Mixing
Why the dimension of the range of linear map is smaller than the dimension of $W$?
$begingroup$
I am working on the proof of a map to a smaller dimensional space is not injective.
Proof:
Let $T in mathcal{L}(V,W)$. Then
$$text {dim null } T = text {dim } V - text {dim range } T geq text {dim } V - text {dim } W gt 0$$
I don't understand why $text {dim } V - text {dim range } T geq text {dim } V - text {dim } W$. Or why $ text {dim range } T leq text {dim } W$?
I guess it is because as range $T$ is coming from $V$, at most you can get the dimension of $V$. As dim $V$ is less than dim $W$, dim range $T$ is also less than dim $W$. I don't know if it is correct, and some better way to look at it.
linear-algebra linear-transformations
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
asked Jan 22 at 22:43
JOHN JOHN
4209
$endgroup$
add a comment |
$begingroup$
I am working on the proof of a map to a smaller dimensional space is not injective.
Proof:
Let $T in mathcal{L}(V,W)$. Then
$$text {dim null } T = text {dim } V - text {dim range } T geq text {dim } V - text {dim } W gt 0$$
I don't understand why $text {dim } V - text {dim range } T geq text {dim } V - text {dim } W$. Or why $ text {dim range } T leq text {dim } W$?
I guess it is because as range $T$ is coming from $V$, at most you can get the dimension of $V$. As dim $V$ is less than dim $W$, dim range $T$ is also less than dim $W$. I don't know if it is correct, and some better way to look at it.
linear-algebra linear-transformations
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
asked Jan 22 at 22:43
JOHN JOHN
4209
$endgroup$
add a comment |
$begingroup$
I am working on the proof of a map to a smaller dimensional space is not injective.
Proof:
Let $T in mathcal{L}(V,W)$. Then
$$text {dim null } T = text {dim } V - text {dim range } T geq text {dim } V - text {dim } W gt 0$$
I don't understand why $text {dim } V - text {dim range } T geq text {dim } V - text {dim } W$. Or why $ text {dim range } T leq text {dim } W$?
I guess it is because as range $T$ is coming from $V$, at most you can get the dimension of $V$. As dim $V$ is less than dim $W$, dim range $T$ is also less than dim $W$. I don't know if it is correct, and some better way to look at it.
linear-algebra linear-transformations
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
asked Jan 22 at 22:43
JOHN JOHN
4209
$endgroup$
I am working on the proof of a map to a smaller dimensional space is not injective.
Proof:
Let $T in mathcal{L}(V,W)$. Then
$$text {dim null } T = text {dim } V - text {dim range } T geq text {dim } V - text {dim } W gt 0$$
I don't understand why $text {dim } V - text {dim range } T geq text {dim } V - text {dim } W$. Or why $ text {dim range } T leq text {dim } W$?
I guess it is because as range $T$ is coming from $V$, at most you can get the dimension of $V$. As dim $V$ is less than dim $W$, dim range $T$ is also less than dim $W$. I don't know if it is correct, and some better way to look at it.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
asked Jan 22 at 22:43
JOHN JOHN
4209
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
asked Jan 22 at 22:43
JOHN JOHN
4209
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
edited Jan 22 at 22:48
José Carlos Santos
166k22132235
166k22132235
asked Jan 22 at 22:43
JOHN JOHN
4209
asked Jan 22 at 22:43
JOHN JOHN
4209
asked Jan 22 at 22:43
JOHN JOHN
4209
4209
add a comment |
add a comment |
4 Answers
4
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oldest
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$begingroup$
By definition, $text{range} (T)$ is a subspace of $W$.
$$text{range} (T) le W$$
So,
$$text{dim} ,text{range} (T) le text{dim} W$$
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
$endgroup$
add a comment |
$begingroup$
It's because $operatorname{range}T$ is a subspace of $W$. Therefore, $dimoperatorname{range}Tleqslantdim W$.
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
If $text{rank}(T) > dim{W}$, then you can find a linearly independent set in $W$ that has more elements than the dimension of $W$!
answered Jan 23 at 0:07
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
Let me address what I think is giving you more of a problem thinking about this:
In vector spaces we have the notion of a basis; that is a set of vectors which are linearly independent and span the whole space. The smallest cardinality (size) of such a basis, which is invariant to the space, is called the dimension.
