For finite dimensional vector spaces $V$ and $W$, let $i_V: V \rightarrow V^{**}$ and $i_W: W \rightarrow W^{**}$ be natural isomorphisms. Show that for any linear transformation $f : V \rightarrow W$, $i_W\circ f = f^{**}\circ i_V$. And in addition, show that there do not exist isomorphisms $i_V : V \rightarrow V^*$ and $i_W : W \rightarrow W^*$ such that for any linear transformation $f: V\rightarrow W$, such that $f^{*}\circ i_W\circ f = i_V$.

Questions: For the first part of the question, are $f$ and $f^{**}$ related in any way? Similarly, for the second part of the question, are $f$ and $f^*$ related? In addition, can I assume a basis for $V$ and a basis for $W$ to solve these questions? Any hints? For the second part, where to start?


  1. Part $1$: For an arbitrary element $v\in V$, the left hand side is $i_W[f(v)\in W]\in W^{**}$. That is, for an arbitrary $\psi\in W^*$, $i_W[f(v)\in W](\psi) =\psi[f(v)] \in\mathbb R$. Now consider the right hand side, for an arbitrary $\phi\in V^*$, $i_V(v)(\phi)=\phi(v)\in V^{**}$; then $f^{**}[i_V(v)]\in W^{**}.$ Hence, $f^{**}[i_V(v)](\psi)= i_V(v)[f^*(\psi)\in V^*]=f^*(\psi)(v)\in\mathbb R.$ I should now have $f^*(\psi)(v) = \psi[f(v)] $ simply by definition of $f^*$.
  2. Part $2$: Assume there exist the two isomorphism $i_V$ and $i_W$. Let $V=W=\mathbb R^2$ and set $f$ to be $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$ Consider the left hand side, for an arbitrary $(a, b)\in \mathbb R^2$, $f(a, b) = (b, a)$. Then $i_W(b, a)\in W^*$ and $f^*[i_W(b, a)](a, b) = i_W(b, a)[f(a,b)] = i_W(b, a)(b, a)\in\mathbb R$. As for the right hand side, $i_V(a, b)(a,b)\in\mathbb R$. I should now have $i_V(a, b)(a,b)\neq i_W(b, a)(b, a)$, but how?
  • $\begingroup$ $i_V$ is an isomorphism only $\dim V<\infty$. And don't use bases $\endgroup$ – Hagen von Eitzen May 12 '14 at 10:29
  • $\begingroup$ @HagenvonEitzen Yes, I updated that point. $\endgroup$ – LaTeXFan May 12 '14 at 10:31
  • $\begingroup$ @Hagen: Or sometimes when $\dim V=\infty$ and the axiom of choice fails! :-) $\endgroup$ – Asaf Karagila May 22 '14 at 13:31
  • $\begingroup$ Whatever happened to the bounty for this problem? It originally asked for "more detail," and I would think that either my answer or Rene's ought to satisfy anyone who could ask the original question. Was there something missing? $\endgroup$ – John Hughes May 25 '14 at 17:40

The maps $f^\star$ and $f^{\star\star}$ are the dual and double-dual of $f$, respectively. So, "Yes, they're related to $f$".

In particular, for $\phi$ in $W^\star$, the definition of $f^\star$ is something like this:

$f^\star(\phi)$ is supposed to be an element of $V^\star$, so it takes elements of $V$ to real numbers. So for $v \in V$, what's $f^\star(\phi)(v)$? it's just $$ f^\star(\phi)(v) = \phi(f(v)). $$

The double dual is this applied twice.

You could assume bases for $V$ and $W$, but it's not needed, as Hagen suggests.

