Show equivalence of propositions

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Equivalence
In summary: 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
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

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)
 
Physics news on Phys.org
  • #2
mathmari said:
Hey! :eek:

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)
 
  • #3
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)
 
  • #4
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)
 
  • #5
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)
 
  • #6
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)
 
  • #7
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)
 
  • #8
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)
 
  • #9
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.
 

1. What is the definition of "show equivalence of propositions"?

"Show equivalence of propositions" refers to the process of proving that two logical statements or propositions are equivalent, meaning that they have the same truth value in all possible scenarios.

2. How is the equivalence of propositions determined?

The equivalence of propositions is determined through the use of logical equivalences and truth tables. Logical equivalences are established through the use of logical rules and laws, while truth tables provide a systematic way to evaluate the truth values of propositions in all possible scenarios.

3. Why is it important to show equivalence of propositions?

Showcasing the equivalence of propositions is crucial in logical reasoning and problem solving. It allows for the simplification and transformation of complex propositions, making them easier to understand and work with. It also helps to identify the relationships between different propositions and can aid in finding solutions to logical problems.

4. What are some common methods used to show equivalence of propositions?

Some common methods used to demonstrate the equivalence of propositions include the use of logical equivalences, truth tables, and proof by contradiction. Other techniques such as direct proof and proof by contrapositive may also be used depending on the specific propositions being evaluated.

5. Can two propositions be equivalent even if they have different forms?

Yes, two propositions can be equivalent even if they have different forms. This is because logical equivalences are based on the meaning and relationship between propositions, rather than their specific structure or form. As long as two propositions have the same truth value in all possible scenarios, they can be considered equivalent.

Similar threads

MHB If V=Lin(v_1, .... , v_k) then does it follow that W=Lin(φ(v_1), .... , φ(v_k)) ?
    • Linear and Abstract Algebra
Replies
24
Views
1K
MHB Checking if $\beta \in \text{Isom}(\mathbb{R}^2)$
    • Linear and Abstract Algebra
Replies
15
Views
1K
MHB Kernel of Linear Map: Show $\ker \phi$ Equation
    • Linear and Abstract Algebra
Replies
5
Views
1K
MHB Show that the subspace U is ϕ^z-invariant
    • Linear and Abstract Algebra
Replies
4
Views
841
MHB Show that these elements are linearly independent
    • Linear and Abstract Algebra
Replies
12
Views
1K
A A problem in multilinear algebra
    • Linear and Abstract Algebra
Replies
2
Views
866
MHB Show that φ(v)=λv for a vector v and a coefficient λ
    • Linear and Abstract Algebra
Replies
1
Views
840
MHB Exploring Affine Bases in $\mathbb{R}^n$
    • Linear and Abstract Algebra
Replies
15
Views
1K
MHB Eigenspace is a subspace of V - ψ is diagonalizable
    • Linear and Abstract Algebra
2
Replies
39
Views
2K
MHB Kernel of Linear Map $\theta$ in $\mathbb{F}_{q^n}$
    • Linear and Abstract Algebra
Replies
24
Views
3K

Hot Threads

Recent Insights

Back
Top