Mixing
Theoretical question about span and dimension
$begingroup$
I have a theoretical question about Span, based on a question in my textbook. My answer started the same as that given in the answer key, but then diverges. I am not sure if my original answer is valid, but it is important to know, so that I gain deeper understanding of the subject matter.
$V$ is a vector space over field $F$
$A={u_1,u_2,u_3} subseteq V$, and $A$ is linearly independent.
$C={u_1+u_2,u_2+u_3,u_3+u_1} subseteq V$
I am asked to prove that $SpA = SpC$.
Clearly the elements in C are linear combinations of the elements in $A$, and therefore part of $SpA$, and we can easily show that any linear combination of C is also a part of $SpA$ (I have omitted the details/proof, because they are self-evident).
The answer in the textbook then says that $dimSpA=3$ (because the elements in A are linearly independent), and it proves that the elements in C are linearly independent, thus $dimSpC=3$. And as $dimSpA=dimSpC$, $SpC = SpA$.
My own answer continued differently:
I showed that all elements in A are linear combinations of the elements in C:
$(u_1=frac{1}{2}(u_1+u_2)-frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$,
$(u_2=frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)-frac{1}{2}(u_3+u_1)$
$(u_3=-frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$
Therefore, $A subseteq SpC$, and thus per definition all linear combinations of $A$ are a part of $SpC$, i.e. $SpA subseteq SpC$. We already proved that $spC subseteq SpA$. If $spC subseteq SpA$ and $SpA subseteq SpC$, then $SpA = SpC$.
Is my reasoning valid, and if not, why not?
Thank you!
linear-algebra
edited Jan 28 at 11:40
Peter
49.1k1240136
asked Jan 28 at 11:31
daltadalta
1508
$endgroup$
add a comment |
$begingroup$
I have a theoretical question about Span, based on a question in my textbook. My answer started the same as that given in the answer key, but then diverges. I am not sure if my original answer is valid, but it is important to know, so that I gain deeper understanding of the subject matter.
$V$ is a vector space over field $F$
$A={u_1,u_2,u_3} subseteq V$, and $A$ is linearly independent.
$C={u_1+u_2,u_2+u_3,u_3+u_1} subseteq V$
I am asked to prove that $SpA = SpC$.
Clearly the elements in C are linear combinations of the elements in $A$, and therefore part of $SpA$, and we can easily show that any linear combination of C is also a part of $SpA$ (I have omitted the details/proof, because they are self-evident).
The answer in the textbook then says that $dimSpA=3$ (because the elements in A are linearly independent), and it proves that the elements in C are linearly independent, thus $dimSpC=3$. And as $dimSpA=dimSpC$, $SpC = SpA$.
My own answer continued differently:
I showed that all elements in A are linear combinations of the elements in C:
$(u_1=frac{1}{2}(u_1+u_2)-frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$,
$(u_2=frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)-frac{1}{2}(u_3+u_1)$
$(u_3=-frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$
Therefore, $A subseteq SpC$, and thus per definition all linear combinations of $A$ are a part of $SpC$, i.e. $SpA subseteq SpC$. We already proved that $spC subseteq SpA$. If $spC subseteq SpA$ and $SpA subseteq SpC$, then $SpA = SpC$.
Is my reasoning valid, and if not, why not?
Thank you!
linear-algebra
edited Jan 28 at 11:40
Peter
49.1k1240136
asked Jan 28 at 11:31
daltadalta
1508
$endgroup$
1
$begingroup$
Your solution is valid as well.
$endgroup$
– Peter
Jan 28 at 11:38
$begingroup$
@peter Sorry: my mistake… Initially this was a much longer question involving other sets of vectors as well. I got those parts of the question right, so I left those sets of vectors out, but got confused in the process… I have edited accordingly.
$endgroup$
– dalta
Jan 28 at 11:40
add a comment |
$begingroup$
I have a theoretical question about Span, based on a question in my textbook. My answer started the same as that given in the answer key, but then diverges. I am not sure if my original answer is valid, but it is important to know, so that I gain deeper understanding of the subject matter.
$V$ is a vector space over field $F$
$A={u_1,u_2,u_3} subseteq V$, and $A$ is linearly independent.
$C={u_1+u_2,u_2+u_3,u_3+u_1} subseteq V$
I am asked to prove that $SpA = SpC$.
Clearly the elements in C are linear combinations of the elements in $A$, and therefore part of $SpA$, and we can easily show that any linear combination of C is also a part of $SpA$ (I have omitted the details/proof, because they are self-evident).
The answer in the textbook then says that $dimSpA=3$ (because the elements in A are linearly independent), and it proves that the elements in C are linearly independent, thus $dimSpC=3$. And as $dimSpA=dimSpC$, $SpC = SpA$.