For the second part, I believe that one of the two following approaches will get you somewhere:

  1. Pick $V = W = \mathbb R^2$. Go ahead and pick bases, and suppose that you've got the isomorphisms. Now consider maps $f$ of the form $$ \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \\ \text { and } \\ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}. $$

  2. Pick $ V = \mathbb R$ and $W = \mathbb R^2$ (or vice versa) and consider either the maps $1 \mapsto a e_1 + b e_2$ or the maps $a e_1 + b e_2 \mapsto ap + bq$, accordingly. I think that one of these will lead you to a contradiction, but I'm not certain. Addendum: solution to part 2: $$ \newcommand{\vstar}{V^{\star}} \newcommand{\wstar}{W^{\star}} \newcommand{\fstar}{f^{\star}} \newcommand{\iv}{i_V} \newcommand{\iw}{i_W} \newcommand{\Rtwo}{{\mathbb R^2}} \newcommand{\aij}{a_{ij}} \newcommand{\bij}{b_{ij}} \newcommand{\cij}{c_{ij}} $$ $$ \newcommand{\Amat}{\mathbf A} \newcommand{\Bmat}{\mathbf B} \newcommand{\Cmat}{\mathbf C} \newcommand{\vvec}{\mathbf v} \newcommand{\evec}{\mathbf e} \newcommand{\ei}{\mathbf e_i} \newcommand{\ej}{\mathbf e_j} \newcommand{\ek}{\mathbf e_k} \newcommand{\el}{\mathbf e_\ell} $$

For part 2, let’s work with $V = W = \Rtwo$ and the standard basis $\evec_1, \evec_2$ for $\Rtwo$. All indices in what follows will be over the set $\{1, 2\}$, so when I write “Let $\phi_i = \iv(\ei)$”, I mean “Let $\phi_1 = \iv(\evec_1)$ and $\phi_2 = \iv(\evec_2)$.” Clear?

I need one small observation before I start: for any nonzero vector $\vvec \in V = \Rtwo$, there’s a functional $\phi$ with $\phi(\vvec) \ne 0$. (For instance, using dot-products, we could define $\phi(\mathbf x) = \vvec \cdot \mathbf x$.)

The proof is by contradiction, i.e., I’ll assume the existence of $i_V$ and $i_W$ and derive a contradiction.

OK. Let $$ \phi_i = \iv(\ei) \\ \psi_i = \iw(\ei). $$ These constitute bases $\Phi$ and $\Psi$ for $\vstar$ and $\wstar$, respectively. Why are they bases? Because each is the image, under an isomorphism, of a basis for $\Rtwo$.

Let $$ a_{ij} = \phi_i (\ej). $$ Then the numbers $a_{ij}$ are the entries of a matrix $\Amat$, which is nonsingular. Proof: suppose that $\Amat\vvec = 0$, but $\vvec \ne 0$. Then for each $i$, $$ \sum_j a_{ij}v_j = 0 \\ \sum_j \phi_i(\ej)v_j = 0 \\ \phi_i(\sum_j v_j\ej) = 0 \\ \phi_i(\vvec) = 0 $$ In other words, both $\phi_1$ and $\phi_2$ kill $\vvec$. Since the $\phi$s form a basis, every element of $\vstar$ would kill $\vvec$. This contradicts the observation above.

Similarly, the numbers $\bij = \psi_i(\ej)$ form a nonsingular matrix $\Bmat$.

Finally, we can write $f(\ei) = \sum_j \cij \ej$. The matrix $C$ for the transformation $f$ is not necessarily nonsingular, however.

With these conventions, we can say something about the matrices of $\iv, \iw, f, $ and $\fstar$ with respect to the various bases. I’ll write $T|_{K,L}$ for the matrix of the transformation $T$ with repect to the bases $K$ and $L$ of its domain and codomain, respectively. I’ll use $E$ to denote the basis $\{\evec_1, \evec_2 \}$ of $V = W = \Rtwo$.

