Show equivalence of propositions

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SUMMARY

This discussion establishes the equivalence of two propositions regarding a linear map $\Phi: V \rightarrow W$ between finite-dimensional $\mathbb{K}$-vector spaces. The first proposition states that $\Phi$ is surjective, while the second asserts that for every linear form $\beta: W \rightarrow \mathbb{K}$, if $\beta \circ \Phi: V \rightarrow \mathbb{K}$ is the zero map, then $\beta$ must also be the zero map. The proof demonstrates that assuming the non-surjectivity of $\Phi$ leads to a contradiction, thereby confirming the equivalence of the two propositions.

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mathmari
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Hey! :o

Let $\mathbb{K}$ be a field, $V,W$ finite dimensional $\mathbb{K}$-vector spaces and $\Phi:V\rightarrow W$ a linear map.

I want to show that the following two propositions are equivalent:
  1. $\Phi$ is surjective
  2. For each linear form $\beta:W\rightarrow \mathbb{K}$ it holds:
    Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.
I have done the following:

$(1)\Rightarrow (2)$

The map $\Phi$ is surjective.

Let $\beta : W\rightarrow \mathbb{K}$ be a linear form.

If $\beta\circ \Phi:V\rightarrow \mathbb{K}$ is the zero map, then it holds that $(\beta \circ \Phi)(v)=0$ for all $v\in V$.

We assume that $\beta$ is not the zero map. Then there is a $w\in W$ such that $\beta (w)\neq 0$.

Since $\Phi$ is surjective, there is a $v\in V$ with $\Phi (v)=w$, for this $v$ it holds \begin{equation*}\beta (\Phi (v))=\beta (w) \Rightarrow (\beta \circ \Phi )(v)=\beta (w)\Rightarrow 0=\beta (w)\neq 0\end{equation*} a contradiction.

So the assumption is wrong, therefore $\beta$ must be the zero map.


Is this direction correct? Could we improve something? (Wondering)
$(2)\Rightarrow (1)$

Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.

We assume that $\Phi$ is not surjective.

Could you give me a hint for that direction? (Wondering)
 
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mathmari said:
Hey! :o

Let $\mathbb{K}$ be a field, $V,W$ finite dimensional $\mathbb{K}$-vector spaces and $\Phi:V\rightarrow W$ a linear map.

I want to show that the following two propositions are equivalent:
  1. $\Phi$ is surjective
  2. For each linear form $\beta:W\rightarrow \mathbb{K}$ it holds:
    Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.

I have done the following:

$(1)\Rightarrow (2)$

The map $\Phi$ is surjective.

Let $\beta : W\rightarrow \mathbb{K}$ be a linear form.

If $\beta\circ \Phi:V\rightarrow \mathbb{K}$ is the zero map, then it holds that $(\beta \circ \Phi)(v)=0$ for all $v\in V$.

We assume that $\beta$ is not the zero map. Then there is a $w\in W$ such that $\beta (w)\neq 0$.

Since $\Phi$ is surjective, there is a $v\in V$ with $\Phi (v)=w$, for this $v$ it holds \begin{equation*}\beta (\Phi (v))=\beta (w) \Rightarrow (\beta \circ \Phi )(v)=\beta (w)\Rightarrow 0=\beta (w)\neq 0\end{equation*} a contradiction.

So the assumption is wrong, therefore $\beta$ must be the zero map.


Is this direction correct? Could we improve something?

Hey mathmari!

It looks just fine to me. (Nod)

mathmari said:
$(2)\Rightarrow (1)$

Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.

We assume that $\Phi$ is not surjective.

Could you give me a hint for that direction?

If $\Phi$ is not surjective, then there must be a $w$ in $W$ that does not have an original in $V$.
Then we can choose a $\beta$ such that $\beta(w)\ne 0$ can't we? (Thinking)
 
I like Serena said:
If $\Phi$ is not surjective, then there must be a $w$ in $W$ that does not have an original in $V$.
Then we can choose a $\beta$ such that $\beta(w)\ne 0$ can't we? (Thinking)

Why can we do that? (Wondering)
 
mathmari said:
Why can we do that? (Wondering)

We start with: For each linear form β:W→K it holds: Is β∘Φ:V→K the zero map, then β is the zero map.

Now suppose we can find a β such that β∘Φ is the zero map, but β is not.
Doesn't that mean that we have a contradiction meaning the assumption of surjectivity must be wrong? (Wondering)
 
I like Serena said:
We start with: For each linear form β:W→K it holds: Is β∘Φ:V→K the zero map, then β is the zero map.

