Mixing
Can zero be any dimension?
$begingroup$
When I am working on the problem:
Suppose $V$ and $W$ are finite-dimensional with $text{dim }V geq text{dim }W geq 2$. Show that ${T in mathcal{L}(V,W): T text{ is not surjective}}$ is not a subspace of $mathcal{L}(V,W)$.
So, I naturally want to check if $0 in {T in mathcal{L}(V,W): T text{ is not surjective}}$. Am I correct?
My first thought was the zero map is surjective because it maps any dimension, say 2, 3, or 4 to zero in 2,3,or 4 dimension. So, range of $T$ is equal to $W$. I think there are some problem in this statement that only compare the dimension to determine whether T and W are equal. But I am not sure.
If range of $T$ is equal to $W$, then it is surjective. So it violates the additive identity properties of suspaces?
linear-algebra linear-transformations
asked Jan 26 at 7:34
JOHN JOHN
4279
$endgroup$
add a comment |
$begingroup$
When I am working on the problem:
Suppose $V$ and $W$ are finite-dimensional with $text{dim }V geq text{dim }W geq 2$. Show that ${T in mathcal{L}(V,W): T text{ is not surjective}}$ is not a subspace of $mathcal{L}(V,W)$.
So, I naturally want to check if $0 in {T in mathcal{L}(V,W): T text{ is not surjective}}$. Am I correct?
My first thought was the zero map is surjective because it maps any dimension, say 2, 3, or 4 to zero in 2,3,or 4 dimension. So, range of $T$ is equal to $W$. I think there are some problem in this statement that only compare the dimension to determine whether T and W are equal. But I am not sure.
If range of $T$ is equal to $W$, then it is surjective. So it violates the additive identity properties of suspaces?
linear-algebra linear-transformations
asked Jan 26 at 7:34
JOHN JOHN
4279
$endgroup$
$begingroup$
The dimension of the range of the zero map is $0$. The dimension of $W$ is not zero. Thus the two can not be equal. Generally speaking the zero map is never surjective if it maps to a nontrivial space.
$endgroup$
– maxmilgram
Jan 26 at 7:45
add a comment |
$begingroup$
When I am working on the problem:
Suppose $V$ and $W$ are finite-dimensional with $text{dim }V geq text{dim }W geq 2$. Show that ${T in mathcal{L}(V,W): T text{ is not surjective}}$ is not a subspace of $mathcal{L}(V,W)$.
So, I naturally want to check if $0 in {T in mathcal{L}(V,W): T text{ is not surjective}}$. Am I correct?
My first thought was the zero map is surjective because it maps any dimension, say 2, 3, or 4 to zero in 2,3,or 4 dimension. So, range of $T$ is equal to $W$. I think there are some problem in this statement that only compare the dimension to determine whether T and W are equal. But I am not sure.
If range of $T$ is equal to $W$, then it is surjective. So it violates the additive identity properties of suspaces?
linear-algebra linear-transformations
asked Jan 26 at 7:34
JOHN JOHN
4279
$endgroup$
When I am working on the problem:
Suppose $V$ and $W$ are finite-dimensional with $text{dim }V geq text{dim }W geq 2$. Show that ${T in mathcal{L}(V,W): T text{ is not surjective}}$ is not a subspace of $mathcal{L}(V,W)$.
So, I naturally want to check if $0 in {T in mathcal{L}(V,W): T text{ is not surjective}}$. Am I correct?
My first thought was the zero map is surjective because it maps any dimension, say 2, 3, or 4 to zero in 2,3,or 4 dimension. So, range of $T$ is equal to $W$. I think there are some problem in this statement that only compare the dimension to determine whether T and W are equal. But I am not sure.
If range of $T$ is equal to $W$, then it is surjective. So it violates the additive identity properties of suspaces?
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 26 at 7:34
JOHN JOHN
4279
asked Jan 26 at 7:34
JOHN JOHN
4279
edited Jan 26 at 7:37
JOHN
asked Jan 26 at 7:34
JOHN JOHN
4279
asked Jan 26 at 7:34
JOHN JOHN
4279
asked Jan 26 at 7:34
JOHN JOHN
4279
4279
$begingroup$
The dimension of the range of the zero map is $0$. The dimension of $W$ is not zero. Thus the two can not be equal. Generally speaking the zero map is never surjective if it maps to a nontrivial space.
$endgroup$
– maxmilgram
Jan 26 at 7:45
add a comment |
$begingroup$
The dimension of the range of the zero map is $0$. The dimension of $W$ is not zero. Thus the two can not be equal. Generally speaking the zero map is never surjective if it maps to a nontrivial space.
