Welcome to our community

Be a part of something great, join today!

[SOLVED] Linear Transformations & Dual Space Problem

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
Yes, this is correct.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621