Now suppose we can find a β such that β∘Φ is the zero map, but β is not.
Doesn't that mean that we have a contradiction meaning the assumption of surjectivity must be wrong? (Wondering)

I got stuck right now. At the direction $(2)\Rightarrow (1)$ do we suppose that the statement (2) "For each linear form β:W→K it holds: Is β∘Φ:V→K the zero map, then β is the zero map." is not true? (Wondering)
 
mathmari said:
I got stuck right now. At the direction $(2)\Rightarrow (1)$ do we suppose that the statement (2) "For each linear form β:W→K it holds: Is β∘Φ:V→K the zero map, then β is the zero map." is not true? (Wondering)

Nope.
We assume that statement (2) is true.
Then we assume that $\Phi$ is not surjective.
If we can prove that that leads to a contradiction in (2), we're done.
And we have a contradiction if we can find a linear β for which the implication is false. (Thinking)
 
I like Serena said:
Nope.
We assume that statement (2) is true.
Then we assume that $\Phi$ is not surjective.
If we can prove that that leads to a contradiction in (2), we're done.
And we have a contradiction if we can find a linear β for which the implication is false. (Thinking)

We assume that $\Phi$ is not surjective, so there is a $w\in W$ that does not have an original in $V$.

That means that $\Phi(v)\neq w$ for all $v\in V$.

Then we apply here $\beta$ and we get $\beta (\Phi (v))\neq \beta (w) \Rightarrow (\beta \circ \Phi) (v)\neq \beta (w)$.

We have that if $\beta\circ \Phi$ is the zero map, then $\beta$ is the zero map.

But if $\beta\circ \Phi$ is the zero map then we also have that $0\neq \beta (w)$ and so $\beta$ is not the zero map. A contradiction. Is everything correct? (Wondering)
 
mathmari said:
We assume that $\Phi$ is not surjective, so there is a $w\in W$ that does not have an original in $V$.

That means that $\Phi(v)\neq w$ for all $v\in V$.

Then we apply here $\beta$ and we get $\beta (\Phi (v))\neq \beta (w) \Rightarrow (\beta \circ \Phi) (v)\neq \beta (w)$.

We can't just say that $\beta (\Phi (v))\neq \beta (w)$ can we?

We do have that $\beta (\Phi (v))=0$, but $\beta (w)$ can still be $0$ as well.
What we need is a $\beta$ that is not the zero-map. (Thinking)
 
I like Serena said:
We do have that $\beta (\Phi (v))=0$, but $\beta (w)$ can still be $0$ as well.

Why does this hold? I got stuck right now. (Wondering)
 
  • #10
mathmari said:
Why does this hold? I got stuck right now. (Wondering)

Let's draw a picture.
\begin{tikzpicture}[>=stealth',
bullet/.style={fill=black, circle, minimum width=1pt, inner sep=1pt},
projection/.style={ ->,thick,shorten <=2pt,shorten >=2pt },
every fit/.style={ellipse,draw,inner sep=0pt}]

%preamble \usepackage{amsfonts}
%preamble \usetikzlibrary{fit,shapes,arrows}

\foreach \y/\l in {0/0, 1/v_1,2/v_2,3/v_3, 4/v_4}
\node[bullet,label=left:$\l$] (v\y) at (0,\y) {};
\foreach \y/\l in {0/0, 1/w_1,2/w_2,3/w_3}
\node[bullet,label=above:$\l$] (w\y) at (4,\y) {};
\foreach \y/\l in {0/0, 1/1,2/2}
\node[bullet,label=right:$\l$] (k\y) at (8,\y) {};

\node[draw,fit=(v0) (v4),minimum width=1.5cm,label=above:$V$] {} ;
\node[draw,fit=(w0) (w3),minimum width=1.5cm,label=above:$W$] {} ;
\node[draw,fit=(k0) (k2),minimum width=1.5cm,label=above:$\mathbb K$] {} ;

\draw[projection] (v4) -- node[above] {$\Phi$} (w1);
\draw[projection] (v3) -- (w1);
\draw[projection] (v2) -- (w1);
\draw[projection] (v1) -- (w0);
\draw[projection] (v0) -- (w0);
\draw[projection] (w3) -- node[above] {$\beta$} (k0);
\draw[projection] (w2) -- (k0);
\draw[projection] (w1) -- (k0);
\draw[projection] (w0) -- (k0);
\end{tikzpicture}
This is an example where $\Phi$ is not surjective and $\beta$ is the zero map.
The set $\{v_i\}$ is a basis of $V$ and $\{w_j\}$ is a basis of $W$.

