Mixing
One-dimensional vector space and it's subspaces
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Consider a vector space $V$ , show that: "$dimV=1Leftrightarrow V$ has exactly two subspaces".
Well I can understand that since $dimV=1$ the only subspaces are the zero and it self, but how can i prove that? How can I prove the equivalence? Thanks in advance.
vector-spaces
asked Jan 20 at 18:37
Jim ArtJim Art
5219
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add a comment |
$begingroup$
Consider a vector space $V$ , show that: "$dimV=1Leftrightarrow V$ has exactly two subspaces".
Well I can understand that since $dimV=1$ the only subspaces are the zero and it self, but how can i prove that? How can I prove the equivalence? Thanks in advance.
vector-spaces
asked Jan 20 at 18:37
Jim ArtJim Art
5219
$endgroup$
add a comment |
$begingroup$
Consider a vector space $V$ , show that: "$dimV=1Leftrightarrow V$ has exactly two subspaces".
Well I can understand that since $dimV=1$ the only subspaces are the zero and it self, but how can i prove that? How can I prove the equivalence? Thanks in advance.
vector-spaces
asked Jan 20 at 18:37
Jim ArtJim Art
5219
$endgroup$
Consider a vector space $V$ , show that: "$dimV=1Leftrightarrow V$ has exactly two subspaces".
Well I can understand that since $dimV=1$ the only subspaces are the zero and it self, but how can i prove that? How can I prove the equivalence? Thanks in advance.
vector-spaces
vector-spaces
asked Jan 20 at 18:37
Jim ArtJim Art
5219
asked Jan 20 at 18:37
Jim ArtJim Art
5219
asked Jan 20 at 18:37
Jim ArtJim Art
5219
asked Jan 20 at 18:37
Jim ArtJim Art
5219
asked Jan 20 at 18:37
Jim ArtJim Art
5219
5219
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that dim$V = 1$. Let $V = span{x}$. If $U leq V$ is a subspace, either $U$ is zero, or has a multiple of $x$, in which case by linearity it is $V$. So $V$ can only have $2$ subspaces indeed.
On the other hand, assume that the linear space $V$ has only two subspaces; then naturally these are $V$ and the zero subspace. If dim$V > 1$ then let ${x_n}$ be a linearly independent set of more than 1 vector. Selecting two of these and taking the span of each yields two non-zero subspaces, which must then both be $V$. Hence these aren't linearly independent, and so dim$V = 1$.
answered Jan 20 at 18:45
MariahMariah
1,5561718
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$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
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@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
add a comment |
$begingroup$
Hint:
Its dimension is 1, so it has a basis consisting of one element. You use this to prove the equivalence.
answered Jan 20 at 18:39
MetricMetric
1,23649
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2 Answers
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2 Answers
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$begingroup$
Suppose that dim$V = 1$. Let $V = span{x}$. If $U leq V$ is a subspace, either $U$ is zero, or has a multiple of $x$, in which case by linearity it is $V$. So $V$ can only have $2$ subspaces indeed.
On the other hand, assume that the linear space $V$ has only two subspaces; then naturally these are $V$ and the zero subspace. If dim$V > 1$ then let ${x_n}$ be a linearly independent set of more than 1 vector. Selecting two of these and taking the span of each yields two non-zero subspaces, which must then both be $V$. Hence these aren't linearly independent, and so dim$V = 1$.
answered Jan 20 at 18:45
MariahMariah
1,5561718
$endgroup$
$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
$begingroup$
@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
add a comment |
$begingroup$
Suppose that dim$V = 1$. Let $V = span{x}$. If $U leq V$ is a subspace, either $U$ is zero, or has a multiple of $x$, in which case by linearity it is $V$. So $V$ can only have $2$ subspaces indeed.
On the other hand, assume that the linear space $V$ has only two subspaces; then naturally these are $V$ and the zero subspace. If dim$V > 1$ then let ${x_n}$ be a linearly independent set of more than 1 vector. Selecting two of these and taking the span of each yields two non-zero subspaces, which must then both be $V$. Hence these aren't linearly independent, and so dim$V = 1$.
answered Jan 20 at 18:45
MariahMariah
1,5561718
$endgroup$
$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
$begingroup$
@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
add a comment |
$begingroup$
Suppose that dim$V = 1$. Let $V = span{x}$. If $U leq V$ is a subspace, either $U$ is zero, or has a multiple of $x$, in which case by linearity it is $V$. So $V$ can only have $2$ subspaces indeed.
On the other hand, assume that the linear space $V$ has only two subspaces; then naturally these are $V$ and the zero subspace. If dim$V > 1$ then let ${x_n}$ be a linearly independent set of more than 1 vector. Selecting two of these and taking the span of each yields two non-zero subspaces, which must then both be $V$. Hence these aren't linearly independent, and so dim$V = 1$.
answered Jan 20 at 18:45
MariahMariah
1,5561718
$endgroup$
Suppose that dim$V = 1$. Let $V = span{x}$. If $U leq V$ is a subspace, either $U$ is zero, or has a multiple of $x$, in which case by linearity it is $V$. So $V$ can only have $2$ subspaces indeed.
On the other hand, assume that the linear space $V$ has only two subspaces; then naturally these are $V$ and the zero subspace. If dim$V > 1$ then let ${x_n}$ be a linearly independent set of more than 1 vector. Selecting two of these and taking the span of each yields two non-zero subspaces, which must then both be $V$. Hence these aren't linearly independent, and so dim$V = 1$.
answered Jan 20 at 18:45
MariahMariah
1,5561718
answered Jan 20 at 18:45
MariahMariah
1,5561718
answered Jan 20 at 18:45
MariahMariah
1,5561718
answered Jan 20 at 18:45
MariahMariah
1,5561718
1,5561718
$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
$begingroup$
@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
add a comment |
$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
$begingroup$
@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
$begingroup$
It's clear. Thanks a lot Mariah
$endgroup$
– Jim Art
Jan 21 at 8:45
$begingroup$
@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
$begingroup$
@JimArt you're welcome - you should accept one of the answers if you feel they are adequate
$endgroup$
– Mariah
Jan 21 at 9:50
add a comment |
$begingroup$
Hint:
Its dimension is 1, so it has a basis consisting of one element. You use this to prove the equivalence.
answered Jan 20 at 18:39
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
Hint:
Its dimension is 1, so it has a basis consisting of one element. You use this to prove the equivalence.
answered Jan 20 at 18:39
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
Hint:
Its dimension is 1, so it has a basis consisting of one element. You use this to prove the equivalence.
answered Jan 20 at 18:39
MetricMetric
1,23649
$endgroup$
Hint:
Its dimension is 1, so it has a basis consisting of one element. You use this to prove the equivalence.
answered Jan 20 at 18:39
MetricMetric
1,23649
answered Jan 20 at 18:39
MetricMetric
1,23649
answered Jan 20 at 18:39
MetricMetric
1,23649
answered Jan 20 at 18:39
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
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