Mixing
Show that the range of a linear transformation is a subspace
$begingroup$
Let $T : mathbb V to mathbb W$ be a linear transformation from a vector space $mathbb V$ into a vector space $mathbb W$. Prove that the range of $T$ is a subspace of $mathbb W$.
OK here is my attempt...
If we let $x$ and $y$ be vectors in $mathbb V$, then the transformation of these vectors will look like this... $T(x)$ and $T(y)$.
If we let $mathbb V$ be a vector space in $mathbb R^3$ and $mathbb W$ be a vector space in $mathbb R^2$, then
$$
T begin{pmatrix} x_1\ x_2 \ x_3
end{pmatrix} = Tbegin{pmatrix} x_1 + 2x_2 \ 3x_3 + 4 end{pmatrix}.
$$
Now if we tried to row reduce the matrix $begin{pmatrix} 1 & 2 \ 3 & 4 end{pmatrix}$
we would get $begin{pmatrix} 1 & 2 \ 0 & 1 end{pmatrix} .$
SO the range of $T$ are the linear combinations of the pivot colums of the matrix above.
This is as much as I can do by myself. But now that i think about it, I believe this is wrong because the linear combinations of the pivot columns will give out any vector in $mathbb R^2$, and not in the subspace of $mathbb W$.
Any help will be appreciated.
linear-algebra matrices vector-spaces vectors linear-transformations
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
$endgroup$
add a comment |
$begingroup$
Let $T : mathbb V to mathbb W$ be a linear transformation from a vector space $mathbb V$ into a vector space $mathbb W$. Prove that the range of $T$ is a subspace of $mathbb W$.
OK here is my attempt...
If we let $x$ and $y$ be vectors in $mathbb V$, then the transformation of these vectors will look like this... $T(x)$ and $T(y)$.
If we let $mathbb V$ be a vector space in $mathbb R^3$ and $mathbb W$ be a vector space in $mathbb R^2$, then
$$
T begin{pmatrix} x_1\ x_2 \ x_3
end{pmatrix} = Tbegin{pmatrix} x_1 + 2x_2 \ 3x_3 + 4 end{pmatrix}.
$$
Now if we tried to row reduce the matrix $begin{pmatrix} 1 & 2 \ 3 & 4 end{pmatrix}$
we would get $begin{pmatrix} 1 & 2 \ 0 & 1 end{pmatrix} .$
SO the range of $T$ are the linear combinations of the pivot colums of the matrix above.
This is as much as I can do by myself. But now that i think about it, I believe this is wrong because the linear combinations of the pivot columns will give out any vector in $mathbb R^2$, and not in the subspace of $mathbb W$.
Any help will be appreciated.
linear-algebra matrices vector-spaces vectors linear-transformations
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
$endgroup$
3
$begingroup$
You need to prove this result for general vector spaces $V$ and $W$. Do you know the characterization of a linear subspace?
$endgroup$
– Dave
Apr 7 '18 at 19:25
1
$begingroup$
It's better to use descriptive titles. I edited the title to make it more relevant in search results, but @manooooh changed it back to something less useful.....
$endgroup$
– G Tony Jacobs
Apr 7 '18 at 20:06
add a comment |
$begingroup$
Let $T : mathbb V to mathbb W$ be a linear transformation from a vector space $mathbb V$ into a vector space $mathbb W$. Prove that the range of $T$ is a subspace of $mathbb W$.
OK here is my attempt...
If we let $x$ and $y$ be vectors in $mathbb V$, then the transformation of these vectors will look like this... $T(x)$ and $T(y)$.
If we let $mathbb V$ be a vector space in $mathbb R^3$ and $mathbb W$ be a vector space in $mathbb R^2$, then
$$
T begin{pmatrix} x_1\ x_2 \ x_3
end{pmatrix} = Tbegin{pmatrix} x_1 + 2x_2 \ 3x_3 + 4 end{pmatrix}.
$$
Now if we tried to row reduce the matrix $begin{pmatrix} 1 & 2 \ 3 & 4 end{pmatrix}$
we would get $begin{pmatrix} 1 & 2 \ 0 & 1 end{pmatrix} .$
SO the range of $T$ are the linear combinations of the pivot colums of the matrix above.