Now in your case, once you impose the image of $V$ in $W$ (and get a linear space in itself) and ask what the dimension of this image is, you're asking what is the minimal number of vectors I need to span the image. This is by definition less than the required number to span the space $W$ itself. Is this making the question more intuitive?
answered Jan 22 at 22:54
MariahMariah
1,5611718
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition, $text{range} (T)$ is a subspace of $W$.
$$text{range} (T) le W$$
So,
$$text{dim} ,text{range} (T) le text{dim} W$$
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
$endgroup$
add a comment |
$begingroup$
By definition, $text{range} (T)$ is a subspace of $W$.
$$text{range} (T) le W$$
So,
$$text{dim} ,text{range} (T) le text{dim} W$$
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
$endgroup$
add a comment |
$begingroup$
By definition, $text{range} (T)$ is a subspace of $W$.
$$text{range} (T) le W$$
So,
$$text{dim} ,text{range} (T) le text{dim} W$$
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
$endgroup$
By definition, $text{range} (T)$ is a subspace of $W$.
$$text{range} (T) le W$$
So,
$$text{dim} ,text{range} (T) le text{dim} W$$
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
answered Jan 22 at 22:47
preferred_anonpreferred_anon
13k11743
13k11743
add a comment |
add a comment |
$begingroup$
It's because $operatorname{range}T$ is a subspace of $W$. Therefore, $dimoperatorname{range}Tleqslantdim W$.
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
It's because $operatorname{range}T$ is a subspace of $W$. Therefore, $dimoperatorname{range}Tleqslantdim W$.
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
It's because $operatorname{range}T$ is a subspace of $W$. Therefore, $dimoperatorname{range}Tleqslantdim W$.
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
It's because $operatorname{range}T$ is a subspace of $W$. Therefore, $dimoperatorname{range}Tleqslantdim W$.
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 22:47
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
add a comment |
$begingroup$
If $text{rank}(T) > dim{W}$, then you can find a linearly independent set in $W$ that has more elements than the dimension of $W$!
answered Jan 23 at 0:07
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
If $text{rank}(T) > dim{W}$, then you can find a linearly independent set in $W$ that has more elements than the dimension of $W$!
answered Jan 23 at 0:07
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
If $text{rank}(T) > dim{W}$, then you can find a linearly independent set in $W$ that has more elements than the dimension of $W$!
answered Jan 23 at 0:07
MetricMetric
1,25159
$endgroup$
If $text{rank}(T) > dim{W}$, then you can find a linearly independent set in $W$ that has more elements than the dimension of $W$!
answered Jan 23 at 0:07
MetricMetric
1,25159
edited Jan 23 at 3:12
answered Jan 23 at 0:07
MetricMetric
1,25159
answered Jan 23 at 0:07
MetricMetric
1,25159
answered Jan 23 at 0:07
MetricMetric
1,25159
1,25159
add a comment |
add a comment |
$begingroup$
Let me address what I think is giving you more of a problem thinking about this:
In vector spaces we have the notion of a basis; that is a set of vectors which are linearly independent and span the whole space. The smallest cardinality (size) of such a basis, which is invariant to the space, is called the dimension.
Now in your case, once you impose the image of $V$ in $W$ (and get a linear space in itself) and ask what the dimension of this image is, you're asking what is the minimal number of vectors I need to span the image. This is by definition less than the required number to span the space $W$ itself. Is this making the question more intuitive?
answered Jan 22 at 22:54
MariahMariah
1,5611718
$endgroup$
add a comment |
$begingroup$
Let me address what I think is giving you more of a problem thinking about this:
In vector spaces we have the notion of a basis; that is a set of vectors which are linearly independent and span the whole space. The smallest cardinality (size) of such a basis, which is invariant to the space, is called the dimension.
Now in your case, once you impose the image of $V$ in $W$ (and get a linear space in itself) and ask what the dimension of this image is, you're asking what is the minimal number of vectors I need to span the image. This is by definition less than the required number to span the space $W$ itself. Is this making the question more intuitive?
answered Jan 22 at 22:54
MariahMariah
1,5611718
$endgroup$
add a comment |
$begingroup$
Let me address what I think is giving you more of a problem thinking about this:
In vector spaces we have the notion of a basis; that is a set of vectors which are linearly independent and span the whole space. The smallest cardinality (size) of such a basis, which is invariant to the space, is called the dimension.
Now in your case, once you impose the image of $V$ in $W$ (and get a linear space in itself) and ask what the dimension of this image is, you're asking what is the minimal number of vectors I need to span the image. This is by definition less than the required number to span the space $W$ itself. Is this making the question more intuitive?
answered Jan 22 at 22:54
MariahMariah
1,5611718
$endgroup$
Let me address what I think is giving you more of a problem thinking about this:
In vector spaces we have the notion of a basis; that is a set of vectors which are linearly independent and span the whole space. The smallest cardinality (size) of such a basis, which is invariant to the space, is called the dimension.
Now in your case, once you impose the image of $V$ in $W$ (and get a linear space in itself) and ask what the dimension of this image is, you're asking what is the minimal number of vectors I need to span the image. This is by definition less than the required number to span the space $W$ itself. Is this making the question more intuitive?
answered Jan 22 at 22:54
MariahMariah
1,5611718
answered Jan 22 at 22:54
MariahMariah
1,5611718
answered Jan 22 at 22:54
MariahMariah
1,5611718
answered Jan 22 at 22:54
MariahMariah
1,5611718
1,5611718
add a comment |
add a comment |
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