The definitions of $\phi$ and $\psi$ show that $$ \iv|_{E, \Phi} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\ \iw|_{E, \Psi} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\ $$ and the matrix of $f$ is $$ f|_{E, E} = \Cmat. $$

Now, let’s study $\fstar( \iw(f(\ei) )$, i.e., the image of $\ei$ under the left, bottom, and right maps in the diagram

$$ \begin{matrix} V & \stackrel{i_V}{\longrightarrow}& V^{\star}\\ f\downarrow & & \uparrow f^{\star} \\ W & \stackrel{i_W}{\longrightarrow}& W^{\star}\\ \end{matrix} $$

that we're supposing exists and is commutative.

Now $\fstar( \iw(f(\ei) )$ is an element of $\vstar$, so we can study it by looking at its value on an arbitrary vector $\vvec \in V$, say $\vvec = \sum_k v_k \evec_k$. We get \begin{align} \fstar( \iw(f(\ei) )) (\vvec) &= \iw (f(\ei) ) (f(\vvec)) \text{, by defintion of the dual $\fstar$}\\ &= \iw (\sum_j \cij \ej ) (f(\sum_k v_k \ek)) \\ &= \iw (\sum_j \cij \ej ) (\sum_k v_k f(\ek)) \text{, by linearity of $f$}\\ &= \iw (\sum_j \cij \ej ) (\sum_k v_k \sum_\ell c_{k \ell} \el) \text{, by formula for $f$}\\ &= (\sum_j \cij \psi_j ) (\sum_k v_k \sum_\ell c_{k \ell} \el) \text{, by definition of $\psi_i$}\\ &= \sum_{jk\ell} \cij v_k c_{k\ell} \psi_j (\el) \\ &= \sum_{jk\ell} \cij v_k c_{k\ell} b_{j \ell} \text{, definition of $\bij$}\\ &= \sum_{jk\ell} \cij b_{j \ell}c_{k\ell}v_k \text{, rearrangement of factors} \\ &= ((\Cmat\Bmat\Cmat^{t})\vvec)_i \end{align} so that in summary, we get the $i$th entry of $(\Cmat\Bmat\Cmat^{t})\vvec$.

On the other hand, reading across the top of the diagram, this must be equal to \begin{align} \iv((\ei) ) (\vvec) &= \phi_i (\sum_k v_k \ek) \\ &= \sum_k a_{ik} v_k\\ &= (\Amat \vvec)_i \end{align}

Because this is true for $i = 1, 2$, and for any vector $\vvec$, we must have $$ \Amat = \Cmat\Bmat\Cmat^{t} $$ But $\Cmat$ is the matrix for the (arbitrary) transformation $f$. Picking $f = $identity, we get that $\Amat = \Bmat$, but picking $f = 2\mathbf I$, we get $\Amat = 4 \Bmat$, which can only happen if $\Amat = \Bmat = 0$, which is a contradiction, because both are nonsingular.

I apologize for how convoluted this all is. I’m certain that there’s a simpler proof lurking somewhere in there.

| cite | improve this answer | |
  • $\begingroup$ I get $f^*$ now. However, could you explain more about how to get $f^{**}$, please? Say $\phi\in W^*$, I know that $f^*(\phi)\in V^*$, then what is $f^*(f^*(\phi))$? Does it mean now we should end up with something in $W^{**}$? $\endgroup$ – LaTeXFan May 13 '14 at 7:57
  • $\begingroup$ Your question doesn't make sense: you can't take f^\star(f^\star(\phi))$ $\endgroup$ – John Hughes May 14 '14 at 20:12
  • $\begingroup$ Your question doesn't make sense: you can't take $f^\star(f^\star(\phi))$. I'll start from scratch: we want to look at $f^{\star\star}$. That operates on things in $V^{\star\star}$ to produce things in $W^{\star\star}$. Suppose that $\alpha \in V^{\star\star}$. That means that $\alpha$ consumes objects in $V^{\star}$ and produces numbers. So for a typical element $\phi \in V^{\star}$, we have that $\alpha(\phi)$ is a number. Now we want $f^{\star\star}(\alpha)$, which should be an element of $W^{\star\star}$. So if $\psi \in W^{\star}$, then $f^{\star\star}(\alpha)(\psi)$ should be a number. $\endgroup$ – John Hughes May 14 '14 at 20:18
  • $\begingroup$ Can you use the definition of "star" do describe the number suggested in the last comment? Do so. It'll help you untangle things. Moving on, it's sometimes hard to think of an element of something as abstract as $ V^{\star\star}$, so here's an example: let $v \in V$ be any vector. Then we can define an element $\omega_v \in V^{\star\star}$ by the rule $\omega_v(\phi) = \phi(v)$. In other words, $\omega_v$ is the "evaluate at $v$" function. Slightly surprisingly, the map $v \mapsto \omega_v$ is an isomorphism of vector spaces, regardless of dimension, and uses no bases in its definition. Cool! $\endgroup$ – John Hughes May 14 '14 at 20:22
  • $\begingroup$ I think it should be $f^{**}(\alpha)(\psi) = \alpha(f^*(\psi))$. Right? Then how to get a contradiction, please? $\endgroup$ – LaTeXFan May 15 '14 at 7:24