This is as much as I can do by myself. But now that i think about it, I believe this is wrong because the linear combinations of the pivot columns will give out any vector in $mathbb R^2$, and not in the subspace of $mathbb W$.
Any help will be appreciated.
linear-algebra matrices vector-spaces vectors linear-transformations
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
$endgroup$
Let $T : mathbb V to mathbb W$ be a linear transformation from a vector space $mathbb V$ into a vector space $mathbb W$. Prove that the range of $T$ is a subspace of $mathbb W$.
OK here is my attempt...
If we let $x$ and $y$ be vectors in $mathbb V$, then the transformation of these vectors will look like this... $T(x)$ and $T(y)$.
If we let $mathbb V$ be a vector space in $mathbb R^3$ and $mathbb W$ be a vector space in $mathbb R^2$, then
$$
T begin{pmatrix} x_1\ x_2 \ x_3
end{pmatrix} = Tbegin{pmatrix} x_1 + 2x_2 \ 3x_3 + 4 end{pmatrix}.
$$
Now if we tried to row reduce the matrix $begin{pmatrix} 1 & 2 \ 3 & 4 end{pmatrix}$
we would get $begin{pmatrix} 1 & 2 \ 0 & 1 end{pmatrix} .$
SO the range of $T$ are the linear combinations of the pivot colums of the matrix above.
This is as much as I can do by myself. But now that i think about it, I believe this is wrong because the linear combinations of the pivot columns will give out any vector in $mathbb R^2$, and not in the subspace of $mathbb W$.
Any help will be appreciated.
linear-algebra matrices vector-spaces vectors linear-transformations
linear-algebra matrices vector-spaces vectors linear-transformations
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
edited Apr 8 '18 at 7:21
Jeppe Stig Nielsen
3,2981123
3,2981123
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
asked Apr 7 '18 at 19:22
Soon_to_be_code_masterSoon_to_be_code_master
387311
387311
3
$begingroup$
You need to prove this result for general vector spaces $V$ and $W$. Do you know the characterization of a linear subspace?
$endgroup$
– Dave
Apr 7 '18 at 19:25
1
$begingroup$
It's better to use descriptive titles. I edited the title to make it more relevant in search results, but @manooooh changed it back to something less useful.....
$endgroup$
– G Tony Jacobs
Apr 7 '18 at 20:06
add a comment |
3
$begingroup$
You need to prove this result for general vector spaces $V$ and $W$. Do you know the characterization of a linear subspace?
$endgroup$
– Dave
Apr 7 '18 at 19:25
1
$begingroup$
It's better to use descriptive titles. I edited the title to make it more relevant in search results, but @manooooh changed it back to something less useful.....
$endgroup$
– G Tony Jacobs
Apr 7 '18 at 20:06
3
3
$begingroup$
You need to prove this result for general vector spaces $V$ and $W$. Do you know the characterization of a linear subspace?
$endgroup$
– Dave
Apr 7 '18 at 19:25
$begingroup$
You need to prove this result for general vector spaces $V$ and $W$. Do you know the characterization of a linear subspace?
$endgroup$
– Dave
Apr 7 '18 at 19:25
1
1
$begingroup$
It's better to use descriptive titles. I edited the title to make it more relevant in search results, but @manooooh changed it back to something less useful.....
$endgroup$
– G Tony Jacobs
Apr 7 '18 at 20:06
$begingroup$
It's better to use descriptive titles. I edited the title to make it more relevant in search results, but @manooooh changed it back to something less useful.....
$endgroup$
– G Tony Jacobs
Apr 7 '18 at 20:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Let $V$ and $W$ be vector spaces over a fixed field $k$ and $T : V to W$ a morphism. What is the range of $T$? We have that $$T(V) = {T(v) : v in V}.$$ Thus we have to show that $T(V) subseteq W$ is a subspace. Do you know what one has to show for a set being a subspace?
