If every linear equation of a system is a linear combination of another system and vice-versa, then both...
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I'm working on Hoffman and Kunze's Linear Algebra and encountered this theorem but without proof. I know it might seem trivial to some of you, but as I'm a first-year, I'd like to practice my proof-writing skills. Could you review this proof, please?
Lemma. For every linear equation $j$ with solution set $F$ and system of linear equation $A x = b$ with solution set $G$, where $A$ is an $m times n$ matrix of coefficients, $x$ is an $n times 1$ vector of unknowns and $b$ is an $m times 1$ vector of values, if $j$ is a linear combination of $A x = b$, then $G$ is a subset of $F$.
Proof. Suppose $j$ is a linear combination of $Ax = b$. Then for some $1 times m$ vector $c$, $j$ is $cAx = cb$. By equality, for every $x$, if $A x = b$, then $cAx = cb$. That is, by subset definition, $G$ is a subset of $F$. Q.E.D.
Theorem. For every systems of linear equations $R$ and $S$ with respectively solution sets $U$ and $V$, and arbitrary linear equations $r$ and $s$, if $r$ is a linear combination of $s$ and $s$ is a linear combination of $r$, then $U = V$.
Proof. Suppose $r$ is a linear combination of $s$. Then by Lemma, $V$ is a subset of $U$.
Suppose $s$ is a linear combination of $r$. Then by Lemma, $U$ is a subset of $V$. By set equality definition, $U = V$. Q.E.D.
linear-algebra proof-verification
edited Dec 19 '18 at 16:40
Brahadeesh
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asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
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$begingroup$
I'm working on Hoffman and Kunze's Linear Algebra and encountered this theorem but without proof. I know it might seem trivial to some of you, but as I'm a first-year, I'd like to practice my proof-writing skills. Could you review this proof, please?
Lemma. For every linear equation $j$ with solution set $F$ and system of linear equation $A x = b$ with solution set $G$, where $A$ is an $m times n$ matrix of coefficients, $x$ is an $n times 1$ vector of unknowns and $b$ is an $m times 1$ vector of values, if $j$ is a linear combination of $A x = b$, then $G$ is a subset of $F$.
Proof. Suppose $j$ is a linear combination of $Ax = b$. Then for some $1 times m$ vector $c$, $j$ is $cAx = cb$. By equality, for every $x$, if $A x = b$, then $cAx = cb$. That is, by subset definition, $G$ is a subset of $F$. Q.E.D.
Theorem. For every systems of linear equations $R$ and $S$ with respectively solution sets $U$ and $V$, and arbitrary linear equations $r$ and $s$, if $r$ is a linear combination of $s$ and $s$ is a linear combination of $r$, then $U = V$.
Proof. Suppose $r$ is a linear combination of $s$. Then by Lemma, $V$ is a subset of $U$.
Suppose $s$ is a linear combination of $r$. Then by Lemma, $U$ is a subset of $V$. By set equality definition, $U = V$. Q.E.D.
linear-algebra proof-verification
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
$endgroup$
add a comment |
$begingroup$
I'm working on Hoffman and Kunze's Linear Algebra and encountered this theorem but without proof. I know it might seem trivial to some of you, but as I'm a first-year, I'd like to practice my proof-writing skills. Could you review this proof, please?
Lemma. For every linear equation $j$ with solution set $F$ and system of linear equation $A x = b$ with solution set $G$, where $A$ is an $m times n$ matrix of coefficients, $x$ is an $n times 1$ vector of unknowns and $b$ is an $m times 1$ vector of values, if $j$ is a linear combination of $A x = b$, then $G$ is a subset of $F$.
Proof. Suppose $j$ is a linear combination of $Ax = b$. Then for some $1 times m$ vector $c$, $j$ is $cAx = cb$. By equality, for every $x$, if $A x = b$, then $cAx = cb$. That is, by subset definition, $G$ is a subset of $F$. Q.E.D.
Theorem. For every systems of linear equations $R$ and $S$ with respectively solution sets $U$ and $V$, and arbitrary linear equations $r$ and $s$, if $r$ is a linear combination of $s$ and $s$ is a linear combination of $r$, then $U = V$.
Proof. Suppose $r$ is a linear combination of $s$. Then by Lemma, $V$ is a subset of $U$.
Suppose $s$ is a linear combination of $r$. Then by Lemma, $U$ is a subset of $V$. By set equality definition, $U = V$. Q.E.D.
linear-algebra proof-verification
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
$endgroup$
I'm working on Hoffman and Kunze's Linear Algebra and encountered this theorem but without proof. I know it might seem trivial to some of you, but as I'm a first-year, I'd like to practice my proof-writing skills. Could you review this proof, please?
Lemma. For every linear equation $j$ with solution set $F$ and system of linear equation $A x = b$ with solution set $G$, where $A$ is an $m times n$ matrix of coefficients, $x$ is an $n times 1$ vector of unknowns and $b$ is an $m times 1$ vector of values, if $j$ is a linear combination of $A x = b$, then $G$ is a subset of $F$.
Proof. Suppose $j$ is a linear combination of $Ax = b$. Then for some $1 times m$ vector $c$, $j$ is $cAx = cb$. By equality, for every $x$, if $A x = b$, then $cAx = cb$. That is, by subset definition, $G$ is a subset of $F$. Q.E.D.
Theorem. For every systems of linear equations $R$ and $S$ with respectively solution sets $U$ and $V$, and arbitrary linear equations $r$ and $s$, if $r$ is a linear combination of $s$ and $s$ is a linear combination of $r$, then $U = V$.
