Mixing
Vector Subspaces with respect to Linear Functionals
0
$begingroup$
In my Linear Algebra class, my teacher proposed the following: The solution to a linear equation is a vector subspace if and only if the linear equation is homogeneous. My question is if the converse of that statement is true, and if there are any exceptions.
linear-algebra vector-spaces
asked Jan 30 at 18:47
Corsair64Corsair64
1343
$endgroup$
add a comment |
0
$begingroup$
In my Linear Algebra class, my teacher proposed the following: The solution to a linear equation is a vector subspace if and only if the linear equation is homogeneous. My question is if the converse of that statement is true, and if there are any exceptions.
linear-algebra vector-spaces
asked Jan 30 at 18:47
Corsair64Corsair64
1343
$endgroup$
$begingroup$
Wow there's so much wrong with the statement you're reporting your teacher made it's hard to know where to begin with writing an answer here. Linear functionals are by definition linear, and thus homogeneous, something you might call a nonhomogeneous linear functional is probably what is more properly called an affine map. Also linear functionals are not sets, they're functions, although as such they can be identified with sets by identifying them with their graphs. If that's what you mean, then it's probably best to say so explicitly though.
$endgroup$
– jgon
Jan 30 at 18:59
$begingroup$
Also the "converse" of an if and only if statement is equivalent to the original statement, so if an if and only if statement is true, then so is its "converse." Converse in quotes because converses are really for one way implications rather than if and only ifs.
$endgroup$
– jgon
Jan 30 at 19:00
$begingroup$
I'll try to clarify to see if I was just misunderstanding. From what I understand, the solution of a linear function in R^n is a hyperplane. If the constant term of the linear function is zero, then the hyperplane is a subspace of R^n since it preserves vector operations and passes through the origin. I think my teacher may have misused homogenous, as he used it to mean a linear function with zero as the constant term. So, if a linear function has a zero for the constant term, is it's solution necessarily a subspace (of R^n)? And the converse, if the hyperplane is a subspace, is the constant 0
$endgroup$
– Corsair64
Jan 30 at 19:13
1
$begingroup$
I see. It's actually the usage of linear functional that was what was confusing me about your question. I would recommend using "linear equation" instead of linear functional. Here is how I would suggest phrasing the result: "The solutions of a linear equation, $Avec{x}=vec{b}$, form a vector subspace of $Bbb{R}^n$ if and only if the linear equation is homogeneous (i.e. $vec{b}=0$)." I've used vector arrows over the vector variables just to be clear, though usually in mathematics we omit them.
$endgroup$
– jgon
Jan 30 at 19:17
1
$begingroup$
Lastly, if and only if statements don't have converses. If I say $P$ if and only if $Q$, then $P$ and $Q$ are equivalent. This means that they are either both false at the same time or both true at the same time. It is never the case that one is true, but the other isn't. This means that it is also true that $Q$ if and only if $P$ holds. Thus it is also true that a linear equation is homogeneous if and only if its solutions form a vector subspace.
$endgroup$
– jgon
Jan 30 at 19:20
add a comment |
0
0
0
$begingroup$
In my Linear Algebra class, my teacher proposed the following: The solution to a linear equation is a vector subspace if and only if the linear equation is homogeneous. My question is if the converse of that statement is true, and if there are any exceptions.
linear-algebra vector-spaces
asked Jan 30 at 18:47
Corsair64Corsair64
1343
$endgroup$
In my Linear Algebra class, my teacher proposed the following: The solution to a linear equation is a vector subspace if and only if the linear equation is homogeneous. My question is if the converse of that statement is true, and if there are any exceptions.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 30 at 18:47
Corsair64Corsair64
1343
asked Jan 30 at 18:47
Corsair64Corsair64
1343
edited Jan 30 at 19:25
Corsair64
asked Jan 30 at 18:47
Corsair64Corsair64
1343
asked Jan 30 at 18:47
Corsair64Corsair64
1343
asked Jan 30 at 18:47
Corsair64Corsair64
1343
1343
$begingroup$
Wow there's so much wrong with the statement you're reporting your teacher made it's hard to know where to begin with writing an answer here. Linear functionals are by definition linear, and thus homogeneous, something you might call a nonhomogeneous linear functional is probably what is more properly called an affine map. Also linear functionals are not sets, they're functions, although as such they can be identified with sets by identifying them with their graphs. If that's what you mean, then it's probably best to say so explicitly though.
$endgroup$
– jgon
Jan 30 at 18:59
$begingroup$
Also the "converse" of an if and only if statement is equivalent to the original statement, so if an if and only if statement is true, then so is its "converse." Converse in quotes because converses are really for one way implications rather than if and only ifs.