Here's an improved answer to part 2, at least for the case where $V$ and $W$ are finite dimensional vector spaces of different dimensions. Let's suppose $dim V = n$ and $dim W = k$. You're hoping to have $f^{\star} \circ i_W \circ f = i_V$ for every $f$. Well, if $k < n$, then the rank of $f$ is at most $k$, so the rank of the composite map on the left is at most $k$, while the rank of $i_v$ is $n$. That shows that the existence of $i_V$ and $i_W$ in general is impossible.

Assuming that $i_W$ is supposed to depend only on $W$, and not on $V$, we can always pick $V = W \times \mathbb R$, so that the dimension of $V$ is one greater than the dimension of $W$, so that $i_W$ cannot exist.

| cite | improve this answer | |

The first part is to show that $ V\rightarrow V^{**}$ is an natural transformation and the second to show that $ V\rightarrow V^{*}$ is not a natural transformation.

First Part: Define $i_V : V\rightarrow V^{**}$ by $$i_V(x)(\alpha)=\alpha(x)$$ Notation: $x,y\ldots$ denote vectors in $V, W$ and $\alpha, \beta\ldots$ denote linear functionals in $V^{*}, W^{*}$. So $i_V(x)$ is a functional on $V^{*}$ and its value at $\alpha$ is $\alpha(x)$. Now let $f:V\rightarrow W$ be a linear map then define $f^{*}:W^{*}\rightarrow V^{*}$ by $$f^{*}(\alpha)(x)=\alpha(f(x))$$ where $\alpha \in W^{*}$ and $x\in V$.

The equation to be shown is $$i_W\circ f=f^{**}\circ i_V$$ let $x \in V$ then we need the equality $$i_W(f(x))=f^{**}(i_V(x))$$ in $W^{**}$. So let $\alpha \in W^{*}$ then we need the equality in the base field $$i_W(f(x))(\alpha)=f^{**}(i_V(x))(\alpha)$$ Now for the left side $$i_W(f(x))(\alpha)=\alpha(f(x))$$ is immediate and for the other side,

\begin{equation*} \begin{split} f^{**}(i_V(x))(\alpha)&=i_V(x)(f^{*}(\alpha))\\ &=f^{*}(\alpha)(x)\\ &=\alpha(f(x))\\ \end{split} \end{equation*}

Second Part: Let $i_V : V\rightarrow V^{*}$ and $i_W : V\rightarrow W^{*}$ be given isomorphisms. Then they define non-degenerate bilinear forms on $V$ and $W$

via $$(y,x)=i_V(x)(y)$$ for $x,y \in V$ and similarly for $W$. Conversely an non-degenerate bilinear forms defines an isomorphism with the dual.

