2
$\begingroup$

Let $\phi , \sigma\in V^*$ be linear functionals such that $\phi(v)=0 \ \Rightarrow \sigma(v)=0, \forall v\in V.$ Prove $\sigma=\alpha\phi$ for some scalar $\alpha$.

I can conclude from what's given that $\text{ker}\phi\subseteq\text{ker}\sigma$, but not sure how to continue from here.

Any help appreciated.

  • 0
    Are you sure about the statement? If one takes $\Phi = id$ and $\sigma$ to be a functional with non-zero kernel, the condition is satisfied too, but the conclusion does not hold.. – 2017-02-28
  • 0
    @pepa.dvorak Well, that's how it's stated in my textbook. But I can see what you claim.. – 2017-02-28
  • 0
    Are there some conditions in the previous text? What textbook is it? – 2017-02-28
  • 0
    @pepa.dvorak Nope. It's a textbook in my language. – 2017-02-28
  • 1
    @pepa.dvorak $\Phi = id$ is not a linear functional – 2017-02-28

1 Answers 1

2

The statement trivially holds if $\phi = 0$. For the purposes of the proofs below, we assume this is not the case.

Hint: (if you know about quotient spaces): consider the induced maps $\tilde \phi : V/\ker \phi \to \Bbb F$ and $\tilde \sigma : V / \ker \phi \to \Bbb F$. Note that $\tilde \sigma = \alpha \tilde \phi$, and conclude that $\sigma = \alpha \phi$.

Hint: (alternate approach): consider the linear map $T:V \to \Bbb F^2$ given by $T(v) = (\phi(v),\sigma(v))$. If $\phi(v) = 0 \implies \sigma(v) = 0$, then $T$ fails to be onto. Thus, there exists a non-zero map $f: \Bbb F^2 \to \Bbb F$ (given by $f(x_1,x_1) = a_1x_1 + a_2x_2$) such that $f$ is zero over the image of $T$. Use $a_1,a_2$ to reach the desired conclusion.

  • 0
    Why does $T$ fails to be onto, and why is that important for me? How can I conclude from that there exists a non-zero map? And $a2$ must not be zero, correct? – 2017-02-28
  • 0
    $T$ fails to be onto because, for example, we can never have $T(v) = (0,1)$. This is important because now the image of $T$ is a proper subspace of $\Bbb F^2$, which allows us to conclude the existence of a non-zero map. All of this allows us to reach the fact that for all $v$, we have $$ a_1 \phi(v) + a_2 \sigma(v) = 0 $$ for some non-zero coefficients $a_1,a_2$. I claim that this is enough to reach the desired conclusion. – 2017-02-28
  • 0
    If we reach that point and find that $a_2 = 0$, then we have $a_1\phi(v) = 0$, which is to say that $\phi(v) = 0$, for all $v$. But, we have already discussed the case of $\phi = 0$. – 2017-02-28
  • 0
    I'm sorry, but I must be missing something. If $T$ was onto, then $Im(T)$ would not be a subspace? I'm having hard time understanding why do we need it to conclude there exists a nonzero map. – 2017-02-28
  • 0
    $Im(T)$ would be a subspace, but not the entire space. It would be a one-dimensional subspace of $\Bbb F^2$. So, we can find a functional $f$ whose kernel is precisely $Im(T)$. – 2017-02-28