Proof. Suppose $r$ is a linear combination of $s$. Then by Lemma, $V$ is a subset of $U$.
Suppose $s$ is a linear combination of $r$. Then by Lemma, $U$ is a subset of $V$. By set equality definition, $U = V$. Q.E.D.
linear-algebra proof-verification
linear-algebra proof-verification
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
edited Dec 19 '18 at 16:40
Brahadeesh
6,22242361
6,22242361
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
asked Sep 28 '17 at 1:39
repetitiousrepetitiverepetitiousrepetitive
213
213
add a comment |
add a comment |
1 Answer
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The lemma and its proof are correct. The statement of the theorem can be improved. For instance, you want to say that $r$ and $s$ are arbitrary linear equations in the systems $R$ and $S$ respectively. Also, you want to say, "If $r$ is a linear combination of $S$ and $s$ is a linear combination of $R$, then $U = V$."
The proof of the theorem also needs a few more ingredients. Here is one way to complete it.
Proof. Suppose $r$ is a linear combination of $S$. Then by Lemma, $V$ is a subset of the set of solutions of $r$, say $F_r$. Since $r$ was an arbitrary linear equation in the system $R$, $V subseteq bigcap_{r in R} F_r = U$. Repeating the argument the other way around, we get $U subseteq V$. Hence, $U = V$.
Just to make a note, this theorem is actually proved by Hoffman and Kunze. The informal discussion (on page 4, 2nd edition) before the statement of Theorem 1 actually constitutes a proof.
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
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The lemma and its proof are correct. The statement of the theorem can be improved. For instance, you want to say that $r$ and $s$ are arbitrary linear equations in the systems $R$ and $S$ respectively. Also, you want to say, "If $r$ is a linear combination of $S$ and $s$ is a linear combination of $R$, then $U = V$."
The proof of the theorem also needs a few more ingredients. Here is one way to complete it.
Proof. Suppose $r$ is a linear combination of $S$. Then by Lemma, $V$ is a subset of the set of solutions of $r$, say $F_r$. Since $r$ was an arbitrary linear equation in the system $R$, $V subseteq bigcap_{r in R} F_r = U$. Repeating the argument the other way around, we get $U subseteq V$. Hence, $U = V$.
Just to make a note, this theorem is actually proved by Hoffman and Kunze. The informal discussion (on page 4, 2nd edition) before the statement of Theorem 1 actually constitutes a proof.
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
$endgroup$
add a comment |
$begingroup$
The lemma and its proof are correct. The statement of the theorem can be improved. For instance, you want to say that $r$ and $s$ are arbitrary linear equations in the systems $R$ and $S$ respectively. Also, you want to say, "If $r$ is a linear combination of $S$ and $s$ is a linear combination of $R$, then $U = V$."
The proof of the theorem also needs a few more ingredients. Here is one way to complete it.
Proof. Suppose $r$ is a linear combination of $S$. Then by Lemma, $V$ is a subset of the set of solutions of $r$, say $F_r$. Since $r$ was an arbitrary linear equation in the system $R$, $V subseteq bigcap_{r in R} F_r = U$. Repeating the argument the other way around, we get $U subseteq V$. Hence, $U = V$.
Just to make a note, this theorem is actually proved by Hoffman and Kunze. The informal discussion (on page 4, 2nd edition) before the statement of Theorem 1 actually constitutes a proof.
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
$endgroup$
add a comment |
$begingroup$
The lemma and its proof are correct. The statement of the theorem can be improved. For instance, you want to say that $r$ and $s$ are arbitrary linear equations in the systems $R$ and $S$ respectively. Also, you want to say, "If $r$ is a linear combination of $S$ and $s$ is a linear combination of $R$, then $U = V$."
The proof of the theorem also needs a few more ingredients. Here is one way to complete it.
Proof. Suppose $r$ is a linear combination of $S$. Then by Lemma, $V$ is a subset of the set of solutions of $r$, say $F_r$. Since $r$ was an arbitrary linear equation in the system $R$, $V subseteq bigcap_{r in R} F_r = U$. Repeating the argument the other way around, we get $U subseteq V$. Hence, $U = V$.
Just to make a note, this theorem is actually proved by Hoffman and Kunze. The informal discussion (on page 4, 2nd edition) before the statement of Theorem 1 actually constitutes a proof.
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
$endgroup$
The lemma and its proof are correct. The statement of the theorem can be improved. For instance, you want to say that $r$ and $s$ are arbitrary linear equations in the systems $R$ and $S$ respectively. Also, you want to say, "If $r$ is a linear combination of $S$ and $s$ is a linear combination of $R$, then $U = V$."
The proof of the theorem also needs a few more ingredients. Here is one way to complete it.
Proof. Suppose $r$ is a linear combination of $S$. Then by Lemma, $V$ is a subset of the set of solutions of $r$, say $F_r$. Since $r$ was an arbitrary linear equation in the system $R$, $V subseteq bigcap_{r in R} F_r = U$. Repeating the argument the other way around, we get $U subseteq V$. Hence, $U = V$.
Just to make a note, this theorem is actually proved by Hoffman and Kunze. The informal discussion (on page 4, 2nd edition) before the statement of Theorem 1 actually constitutes a proof.
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
edited Jan 12 '18 at 13:33
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
answered Jan 12 '18 at 13:26
BrahadeeshBrahadeesh
6,22242361
6,22242361
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