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Discussion Overview

The discussion revolves around demonstrating the equivalence of two propositions related to a linear map $\Phi$ between finite-dimensional vector spaces. The propositions in question are: (1) $\Phi$ is surjective, and (2) for each 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 scope includes mathematical reasoning and exploration of implications in linear algebra.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • Some participants outline the proof for the direction $(1) \Rightarrow (2)$, asserting that if $\Phi$ is surjective, then any linear form $\beta$ that results in a zero map when composed with $\Phi$ must be the zero map.
  • Others seek clarification on the direction $(2) \Rightarrow (1)$, questioning how to construct a linear form $\beta$ that demonstrates the failure of surjectivity if $\Phi$ is not surjective.
  • A participant suggests that if $\Phi$ is not surjective, there exists a $w \in W$ without a pre-image in $V$, leading to the possibility of choosing a $\beta$ such that $\beta(w) \neq 0$.
  • Another participant emphasizes the need for a non-zero linear form $\beta$ to establish a contradiction if $\beta \circ \Phi$ is the zero map.
  • There is a discussion about the implications of assuming $\beta \circ \Phi$ is the zero map while $\beta$ is not, and whether this leads to a contradiction regarding the surjectivity of $\Phi$.
  • Some participants express confusion about the conditions under which the implications hold, particularly regarding the assumptions made in the proofs.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the direction $(1) \Rightarrow (2)$ but express uncertainty and seek clarification on the direction $(2) \Rightarrow (1)$. The discussion remains unresolved regarding the construction of a suitable linear form $\beta$ to demonstrate the implications.

Contextual Notes

Participants note the importance of selecting appropriate linear forms and the conditions under which the implications hold, highlighting the need for careful reasoning in the proofs. There is also an acknowledgment of potential gaps in understanding the implications of surjectivity and the properties of linear maps.

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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