Now the equation of interest is

$$f^{*}\circ i_W \circ f=i_V.$$ For $x\in V$ this means $$f^{*}(i_W (f(x)))=i_V(x)$$ holds in $V^*$ which in turn means that for any $y \in V$,

$$f^{*}(i_W (f(x)))(y)=i_V(x)(y)$$ or unraveling the definitions again, $$i_W (f(x))(f(y))=i_V(x)(y).$$

which in terms of inner products means that $$(y,x)=(f(y),f(x)).$$

In other words the desired equation is equivalent to saying that $f$ is an orthogonal transformation. Now under all these translations the naturality statement reads: If $V$ and $W$ are two non-degenerate bilinear spaces and $f:V \rightarrow W$ is a linear map then $f$ is orthogonal.

However we see this now as false, it is easy to construct a counterexample in all but the dimension one case. We need a map $f:V \rightarrow W$ which is not othogonal.

Let us note the following fact relative to any non degenerate bilinear space.

If there is such that $(x,x) \neq 0$ then there is $y$, independent from $x$, with $(x,y)=0$. This follows from the following equation $$(x,y-\frac{(x,y)}{(x,x)}x)=(x,y)-\frac{(x,y)}{(x,x)}(x,x)=0.$$

Now to find $f$ first assume that there is $x \in V$ such that $(x,x)\neq 0$, then we have independent $x,y \in V$ with $(x,y)=0$ now there are $u,w \in W$ such that $(u,w)\neq 0$ by the assumption of non degeneracy. So $f(x)=u$ and $f(y)=w$ is an example of a non orthogonal map.

Now assume that $(x,x)=0$ for all $x \in V$. We have linearly independent $x,y$ such that $(x,y)\neq 0$ (because some such pair exists and they cannot be dependent by the assumption of this case). Now note that there are $u,w \in W$ such that $(u,w)=0$, since either $(u,u)=0$ all $u$ or by the argument above there are two orthogonal vectors. So again $f(x)=u$ and $f(y)=w$ is an example of a non orthogonal map.

| cite | improve this answer | |
  • $\begingroup$ Your inner product's not necessarily symmetric, is it? I think you might want $\langle x, y \rangle = i_V(x)(y) + i_V(y)(x)$ to ensure symmetry. $\endgroup$ – John Hughes May 22 '14 at 20:45
  • $\begingroup$ Now that I think of it, it's not even positive definite, is it? Suppose $V = \mathbb R^2$ and $i_V(x) = \{y \mapsto -x \cdot y \}$, where $\cdot$ is the usual dot-product. Then $i_V(x)(x) = -\|x\|^2 \le 0$ in general. I don't see a simple way to repair this. And if $i_V(x)$ is defined to be dot-product-with-$x$-rotated-90-degrees, then you have $i_V(x)(x) = 0$, which makes it distinctly not PD. $\endgroup$ – John Hughes May 22 '14 at 20:49
  • $\begingroup$ Yes what you say is correct. Inner product was a bad choice of terminology on my part I have edited my post to read bilinear form. I am working over an arbitrary field (maybe I need $\char\neq 2$), so I am not trying to get positive definite. However it is non degenerate meaning that if $(x,y)=0$ for all $y$ then $x=0$, this is a consequence of the isomorphism. $\endgroup$ – Rene Schipperus May 22 '14 at 23:18
  • $\begingroup$ The symmetry trick will not work also since $(x,y)=-(y,x)$ is possible. However my proof is still in all essential points correct (I hope). I have added a more detailed construction of $f$ at the end, which takes into account the problems you bring to my attention. $\endgroup$ – Rene Schipperus May 22 '14 at 23:30
  • $\begingroup$ Right -- I didn't meant to critique the general idea of the proof, which is good, but just bring up a small point. I think that you should also correct the word "orthogonal" to something like "orthogonal with respect to this bilinear form", or perhaps better, just say "We get $(x, y) = (f(x), f(y))$," and then show that this cannot be true for every $f$. [Indeed, that's more or less what my proof did, but in coordinates.] $\endgroup$ – John Hughes May 23 '14 at 0:42

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.