High School How Can You Determine if an Operator is Surjective, Injective, or Bijective?

[SeM]

Hi, I found in Kreyszig that if for any $x_1\ and\ x_2\ \in \mathscr{D}(T)$

then an injective operator gives:

$x_1 \ne x_2 \rightarrow Tx_1 \ne Tx_2$

and

$x_1 = x_2 \rightarrow Tx_1 = Tx_2$If one has an operator T, is there an inequality or equality one can deduce from this, in order to check if an operator is surjective/injective or bijective? (In a similar manner to check for boundedness.)

Thanks!

 

[fresh_42]

Hi, I found in Kreyszig that if for any $x_1\ and\ x_2\ \in \mathscr{D}(T)$

then an injective operator gives:

$x_1 \ne x_2 \rightarrow Tx_1 \ne Tx_2$

I wouldn't say gives here, because it is the definition of injectivity:
Different elements map to different points, no multiple hits.

and

$x_1 = x_2 \rightarrow Tx_1 = Tx_2$

This is called well-definition. It distinguishes between functions and relations. Every function has this property. It is not related to injectivity. However, if you turn around the arrow:
$$Tx_1=Tx_2 \Longrightarrow x_1=x_2$$
then it is the definition of an injective function $T$.

If one has an operator T, is there an inequality or equality one can deduce from this, in order to check if an operator is surjective/injective or bijective? (In a similar manner to check for boundedness.)

Thanks!

As this has been already the definition, what other equality are you looking for?

In case of linear functions (operators), which might be given here as you posted in the linear algebra section, then we get:
$$(\;Tx_1=Tx_2 \Longrightarrow x_1=x_2\;) \Longleftrightarrow (\;T(x_1-x_2)=0 \Longrightarrow x_1-x_2=0 \;) \Longleftrightarrow (\;Tx=0 \Longrightarrow x=0\;) \Longleftrightarrow (\;\operatorname{ker}T=0\;)$$
which finally is a single equation.

 

[SeM]

I wouldn't say gives here, because it is the definition of injectivity:

In case of linear functions (operators), which might be given here as you posted in the linear algebra section, then we get:
$$(\;Tx_1=Tx_2 \Longrightarrow x_1=x_2\;) \Longleftrightarrow (\;T(x_1-x_2)=0 \Longrightarrow x_1-x_2=0 \;) \Longleftrightarrow (\;Tx=0 \Longrightarrow x=0\;) \Longleftrightarrow (\;\operatorname{ker}T=0\;)$$
which finally is a single equation.

So if an operator is say id/dx, and the element it acts on is the variable x, we have:

Tx_1 = id/dx x = i

the second variable is y,

Tx_2 = id/dx y = 0

$$(\;Tx \ne Ty \Longrightarrow x \ne y\;) \Longleftrightarrow (\;T(x-y) \ne 0 \Longrightarrow x-y\ne0 \;) \Longleftrightarrow (\;Tx\ne0 \Longrightarrow x\ne0\;) \Longleftrightarrow (\;\operatorname{ker}T\ne0\;)$$

so T is not injective.

 

[fresh_42]

So if an operator is say id/dx, and the element it acts on is the variable x, we have:

Tx_1 = id/dx x = i

the second variable is y,

Tx_2 = id/dx y = 0

$$(\;Tx \ne Ty \Longrightarrow x \ne y\;) \Longleftrightarrow (\;T(x-y) \ne 0 \Longrightarrow x-y\ne0 \;) \Longleftrightarrow (\;Tx\ne0 \Longrightarrow x\ne0\;) \Longleftrightarrow (\;\operatorname{ker}T\ne0\;)$$

so T is not injective.

$T$ is not injective because all polynomials in $y$ are contained in its kernel: $T(p(y))=0$.

However, your negation is wrong.
$$
T \text{ not injective } \Longleftrightarrow \neg \, (Tx_1=Tx_2 \Longrightarrow x_1 = x_2) \Longleftrightarrow (Tx_1=Tx_2 \nRightarrow x_1 = x_2) \Longleftrightarrow \exists \, x_1 \neq x_2 \, : \,Tx_1 =Tx_2
$$
You cannot simply negate the conditions. E.g. if you win in the lottery, you will never work again. The negation is: If you will work in the future, you might or might not have won in the lottery. It simply can't be concluded. What you wrote was: If you don't win in the lottery, you will definitely work again. But this is something different.

 

[mathwonk]

If that implication means that, for all x, if x wins the lottery then x will not work again, then it seems from the previous example that the negation would be there is some x such that x both wins the lottery and then works again afterwards. Does that make sense?

 

[WWGD]

The $A \rightarrow B$ law in T.F Logic has a nice equivalent : " ~A or B " ( sorry , don't know how to Tex Logic signs) , which negates as "A and ~B " where ~ is negation. It is somewhat ( or maybe plenty) artificial , but it is helpful when negating implications. And the proof is also kind of unsatisfactory: just notice the truth-functional equivalence.

 

[fresh_42]

The $A \rightarrow B$ law in T.F Logic has a nice equivalent : " ~A or B " ( sorry , don't know how to Tex Logic signs) , which negates as "A and ~B " where ~ is negation. It is somewhat ( or maybe plenty) artificial , but it is helpful when negating implications. And the proof is also kind of unsatisfactory: just notice the truth-functional equivalence.

\sim or better \lnot

 

[WWGD]

\sim

\Thanks ;).