Can this basic fact in linear algebra be formulated as a completeness result?
Clash Royale CLAN TAG#URR8PPP
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Suppose we have a system of linear equations $S$. If $E$ is some linear equation, we'll say that $E$ is a logical consequence of $S$ if whenever a vector satisfies $S$, it satisfies $E$. Now, in a first course in linear algebra we're taught one simple way to prove that an equation $E$ is a logical consequence of $S$: express it as a linear combination of equations in $S$. But notice that in theory, there could be equations $E$ which are logical consequences of $S$, and yet are not linear combinations of equations in $S$: maybe linear consequence is a strictly stronger notion that mere logical consequence.
Well, it turns out this isn't the case. Taking linear combinations is a sufficiently powerful technique to derive all equations which follow from $S$. Now, this definitely feels like a kind of logical completeness result. We could say that we have the following axioms: the equations of $S$, and the axiom that if two equations are true, any linear combination of them is true. What we've shown is that this set of axioms is complete: it proves all true statements in this formal system.
Can this be formally stated as an actual logical completeness result? Could somebody explain the finer details of how to do that?
linear-algebra logic
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up vote
5
down vote
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Suppose we have a system of linear equations $S$. If $E$ is some linear equation, we'll say that $E$ is a logical consequence of $S$ if whenever a vector satisfies $S$, it satisfies $E$. Now, in a first course in linear algebra we're taught one simple way to prove that an equation $E$ is a logical consequence of $S$: express it as a linear combination of equations in $S$. But notice that in theory, there could be equations $E$ which are logical consequences of $S$, and yet are not linear combinations of equations in $S$: maybe linear consequence is a strictly stronger notion that mere logical consequence.
Well, it turns out this isn't the case. Taking linear combinations is a sufficiently powerful technique to derive all equations which follow from $S$. Now, this definitely feels like a kind of logical completeness result. We could say that we have the following axioms: the equations of $S$, and the axiom that if two equations are true, any linear combination of them is true. What we've shown is that this set of axioms is complete: it proves all true statements in this formal system.
Can this be formally stated as an actual logical completeness result? Could somebody explain the finer details of how to do that?
linear-algebra logic
What end goal would be achieved by obtaining a completeness result beyond what linear combinations would give you?
â ShyPerson
Aug 30 at 3:21
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose we have a system of linear equations $S$. If $E$ is some linear equation, we'll say that $E$ is a logical consequence of $S$ if whenever a vector satisfies $S$, it satisfies $E$. Now, in a first course in linear algebra we're taught one simple way to prove that an equation $E$ is a logical consequence of $S$: express it as a linear combination of equations in $S$. But notice that in theory, there could be equations $E$ which are logical consequences of $S$, and yet are not linear combinations of equations in $S$: maybe linear consequence is a strictly stronger notion that mere logical consequence.
Well, it turns out this isn't the case. Taking linear combinations is a sufficiently powerful technique to derive all equations which follow from $S$. Now, this definitely feels like a kind of logical completeness result. We could say that we have the following axioms: the equations of $S$, and the axiom that if two equations are true, any linear combination of them is true. What we've shown is that this set of axioms is complete: it proves all true statements in this formal system.
Can this be formally stated as an actual logical completeness result? Could somebody explain the finer details of how to do that?
linear-algebra logic
Suppose we have a system of linear equations $S$. If $E$ is some linear equation, we'll say that $E$ is a logical consequence of $S$ if whenever a vector satisfies $S$, it satisfies $E$. Now, in a first course in linear algebra we're taught one simple way to prove that an equation $E$ is a logical consequence of $S$: express it as a linear combination of equations in $S$. But notice that in theory, there could be equations $E$ which are logical consequences of $S$, and yet are not linear combinations of equations in $S$: maybe linear consequence is a strictly stronger notion that mere logical consequence.
Well, it turns out this isn't the case. Taking linear combinations is a sufficiently powerful technique to derive all equations which follow from $S$. Now, this definitely feels like a kind of logical completeness result. We could say that we have the following axioms: the equations of $S$, and the axiom that if two equations are true, any linear combination of them is true. What we've shown is that this set of axioms is complete: it proves all true statements in this formal system.
Can this be formally stated as an actual logical completeness result? Could somebody explain the finer details of how to do that?
linear-algebra logic
edited Aug 29 at 11:53
asked Aug 29 at 9:42
Jack M
17.6k33574
17.6k33574
What end goal would be achieved by obtaining a completeness result beyond what linear combinations would give you?
â ShyPerson
Aug 30 at 3:21
add a comment |Â
What end goal would be achieved by obtaining a completeness result beyond what linear combinations would give you?
â ShyPerson
Aug 30 at 3:21
What end goal would be achieved by obtaining a completeness result beyond what linear combinations would give you?
â ShyPerson
Aug 30 at 3:21
What end goal would be achieved by obtaining a completeness result beyond what linear combinations would give you?
â ShyPerson
Aug 30 at 3:21
add a comment |Â
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What end goal would be achieved by obtaining a completeness result beyond what linear combinations would give you?
â ShyPerson
Aug 30 at 3:21