For subsets $Msubset N$ of a vector space, $langle{Mrangle}subset langle{Nrangle}$.
$begingroup$
Hello I had to prove that $Msubset N$ implies $langle{M}rangle subset langle{M}rangle$ for $M, N$ subsets of a vector space.
My idea looks like the following:
Suppose $ M = {v_1, ..., v_m} $ and $ N = {v_1, ..., v_n} $ $with space 1 leq m leq n space m,n in N$
From that it follows by applying the definition of a linear hull:
$$〈M〉=K cdot v_1 + ... + K cdot v_m$$ and $$〈N〉=K cdot v_1 + ... + K cdot v_n$$
Because m<=n it follows that 〈M〉⊂〈N〉
Apparently that was the wrong way to do it though as i assumed that the sets are finite. Can somebody help me out here? How do you proof this correctly? Thanks in advance!
linear-algebra proof-writing
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
asked Dec 1 '18 at 0:22
D. JohnD. John
283
$endgroup$
add a comment |
$begingroup$
Hello I had to prove that $Msubset N$ implies $langle{M}rangle subset langle{M}rangle$ for $M, N$ subsets of a vector space.
My idea looks like the following:
Suppose $ M = {v_1, ..., v_m} $ and $ N = {v_1, ..., v_n} $ $with space 1 leq m leq n space m,n in N$
From that it follows by applying the definition of a linear hull:
$$〈M〉=K cdot v_1 + ... + K cdot v_m$$ and $$〈N〉=K cdot v_1 + ... + K cdot v_n$$
Because m<=n it follows that 〈M〉⊂〈N〉
Apparently that was the wrong way to do it though as i assumed that the sets are finite. Can somebody help me out here? How do you proof this correctly? Thanks in advance!
linear-algebra proof-writing
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
asked Dec 1 '18 at 0:22
D. JohnD. John
283
$endgroup$
$begingroup$
How is $K$ defined?
$endgroup$
– Anurag A
Dec 1 '18 at 0:28
$begingroup$
K is just a scalar. If my vectorspace was $R^3$ then K would be $R$.
$endgroup$
– D. John
Dec 1 '18 at 0:33
$begingroup$
$M subseteq N subseteq langle N rangle$, so $langle N rangle$ is a subspace containing $M$. Since $langle M rangle$ is the smallest subspace containing $M$ (i.e. it is the intersection of all subspaces containing $M$), we conclude that $langle M rangle subseteq langle N rangle$.
$endgroup$
– Bungo
Dec 1 '18 at 2:13
add a comment |
$begingroup$
Hello I had to prove that $Msubset N$ implies $langle{M}rangle subset langle{M}rangle$ for $M, N$ subsets of a vector space.
My idea looks like the following:
Suppose $ M = {v_1, ..., v_m} $ and $ N = {v_1, ..., v_n} $ $with space 1 leq m leq n space m,n in N$
From that it follows by applying the definition of a linear hull:
$$〈M〉=K cdot v_1 + ... + K cdot v_m$$ and $$〈N〉=K cdot v_1 + ... + K cdot v_n$$
Because m<=n it follows that 〈M〉⊂〈N〉
Apparently that was the wrong way to do it though as i assumed that the sets are finite. Can somebody help me out here? How do you proof this correctly? Thanks in advance!
linear-algebra proof-writing
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
asked Dec 1 '18 at 0:22
D. JohnD. John
283
$endgroup$
Hello I had to prove that $Msubset N$ implies $langle{M}rangle subset langle{M}rangle$ for $M, N$ subsets of a vector space.
My idea looks like the following:
Suppose $ M = {v_1, ..., v_m} $ and $ N = {v_1, ..., v_n} $ $with space 1 leq m leq n space m,n in N$
From that it follows by applying the definition of a linear hull:
$$〈M〉=K cdot v_1 + ... + K cdot v_m$$ and $$〈N〉=K cdot v_1 + ... + K cdot v_n$$
Because m<=n it follows that 〈M〉⊂〈N〉
Apparently that was the wrong way to do it though as i assumed that the sets are finite. Can somebody help me out here? How do you proof this correctly? Thanks in advance!
linear-algebra proof-writing
linear-algebra proof-writing
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
asked Dec 1 '18 at 0:22
D. JohnD. John
283
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
asked Dec 1 '18 at 0:22
D. JohnD. John
283
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
edited Dec 1 '18 at 2:04
anomaly
17.4k42664
17.4k42664
asked Dec 1 '18 at 0:22
D. JohnD. John
283
asked Dec 1 '18 at 0:22
D. JohnD. John
283
asked Dec 1 '18 at 0:22
D. JohnD. John
283
283
$begingroup$
How is $K$ defined?
$endgroup$
– Anurag A
Dec 1 '18 at 0:28
$begingroup$
K is just a scalar. If my vectorspace was $R^3$ then K would be $R$.
$endgroup$
– D. John
Dec 1 '18 at 0:33
$begingroup$
$M subseteq N subseteq langle N rangle$, so $langle N rangle$ is a subspace containing $M$. Since $langle M rangle$ is the smallest subspace containing $M$ (i.e. it is the intersection of all subspaces containing $M$), we conclude that $langle M rangle subseteq langle N rangle$.
$endgroup$
– Bungo
Dec 1 '18 at 2:13
add a comment |
$begingroup$
How is $K$ defined?