My own answer continued differently:
I showed that all elements in A are linear combinations of the elements in C:
$(u_1=frac{1}{2}(u_1+u_2)-frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$,
$(u_2=frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)-frac{1}{2}(u_3+u_1)$
$(u_3=-frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$
Therefore, $A subseteq SpC$, and thus per definition all linear combinations of $A$ are a part of $SpC$, i.e. $SpA subseteq SpC$. We already proved that $spC subseteq SpA$. If $spC subseteq SpA$ and $SpA subseteq SpC$, then $SpA = SpC$.
Is my reasoning valid, and if not, why not?
Thank you!
linear-algebra
edited Jan 28 at 11:40
Peter
49.1k1240136
asked Jan 28 at 11:31
daltadalta
1508
$endgroup$
I have a theoretical question about Span, based on a question in my textbook. My answer started the same as that given in the answer key, but then diverges. I am not sure if my original answer is valid, but it is important to know, so that I gain deeper understanding of the subject matter.
$V$ is a vector space over field $F$
$A={u_1,u_2,u_3} subseteq V$, and $A$ is linearly independent.
$C={u_1+u_2,u_2+u_3,u_3+u_1} subseteq V$
I am asked to prove that $SpA = SpC$.
Clearly the elements in C are linear combinations of the elements in $A$, and therefore part of $SpA$, and we can easily show that any linear combination of C is also a part of $SpA$ (I have omitted the details/proof, because they are self-evident).
The answer in the textbook then says that $dimSpA=3$ (because the elements in A are linearly independent), and it proves that the elements in C are linearly independent, thus $dimSpC=3$. And as $dimSpA=dimSpC$, $SpC = SpA$.
My own answer continued differently:
I showed that all elements in A are linear combinations of the elements in C:
$(u_1=frac{1}{2}(u_1+u_2)-frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$,
$(u_2=frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)-frac{1}{2}(u_3+u_1)$
$(u_3=-frac{1}{2}(u_1+u_2)+frac{1}{2}(u_2+u_3)+frac{1}{2}(u_3+u_1)$
Therefore, $A subseteq SpC$, and thus per definition all linear combinations of $A$ are a part of $SpC$, i.e. $SpA subseteq SpC$. We already proved that $spC subseteq SpA$. If $spC subseteq SpA$ and $SpA subseteq SpC$, then $SpA = SpC$.
Is my reasoning valid, and if not, why not?
Thank you!
linear-algebra
linear-algebra
edited Jan 28 at 11:40
Peter
49.1k1240136
asked Jan 28 at 11:31
daltadalta
1508
edited Jan 28 at 11:40
Peter
49.1k1240136
asked Jan 28 at 11:31
daltadalta
1508
edited Jan 28 at 11:40
Peter
49.1k1240136
edited Jan 28 at 11:40
Peter
49.1k1240136
edited Jan 28 at 11:40
Peter
49.1k1240136
49.1k1240136
asked Jan 28 at 11:31
daltadalta
1508
asked Jan 28 at 11:31
daltadalta
1508
asked Jan 28 at 11:31
daltadalta
1508
1508
1
$begingroup$
Your solution is valid as well.
$endgroup$
– Peter
Jan 28 at 11:38
$begingroup$
@peter Sorry: my mistake… Initially this was a much longer question involving other sets of vectors as well. I got those parts of the question right, so I left those sets of vectors out, but got confused in the process… I have edited accordingly.
$endgroup$
– dalta
Jan 28 at 11:40
add a comment |
1
$begingroup$
Your solution is valid as well.
$endgroup$
– Peter
Jan 28 at 11:38
$begingroup$
@peter Sorry: my mistake… Initially this was a much longer question involving other sets of vectors as well. I got those parts of the question right, so I left those sets of vectors out, but got confused in the process… I have edited accordingly.
$endgroup$
– dalta
Jan 28 at 11:40
1
1
$begingroup$
Your solution is valid as well.
$endgroup$
– Peter
Jan 28 at 11:38
$begingroup$
Your solution is valid as well.
$endgroup$
– Peter
Jan 28 at 11:38
$begingroup$
@peter Sorry: my mistake… Initially this was a much longer question involving other sets of vectors as well. I got those parts of the question right, so I left those sets of vectors out, but got confused in the process… I have edited accordingly.
$endgroup$
– dalta
Jan 28 at 11:40
$begingroup$
@peter Sorry: my mistake… Initially this was a much longer question involving other sets of vectors as well. I got those parts of the question right, so I left those sets of vectors out, but got confused in the process… I have edited accordingly.
$endgroup$
– dalta
Jan 28 at 11:40
add a comment |
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1
$begingroup$
Your solution is valid as well.
$endgroup$
– Peter
Jan 28 at 11:38
$begingroup$
@peter Sorry: my mistake… Initially this was a much longer question involving other sets of vectors as well. I got those parts of the question right, so I left those sets of vectors out, but got confused in the process… I have edited accordingly.
$endgroup$
– dalta
Jan 28 at 11:40