Comment. It is a very good idea to get used to a concept by explicitely considering an example like you did and your idea is right.
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
$endgroup$
add a comment |
$begingroup$
The range of $T$ is a set inside the vector space $W$. To prove that a subset of a vector space is a subspace, you need to show that the subset is non-empty and closed under addition and scalar multiplication. Can you see why the range of $T$ is not empty? Now, pick two arbitrary members of the range of $T$. What can you say about them? How can you prove that their sum (well-defined in $W$) is also in the range of $T$? And similarly for the product with a scalar.
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
$endgroup$
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
add a comment |
$begingroup$
$0in mathbb{V}$. Since $T$ is linear, $T(0_mathbb{V})=0_mathbb{W},$ which implies $0_mathbb{W}in R(T).$ Given any $x,yin R(T),$ which implies $v_x,v_yin mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $cin F, T(cv_x+v_y)=cx+y,$ so $cx+yin R(T).$ This complete the proof.
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $V$ and $W$ be vector spaces over a fixed field $k$ and $T : V to W$ a morphism. What is the range of $T$? We have that $$T(V) = {T(v) : v in V}.$$ Thus we have to show that $T(V) subseteq W$ is a subspace. Do you know what one has to show for a set being a subspace?
Comment. It is a very good idea to get used to a concept by explicitely considering an example like you did and your idea is right.
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
$endgroup$
add a comment |
$begingroup$
Hint: Let $V$ and $W$ be vector spaces over a fixed field $k$ and $T : V to W$ a morphism. What is the range of $T$? We have that $$T(V) = {T(v) : v in V}.$$ Thus we have to show that $T(V) subseteq W$ is a subspace. Do you know what one has to show for a set being a subspace?
Comment. It is a very good idea to get used to a concept by explicitely considering an example like you did and your idea is right.
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
$endgroup$
add a comment |
$begingroup$
Hint: Let $V$ and $W$ be vector spaces over a fixed field $k$ and $T : V to W$ a morphism. What is the range of $T$? We have that $$T(V) = {T(v) : v in V}.$$ Thus we have to show that $T(V) subseteq W$ is a subspace. Do you know what one has to show for a set being a subspace?
Comment. It is a very good idea to get used to a concept by explicitely considering an example like you did and your idea is right.
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
$endgroup$
Hint: Let $V$ and $W$ be vector spaces over a fixed field $k$ and $T : V to W$ a morphism. What is the range of $T$? We have that $$T(V) = {T(v) : v in V}.$$ Thus we have to show that $T(V) subseteq W$ is a subspace. Do you know what one has to show for a set being a subspace?
Comment. It is a very good idea to get used to a concept by explicitely considering an example like you did and your idea is right.
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
answered Apr 7 '18 at 19:29
TheGeekGreekTheGeekGreek
5,06331036
5,06331036
add a comment |
add a comment |
$begingroup$
The range of $T$ is a set inside the vector space $W$. To prove that a subset of a vector space is a subspace, you need to show that the subset is non-empty and closed under addition and scalar multiplication. Can you see why the range of $T$ is not empty? Now, pick two arbitrary members of the range of $T$. What can you say about them? How can you prove that their sum (well-defined in $W$) is also in the range of $T$? And similarly for the product with a scalar.
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
$endgroup$
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
add a comment |
$begingroup$
The range of $T$ is a set inside the vector space $W$. To prove that a subset of a vector space is a subspace, you need to show that the subset is non-empty and closed under addition and scalar multiplication. Can you see why the range of $T$ is not empty? Now, pick two arbitrary members of the range of $T$. What can you say about them? How can you prove that their sum (well-defined in $W$) is also in the range of $T$? And similarly for the product with a scalar.