$endgroup$
– Anurag A
Dec 1 '18 at 0:28
$begingroup$
K is just a scalar. If my vectorspace was $R^3$ then K would be $R$.
$endgroup$
– D. John
Dec 1 '18 at 0:33
$begingroup$
$M subseteq N subseteq langle N rangle$, so $langle N rangle$ is a subspace containing $M$. Since $langle M rangle$ is the smallest subspace containing $M$ (i.e. it is the intersection of all subspaces containing $M$), we conclude that $langle M rangle subseteq langle N rangle$.
$endgroup$
– Bungo
Dec 1 '18 at 2:13
$begingroup$
How is $K$ defined?
$endgroup$
– Anurag A
Dec 1 '18 at 0:28
$begingroup$
How is $K$ defined?
$endgroup$
– Anurag A
Dec 1 '18 at 0:28
$begingroup$
K is just a scalar. If my vectorspace was $R^3$ then K would be $R$.
$endgroup$
– D. John
Dec 1 '18 at 0:33
$begingroup$
K is just a scalar. If my vectorspace was $R^3$ then K would be $R$.
$endgroup$
– D. John
Dec 1 '18 at 0:33
$begingroup$
$M subseteq N subseteq langle N rangle$, so $langle N rangle$ is a subspace containing $M$. Since $langle M rangle$ is the smallest subspace containing $M$ (i.e. it is the intersection of all subspaces containing $M$), we conclude that $langle M rangle subseteq langle N rangle$.
$endgroup$
– Bungo
Dec 1 '18 at 2:13
$begingroup$
$M subseteq N subseteq langle N rangle$, so $langle N rangle$ is a subspace containing $M$. Since $langle M rangle$ is the smallest subspace containing $M$ (i.e. it is the intersection of all subspaces containing $M$), we conclude that $langle M rangle subseteq langle N rangle$.
$endgroup$
– Bungo
Dec 1 '18 at 2:13
add a comment |
2 Answers
2
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$begingroup$
For a vector space $V$ and a subset $Xsubset V$, the space $langle{X}rangle$ is by definition or construction the minimal subspace of $V$ containing $X$. But $langle{Nrangle}$ then also contains $Msubset N$, and so it must also contain $langle{Mrangle}$.
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
$endgroup$
add a comment |
$begingroup$
Since M $subset$ N, every vector of M is also a vector of N. So a linear combination of elements of M can be viewed as a linear combination of elements in N. Thus $<M> subset<N>.$
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
$endgroup$
add a comment |
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2 Answers
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$begingroup$
For a vector space $V$ and a subset $Xsubset V$, the space $langle{X}rangle$ is by definition or construction the minimal subspace of $V$ containing $X$. But $langle{Nrangle}$ then also contains $Msubset N$, and so it must also contain $langle{Mrangle}$.
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
$endgroup$
add a comment |
$begingroup$
For a vector space $V$ and a subset $Xsubset V$, the space $langle{X}rangle$ is by definition or construction the minimal subspace of $V$ containing $X$. But $langle{Nrangle}$ then also contains $Msubset N$, and so it must also contain $langle{Mrangle}$.
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
$endgroup$
add a comment |
$begingroup$
For a vector space $V$ and a subset $Xsubset V$, the space $langle{X}rangle$ is by definition or construction the minimal subspace of $V$ containing $X$. But $langle{Nrangle}$ then also contains $Msubset N$, and so it must also contain $langle{Mrangle}$.
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
$endgroup$
For a vector space $V$ and a subset $Xsubset V$, the space $langle{X}rangle$ is by definition or construction the minimal subspace of $V$ containing $X$. But $langle{Nrangle}$ then also contains $Msubset N$, and so it must also contain $langle{Mrangle}$.
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
edited Dec 1 '18 at 2:22
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
answered Dec 1 '18 at 2:06
anomalyanomaly
17.4k42664
17.4k42664
add a comment |
add a comment |
$begingroup$
Since M $subset$ N, every vector of M is also a vector of N. So a linear combination of elements of M can be viewed as a linear combination of elements in N. Thus $<M> subset<N>.$
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
$endgroup$
add a comment |
$begingroup$
Since M $subset$ N, every vector of M is also a vector of N. So a linear combination of elements of M can be viewed as a linear combination of elements in N. Thus $<M> subset<N>.$
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
$endgroup$
add a comment |
$begingroup$
Since M $subset$ N, every vector of M is also a vector of N. So a linear combination of elements of M can be viewed as a linear combination of elements in N. Thus $<M> subset<N>.$
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
$endgroup$
Since M $subset$ N, every vector of M is also a vector of N. So a linear combination of elements of M can be viewed as a linear combination of elements in N. Thus $<M> subset<N>.$
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
answered Dec 1 '18 at 1:45
Joel PereiraJoel Pereira
73519
73519
add a comment |
add a comment |
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$begingroup$
How is $K$ defined?
$endgroup$
– Anurag A
Dec 1 '18 at 0:28
$begingroup$
K is just a scalar. If my vectorspace was $R^3$ then K would be $R$.
$endgroup$
– D. John
Dec 1 '18 at 0:33
$begingroup$
$M subseteq N subseteq langle N rangle$, so $langle N rangle$ is a subspace containing $M$. Since $langle M rangle$ is the smallest subspace containing $M$ (i.e. it is the intersection of all subspaces containing $M$), we conclude that $langle M rangle subseteq langle N rangle$.
$endgroup$
– Bungo
Dec 1 '18 at 2:13