However, the implication in statement (2) must hold for any $\beta$.
So let's pick a $\beta$ that maps $w_3$ to $1$.
\begin{tikzpicture}[>=stealth',
bullet/.style={fill=black, circle, minimum width=1pt, inner sep=1pt},
projection/.style={ ->,thick,shorten <=2pt,shorten >=2pt },
every fit/.style={ellipse,draw,inner sep=0pt}]

%preamble \usepackage{amsfonts}
%preamble \usetikzlibrary{fit,shapes,arrows}

\foreach \y/\l in {0/0, 1/v_1,2/v_2,3/v_3, 4/v_4}
\node[bullet,label=left:$\l$] (v\y) at (0,\y) {};
\foreach \y/\l in {0/0, 1/w_1,2/w_2,3/w_3}
\node[bullet,label=above:$\l$] (w\y) at (4,\y) {};
\foreach \y/\l in {0/0, 1/1,2/2}
\node[bullet,label=right:$\l$] (k\y) at (8,\y) {};

\node[draw,fit=(v0) (v4),minimum width=1.5cm,label=above:$V$] {} ;
\node[draw,fit=(w0) (w3),minimum width=1.5cm,label=above:$W$] {} ;
\node[draw,fit=(k0) (k2),minimum width=1.5cm,label=above:$\mathbb K$] {} ;

\draw[projection] (v4) -- node[above] {$\Phi$} (w1);
\draw[projection] (v3) -- (w1);
\draw[projection] (v2) -- (w1);
\draw[projection] (v1) -- (w0);
\draw[projection] (v0) -- (w0);
\draw[projection] (w3) -- node[above] {$\beta$} (k1);
\draw[projection] (w2) -- (k0);
\draw[projection] (w1) -- (k0);
\draw[projection] (w0) -- (k0);
\end{tikzpicture}

Now the linear $\beta$ is not the zero map, but we still have that $(\beta\circ\Phi)(V)=\{0\}$ don't we?
Therefore the implication in statement (2) is false.
So it is not for every $\beta$ that the implication is true.
Thus statement (2) is false completing the proof by contradiction. (Thinking)
 
  • #11
Ah I understood that now! (Nerd)

Is the proof using these pictures formal? (Wondering)
 
  • #12
mathmari said:
Ah I understood that now!

Good! (Happy)

mathmari said:
Is the proof using these pictures formal?

I'm afraid not. (Shake)
The proof should be standing on its own.
Pictures should only illustrate what is intended to help people understand a proof. (Nerd)
 
  • #13
I like Serena said:
I'm afraid not. (Shake)
The proof should be standing on its own.
Pictures should only illustrate what is intended to help people understand a proof. (Nerd)

Ah ok.. But how does the formal proof look like in this case? (Wondering)
 
  • #14
mathmari said:
Ah ok.. But how does the formal proof look like in this case? (Wondering)

I think you almost already had it.
Can you finish it? (Wondering)
 
  • #15
I like Serena said:
I think you almost already had it.
Can you finish it? (Wondering)

I got stuck right now. I don't really know how to continue formally as a proof. Could you give me a hint? (Wondering)
 
  • #16
mathmari said:
$(2)\Rightarrow (1)$

Is $\beta\circ\Phi:V\rightarrow \mathbb{K}$ the zero map, then $\beta$ is the zero map.

We assume that $\Phi$ is not surjective.

Could you give me a hint for that direction? (Wondering)

mathmari said:
We assume that $\Phi$ is not surjective, so there is a $w\in W$ that does not have an original in $V$.

That means that $\Phi(v)\neq w$ for all $v\in V$.

Then we apply here $\beta$ and we get $\beta (\Phi (v))\neq \beta (w) \Rightarrow (\beta \circ \Phi) (v)\neq \beta (w)$.

We have that if $\beta\circ \Phi$ is the zero map, then $\beta$ is the zero map.

But if $\beta\circ \Phi$ is the zero map then we also have that $0\neq \beta (w)$ and so $\beta$ is not the zero map. A contradiction. Is everything correct? (Wondering)

How about we make it:

(2)⇒(1)

Is β∘Φ:V→𝕂 the zero map, then β is the zero map.

We assume that $\Phi$ is not surjective, so there is a $w\in W$ that does not have an original in $V$.
That means that $\Phi(v)\neq w$ for all $v\in V$.

We pick $\beta$ such that $\beta (\Phi (V))=\{0\}$ and $\beta(w)\ne 0$.

Consequently the initial implication is false completing the proof by contradiction.
 

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