# [SOLVED]Linear Transformations & Dual Space Problem

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

Here's a question and I'll also write down the answer for which I got zero marks. 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
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
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
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
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
Yes, this is correct.

#### Sudharaka

##### Well-known member
MHB Math Helper
Yes, this is correct.
Thanks for all the help you provided and guiding me through the problem.