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
$endgroup$
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
add a comment |
$begingroup$
The range of $T$ is a set inside the vector space $W$. To prove that a subset of a vector space is a subspace, you need to show that the subset is non-empty and closed under addition and scalar multiplication. Can you see why the range of $T$ is not empty? Now, pick two arbitrary members of the range of $T$. What can you say about them? How can you prove that their sum (well-defined in $W$) is also in the range of $T$? And similarly for the product with a scalar.
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
$endgroup$
The range of $T$ is a set inside the vector space $W$. To prove that a subset of a vector space is a subspace, you need to show that the subset is non-empty and closed under addition and scalar multiplication. Can you see why the range of $T$ is not empty? Now, pick two arbitrary members of the range of $T$. What can you say about them? How can you prove that their sum (well-defined in $W$) is also in the range of $T$? And similarly for the product with a scalar.
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
answered Apr 7 '18 at 19:30
Jeppe Stig NielsenJeppe Stig Nielsen
3,2981123
3,2981123
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
add a comment |
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
Since i know that $W$ is a subspace, that means that the addition of any two vectors will always result in another vector in the same subspace, the same logic applies with the product of a scalar and any vector in $W$. The range of T is just another subspace INSIDE $W$, If we choose two vectors in the range of T and add them up, then the result will also be in the range because the range is inside a subspace (and all the rules of a subspace apply)..... Is this correct?
$endgroup$
– Soon_to_be_code_master
Apr 7 '18 at 20:37
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
$begingroup$
@Soon_to_be_code_master You do not know (or cannot utilize) that the range is a subspace before you have solved the problem. You pick two vectors in the range. You know these two vectors have a sum in $W$ because you know that $W$ is a vector space. Consider that sum. Prove that it is actually inside the range (for this, you must understand what "range" is). Since your two vectors were arbitrary, then you will have proved that the range is closed under addition. Analogously with scalar multiplication.
$endgroup$
– Jeppe Stig Nielsen
Apr 8 '18 at 6:53
add a comment |
$begingroup$
$0in mathbb{V}$. Since $T$ is linear, $T(0_mathbb{V})=0_mathbb{W},$ which implies $0_mathbb{W}in R(T).$ Given any $x,yin R(T),$ which implies $v_x,v_yin mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $cin F, T(cv_x+v_y)=cx+y,$ so $cx+yin R(T).$ This complete the proof.
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
$endgroup$
add a comment |
$begingroup$
$0in mathbb{V}$. Since $T$ is linear, $T(0_mathbb{V})=0_mathbb{W},$ which implies $0_mathbb{W}in R(T).$ Given any $x,yin R(T),$ which implies $v_x,v_yin mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $cin F, T(cv_x+v_y)=cx+y,$ so $cx+yin R(T).$ This complete the proof.
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
$endgroup$
add a comment |
$begingroup$
$0in mathbb{V}$. Since $T$ is linear, $T(0_mathbb{V})=0_mathbb{W},$ which implies $0_mathbb{W}in R(T).$ Given any $x,yin R(T),$ which implies $v_x,v_yin mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $cin F, T(cv_x+v_y)=cx+y,$ so $cx+yin R(T).$ This complete the proof.
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
$endgroup$
$0in mathbb{V}$. Since $T$ is linear, $T(0_mathbb{V})=0_mathbb{W},$ which implies $0_mathbb{W}in R(T).$ Given any $x,yin R(T),$ which implies $v_x,v_yin mathbb {V}$ s.t. $T(v_x)=x, T(v_y)=y,$ but this implies for any $cin F, T(cv_x+v_y)=cx+y,$ so $cx+yin R(T).$ This complete the proof.
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
answered Apr 8 '18 at 11:13
nCmnCm
1,2891723
1,2891723
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$begingroup$
You need to prove this result for general vector spaces $V$ and $W$. Do you know the characterization of a linear subspace?
$endgroup$
– Dave
Apr 7 '18 at 19:25
1
$begingroup$
It's better to use descriptive titles. I edited the title to make it more relevant in search results, but @manooooh changed it back to something less useful.....
$endgroup$
– G Tony Jacobs
Apr 7 '18 at 20:06