MHB Show equivalence of propositions

[mathmari]

Hey!

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)

 

[I like Serena]

Hey!

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)

$(2)\Rightarrow (1)$

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

Ah I understood that now! (Nerd)

Is the proof using these pictures formal? (Wondering)

 

[I like Serena]

Ah I understood that now!

Good! (Happy)

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

$(2)\Rightarrow (1)$

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)

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.