Mixing
T/F: The Range of a Linear Transformation must be a subset of the domain.
$begingroup$
I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.
linear-algebra linear-transformations
asked Jan 29 at 18:03
TrebondTrebond
463
$endgroup$
add a comment |
$begingroup$
I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.
linear-algebra linear-transformations
asked Jan 29 at 18:03
TrebondTrebond
463
$endgroup$
2
$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06
2
$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06
3
$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06
add a comment |
$begingroup$
I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.
linear-algebra linear-transformations
asked Jan 29 at 18:03
TrebondTrebond
463
$endgroup$
I am having difficulty understanding why it is false. As I understand, the range is the output of everything from the domain into the transformation function, or the result of the domain being outputted to the co-domain via the transformation. Essentially, I cannot think of a counterexample where the range of a linear transformation is not in the domain. I would appreciate if anyone could shed some light on this basic concept that I am struggling with.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 29 at 18:03
TrebondTrebond
463
asked Jan 29 at 18:03
TrebondTrebond
463
asked Jan 29 at 18:03
TrebondTrebond
463
asked Jan 29 at 18:03
TrebondTrebond
463
asked Jan 29 at 18:03
TrebondTrebond
463
463
2
$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06
2
$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06
3
$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06
add a comment |
2
$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06
2
$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06
3
$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06
2
2
$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06
$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06
2
2
$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06
$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06
3
3
$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06
$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the following example of linear transformation,
$T:mathbb R to {mathbb R}^2$
$T(x)= (x,0)$
Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
$endgroup$
add a comment |
$begingroup$
Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
Consider the following example of linear transformation,
$T:mathbb R to {mathbb R}^2$
$T(x)= (x,0)$
Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
$endgroup$
add a comment |
$begingroup$
Consider the following example of linear transformation,
$T:mathbb R to {mathbb R}^2$
$T(x)= (x,0)$
Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
$endgroup$
add a comment |
$begingroup$
Consider the following example of linear transformation,
$T:mathbb R to {mathbb R}^2$
$T(x)= (x,0)$
Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
$endgroup$
Consider the following example of linear transformation,
$T:mathbb R to {mathbb R}^2$
$T(x)= (x,0)$
Note that Range of $T$ is {$(x,0):x in mathbb R$} which is not a subset of the domain (i.e. $mathbb R$).
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
answered Jan 29 at 18:15
Mayuresh LMayuresh L
1,331323
1,331323
add a comment |
add a comment |
$begingroup$
Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Take $Fcolonmathbb{R}longrightarrowmathbb{R}[x]$ defined by $F(lambda)=lambda x$. Its range is a set of polynomials, and therefore it is not a subset of $mathbb R$.
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 29 at 18:11
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
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2
$begingroup$
Because the range of any function is always a subset of the codomain. Thus, if you have a linear transformation $;T: Vto W;$ , its range is a subset (in fact, a subspace) of $;W;$ .
$endgroup$
– DonAntonio
Jan 29 at 18:06
2
$begingroup$
The range of a function is a subset of the codomain.
$endgroup$
– stressed out
Jan 29 at 18:06
3
$begingroup$
Think of a linear transformation from $mathbb{R}^4 rightarrow mathbb{R}^3$ through a matrix of transformation $A$. $mathbb{R}^3$ is not a subset of $mathbb{R}^4$.
$endgroup$
– Hyperion
Jan 29 at 18:06