$endgroup$
– maxmilgram
Jan 26 at 7:45
$begingroup$
The dimension of the range of the zero map is $0$. The dimension of $W$ is not zero. Thus the two can not be equal. Generally speaking the zero map is never surjective if it maps to a nontrivial space.
$endgroup$
– maxmilgram
Jan 26 at 7:45
$begingroup$
The dimension of the range of the zero map is $0$. The dimension of $W$ is not zero. Thus the two can not be equal. Generally speaking the zero map is never surjective if it maps to a nontrivial space.
$endgroup$
– maxmilgram
Jan 26 at 7:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The range of the zero map ${0}$ and therefore it is surjective when and only when $W={0}$. So, since you are assuming that $dim Wgeqslant 2$, the zero map is not surjective.
Concerning the question from the title of your question: $dim{0}=0$.
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
2
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
add a comment |
$begingroup$
The zero map from $mathbb{R}^{dim V}$ to $mathbb{R}^{dim W}$ is not surjective, unless $dim W=0$. So $0$ is in the subset you are examining. And this is not the avenue to pursue for proving this subset is not a subspace.
Instead, you can show a lack of additive closure. It will be possible to take two non-surjective transformations, add them together, and get a surjective transformation.
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
The range of the zero map ${0}$ and therefore it is surjective when and only when $W={0}$. So, since you are assuming that $dim Wgeqslant 2$, the zero map is not surjective.
Concerning the question from the title of your question: $dim{0}=0$.
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
2
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
add a comment |
$begingroup$
The range of the zero map ${0}$ and therefore it is surjective when and only when $W={0}$. So, since you are assuming that $dim Wgeqslant 2$, the zero map is not surjective.
Concerning the question from the title of your question: $dim{0}=0$.
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
2
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
add a comment |
$begingroup$
The range of the zero map ${0}$ and therefore it is surjective when and only when $W={0}$. So, since you are assuming that $dim Wgeqslant 2$, the zero map is not surjective.
Concerning the question from the title of your question: $dim{0}=0$.
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
The range of the zero map ${0}$ and therefore it is surjective when and only when $W={0}$. So, since you are assuming that $dim Wgeqslant 2$, the zero map is not surjective.
Concerning the question from the title of your question: $dim{0}=0$.
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
answered Jan 26 at 7:43
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
2
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
add a comment |
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
2
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
$begingroup$
So, the dimension of zero vector can only be $0$? e.g. dimension of $(0,0,0,0)$ is $0$?
$endgroup$
– JOHN
Jan 26 at 7:51
2
2
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
$begingroup$
A vector has no dimension. Vector spaces do.
$endgroup$
– José Carlos Santos
Jan 26 at 7:54
add a comment |
$begingroup$
The zero map from $mathbb{R}^{dim V}$ to $mathbb{R}^{dim W}$ is not surjective, unless $dim W=0$. So $0$ is in the subset you are examining. And this is not the avenue to pursue for proving this subset is not a subspace.
Instead, you can show a lack of additive closure. It will be possible to take two non-surjective transformations, add them together, and get a surjective transformation.
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
The zero map from $mathbb{R}^{dim V}$ to $mathbb{R}^{dim W}$ is not surjective, unless $dim W=0$. So $0$ is in the subset you are examining. And this is not the avenue to pursue for proving this subset is not a subspace.
Instead, you can show a lack of additive closure. It will be possible to take two non-surjective transformations, add them together, and get a surjective transformation.
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
The zero map from $mathbb{R}^{dim V}$ to $mathbb{R}^{dim W}$ is not surjective, unless $dim W=0$. So $0$ is in the subset you are examining. And this is not the avenue to pursue for proving this subset is not a subspace.
Instead, you can show a lack of additive closure. It will be possible to take two non-surjective transformations, add them together, and get a surjective transformation.
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
$endgroup$
The zero map from $mathbb{R}^{dim V}$ to $mathbb{R}^{dim W}$ is not surjective, unless $dim W=0$. So $0$ is in the subset you are examining. And this is not the avenue to pursue for proving this subset is not a subspace.
Instead, you can show a lack of additive closure. It will be possible to take two non-surjective transformations, add them together, and get a surjective transformation.
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
answered Jan 26 at 8:20
alex.jordanalex.jordan
39.5k560122
39.5k560122
add a comment |
add a comment |
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$begingroup$
The dimension of the range of the zero map is $0$. The dimension of $W$ is not zero. Thus the two can not be equal. Generally speaking the zero map is never surjective if it maps to a nontrivial space.
$endgroup$
– maxmilgram
Jan 26 at 7:45