMoshe I think the second claim in the answer shows that your case is taken care of. | |
| Nov 19, 2013 at 21:46 | comment | added | Shay Ben Moshe | Unfortunately this proof does not include the case when $\aleph_0\leq \dim V<|F|$. Here is a proof that includes this case math.stackexchange.com/a/35863/10976. | |
| Sep 1, 2011 at 6:36 | comment | added | Joshua P. Swanson | To reply to my previous comment, the full inner product space assumptions are unnecessary. The usual dot product on $h^n$ is a non-degenerate bilinear form, and the sum of the dimensions of the orthogonal complement of a subspace and of that subspace is still the dimension of the ambient space in this setting, as detailed at maths.bris.ac.uk/~maxmr/la2/notes_5.pdf (see Proposition 5.9). Everything works out then! | |
| Aug 31, 2011 at 14:06 | comment | added | Joshua P. Swanson | @KConrad: Since $h^n$ isn't an inner product space, how does orthogonality work that way? | |
| May 8, 2011 at 23:42 | comment | added | KConrad | @Andrea, @Arturo: functionals in $W^*$ need not have finite support. I would argue instead as follows. Given $h$-lin. indep. $f_1,...,f_n$ in $W^*$, consider $h$-linear map $W \rightarrow h^n$ where $w \mapsto (f_1(w),...,f_r(w))$. To show this is onto we show the only vector in $h^n$ orthogonal to image is $(0,...,0)$. If $(c_1,...,c_n)$ in $h^n$ is orthogonal to image then for all $w$ in $W$, $c_1f_1(w) + ... + c_nf_(w) = 0$. Thus $c_1f_1 + ... + c_nf_n$ is 0 in $W^*$, so each $c_i$ is 0 by $h$-lin. indep. of the $f_i$'s. Now use $w_j$ giving image $(0,...,1,...,0)$ with 1 in j-th slot. | |
| Sep 25, 2010 at 18:11 | comment | added | Andrea Ferretti | @Arturo: As the functional have finite (joint) support, you can reduce to the case of a finite-dimensional vector space, where the result is well-known from standard linear algebra (namely, it follows from the fact that $W \cong W^{**}$ in the finite-dimensional case). By the way, my name is Andrea, not Andrew. | |
| Sep 24, 2010 at 20:41 | comment | added | Arturo Magidin | @Andrew: Sorry to resurrect an old answer you probably haven't thought about since January; I was looking for an easy proof of precisely the statement in the title. This looks very nice, but I'm having trouble seeing how to justify the existence in W of elements that are dual to a given (finite) set of functionals. I assume you mean that given $f_1,\ldots,f_n$ distinct elements of $h^E$ with finite support, you will find elements $v_j$ in $W$ (or $V$) with $f_i(v_j)=\delta_{ij}$. But I don't see how to justify that such elements exist. | |
| Mar 20, 2010 at 20:59 | vote | accept | Harry Gindi | ||
| Jan 29, 2010 at 21:45 | comment | added | Pete L. Clark | I nominate this answer for a hypothetical future "Best of MO" collection. I find it almost magical, and with a moral -- don't just stick with the field you're given! -- that I find very appealing. | |
| Jan 29, 2010 at 19:20 | comment | added | Andrea Ferretti | Take a linear combination = 0 with coefficients in k. That involves finitely many functionals which are independent over h. Take elements of W which are dual to this functionals. Now evaluate the linear combination at each of these vectors to find that all coefficients are zero. | |
| Jan 29, 2010 at 18:40 | comment | added | Pace Nielsen | You can extend functionals, but you must also show that if they were linearly independent over the base then they remain linearly independent after the extension. Once you fill in that detail, I think you have a full proof. | |
| Jan 29, 2010 at 18:07 | comment | added | Mark Meckes | Of course, as I thought I was just being dense. Thanks. | |
| Jan 29, 2010 at 17:26 | comment | added | Andrea Ferretti | W and V have the same basis E, but over different fields of definition. A functional on W corresponds to a function from E to h, which is in particular a function from E to k. This gives your desired extension. | |
| Jan 29, 2010 at 16:55 | comment | added | Mark Meckes | "Every functional from W to h extends to a functional from V to k": this should be obvious but I'm not managing to fill in the details. Any hint? | |
| Jan 29, 2010 at 15:42 | comment | added | Ady | @Harry Gindi An interesting related problem: math.leidenuniv.nl/~naw/home/ps/pdf/2009-3.pdf (Problem C). | |
| Jan 29, 2010 at 13:38 | comment | added | Andrea Ferretti | Well, if you look at the equivalent statement dim V* <= dim V => card V* <= card V, it should be obvious. Since the hypothesis implies that V* is isomorphic to a subspace of V. | |
| Jan 29, 2010 at 12:55 | comment | added | Pete L. Clark | It looks correct to me, and if so, is a fantastic proof. | |
| Jan 29, 2010 at 12:45 | history | answered | Andrea Ferretti | CC BY-SA 2.5 |