MHB Linear Transformations & Dual Space Problem

[Sudharaka]

Hi everyone, :)

Here's a question and I'll also write down the answer for which I got zero marks. :p I would really appreciate if you can find where I went wrong.

Question: Let \(\phi,\,\psi\in V^{*}\) be two linear functions on a vector space \(V\) such that \(\phi(x)\,\psi(x)=0\) for all \(x\in V\). Prove that either \(\phi=0\mbox{ or }\psi=0\).

Note: \(V^{*}\) is the dual space of \(V\).

My Answer: Note that both \(\phi(x)\) and \(\psi(x)\) are elements of a field (the underlying field of the vector space \(F\)). Since every field is an integral domain it has no zero divisors. Hence, \(\phi(x)\,\psi(x)=0\Rightarrow \phi=0\mbox{ or }\psi=0\).

 

[Evgeny.Makarov]

Assume $\phi(x)\psi(x)=0$ for all $x\in V$, $\phi(x_1)\ne0$ and $\psi(x_2)\ne0$. Show that $\phi(x_2)=0$ and consider $\phi(x_1+x_2)\psi(x_1+x_2)$.

The dual space is not immediately a field. For example, for $V=\mathbb{R}^3$, $V^*$ is also $\mathbb{R}^3$: any $\phi((x_1,x_2,x_3))$ has the form $a_1x_1+a_2x_2+a_3x_3$ for some $a_1,a_2,a_3$.

 

[Sudharaka]

Assume $\phi(x)\psi(x)=0$ for all $x\in V$, $\phi(x_1)\ne0$ and $\psi(x_2)\ne0$. Show that $\phi(x_2)=0$ and consider $\phi(x_1+x_2)\psi(x_1+x_2)$.

The dual space is not immediately a field. For example, for $V=\mathbb{R}^3$, $V^*$ is also $\mathbb{R}^3$: any $\phi((x_1,x_2,x_3))$ has the form $a_1x_1+a_2x_2+a_3x_3$ for some $a_1,a_2,a_3$.

Thanks very much for the reply, but I'm not sure if I am getting you here. I know that the dual space is not a field in general, but the thing is the dual space consist of linear all the transformations from \(V\) to it's underlying field \(F\). That is maps of the form, \(f:\,V\rightarrow F\). So if we take any linear transformation from the dual space, \(V^{*}\equiv L(V,\, F)\), the images of that linear map lie in the field \(F\). Am I correct up to this point? :)

 

[Evgeny.Makarov]

the dual space consist of linear all the transformations from \(V\) to it's underlying field \(F\). That is maps of the form, \(f:\,V\rightarrow F\). So if we take any linear transformation from the dual space, \(V^{*}\equiv L(V,\, F)\), the images of that linear map lie in the field \(F\). Am I correct up to this point?

Yes. This implies, of course, that if $\phi(x)\psi(x)=0$ for some $x$, then either $\phi$ or $\psi$ is 0 on that $x$. Formally,
\[\forall x\,(\phi(x)\psi(x)=0\to\phi(x)=0\lor\psi(x)=0)\]
whereas the question asked to prove
\[(\forall x\,\phi(x)\psi(x)=0)\to(\forall x\,\phi(x)=0)\lor(\forall x\,\psi(x)=0)\]

 

[Sudharaka]

Yes. This implies, of course, that if $\phi(x)\psi(x)=0$ for some $x$, then either $\phi$ or $\psi$ is 0 on that $x$. Formally,
\[\forall x\,(\phi(x)\psi(x)=0\to\phi(x)=0\lor\psi(x)=0)\]
whereas the question asked to prove
\[(\forall x\,\phi(x)\psi(x)=0)\to(\forall x\,\phi(x)=0)\lor(\forall x\,\psi(x)=0)\]

Thanks again. This clarifies everything. I see the difference now. :) Let me write down the proof along the lines you have indicated in post #2.

Let \(\phi(x)\psi(x)=0\) for all \(x\). Assume that there exist \(x_1\) and \(x_2\) such that \(\phi(x_1)\neq 0\) and \(\psi(x_2)\neq 0\). Then we know that, \(\phi(x_2)\psi(x_2)=0\Rightarrow \phi(x_2)=0\) (since \(\psi(x_2)\neq 0\) and \(F\) is a Field so it can't have zero divisors).

Now consider, \(0=\phi(x_1+x_2)\psi(x_1+x_2)= \phi(x_1)\psi(x_1)+\phi(x_2)\psi(x_2)+ \phi(x_1) \psi(x_2)+ \phi(x_2)\psi(x_1)=\phi(x_1) \psi(x_2)\)

Hence we get, \(\phi(x_1) \psi(x_2)=0\) which implies that \(\phi(x_1)=0\mbox{ or }\psi(x_2)=0\) which is a contradiction. Therefore either \(\psi=0\) or \(\phi =0\)

Now I should be correct. Aren't I? :)

 

[Evgeny.Makarov]

Yes, this is correct.

 

[Sudharaka]

Yes, this is correct.

Thanks for all the help you provided and guiding me through the problem. :)