$endgroup$
– jgon
Jan 30 at 19:00
$begingroup$
I'll try to clarify to see if I was just misunderstanding. From what I understand, the solution of a linear function in R^n is a hyperplane. If the constant term of the linear function is zero, then the hyperplane is a subspace of R^n since it preserves vector operations and passes through the origin. I think my teacher may have misused homogenous, as he used it to mean a linear function with zero as the constant term. So, if a linear function has a zero for the constant term, is it's solution necessarily a subspace (of R^n)? And the converse, if the hyperplane is a subspace, is the constant 0
$endgroup$
– Corsair64
Jan 30 at 19:13
1
$begingroup$
I see. It's actually the usage of linear functional that was what was confusing me about your question. I would recommend using "linear equation" instead of linear functional. Here is how I would suggest phrasing the result: "The solutions of a linear equation, $Avec{x}=vec{b}$, form a vector subspace of $Bbb{R}^n$ if and only if the linear equation is homogeneous (i.e. $vec{b}=0$)." I've used vector arrows over the vector variables just to be clear, though usually in mathematics we omit them.
$endgroup$
– jgon
Jan 30 at 19:17
1
$begingroup$
Lastly, if and only if statements don't have converses. If I say $P$ if and only if $Q$, then $P$ and $Q$ are equivalent. This means that they are either both false at the same time or both true at the same time. It is never the case that one is true, but the other isn't. This means that it is also true that $Q$ if and only if $P$ holds. Thus it is also true that a linear equation is homogeneous if and only if its solutions form a vector subspace.
$endgroup$
– jgon
Jan 30 at 19:20
add a comment |
$begingroup$
Wow there's so much wrong with the statement you're reporting your teacher made it's hard to know where to begin with writing an answer here. Linear functionals are by definition linear, and thus homogeneous, something you might call a nonhomogeneous linear functional is probably what is more properly called an affine map. Also linear functionals are not sets, they're functions, although as such they can be identified with sets by identifying them with their graphs. If that's what you mean, then it's probably best to say so explicitly though.
$endgroup$
– jgon
Jan 30 at 18:59
$begingroup$
Also the "converse" of an if and only if statement is equivalent to the original statement, so if an if and only if statement is true, then so is its "converse." Converse in quotes because converses are really for one way implications rather than if and only ifs.
$endgroup$
– jgon
Jan 30 at 19:00
$begingroup$
I'll try to clarify to see if I was just misunderstanding. From what I understand, the solution of a linear function in R^n is a hyperplane. If the constant term of the linear function is zero, then the hyperplane is a subspace of R^n since it preserves vector operations and passes through the origin. I think my teacher may have misused homogenous, as he used it to mean a linear function with zero as the constant term. So, if a linear function has a zero for the constant term, is it's solution necessarily a subspace (of R^n)? And the converse, if the hyperplane is a subspace, is the constant 0
$endgroup$
– Corsair64
Jan 30 at 19:13
1
$begingroup$
I see. It's actually the usage of linear functional that was what was confusing me about your question. I would recommend using "linear equation" instead of linear functional. Here is how I would suggest phrasing the result: "The solutions of a linear equation, $Avec{x}=vec{b}$, form a vector subspace of $Bbb{R}^n$ if and only if the linear equation is homogeneous (i.e. $vec{b}=0$)." I've used vector arrows over the vector variables just to be clear, though usually in mathematics we omit them.
$endgroup$
– jgon
Jan 30 at 19:17
1
$begingroup$
Lastly, if and only if statements don't have converses. If I say $P$ if and only if $Q$, then $P$ and $Q$ are equivalent. This means that they are either both false at the same time or both true at the same time. It is never the case that one is true, but the other isn't. This means that it is also true that $Q$ if and only if $P$ holds. Thus it is also true that a linear equation is homogeneous if and only if its solutions form a vector subspace.
$endgroup$
– jgon
Jan 30 at 19:20
$begingroup$
Wow there's so much wrong with the statement you're reporting your teacher made it's hard to know where to begin with writing an answer here. Linear functionals are by definition linear, and thus homogeneous, something you might call a nonhomogeneous linear functional is probably what is more properly called an affine map. Also linear functionals are not sets, they're functions, although as such they can be identified with sets by identifying them with their graphs. If that's what you mean, then it's probably best to say so explicitly though.
$endgroup$
– jgon
Jan 30 at 18:59
$begingroup$
Wow there's so much wrong with the statement you're reporting your teacher made it's hard to know where to begin with writing an answer here. Linear functionals are by definition linear, and thus homogeneous, something you might call a nonhomogeneous linear functional is probably what is more properly called an affine map. Also linear functionals are not sets, they're functions, although as such they can be identified with sets by identifying them with their graphs. If that's what you mean, then it's probably best to say so explicitly though.
$endgroup$
– jgon
Jan 30 at 18:59
$begingroup$
Also the "converse" of an if and only if statement is equivalent to the original statement, so if an if and only if statement is true, then so is its "converse." Converse in quotes because converses are really for one way implications rather than if and only ifs.
$endgroup$
– jgon
Jan 30 at 19:00
$begingroup$
Also the "converse" of an if and only if statement is equivalent to the original statement, so if an if and only if statement is true, then so is its "converse." Converse in quotes because converses are really for one way implications rather than if and only ifs.
$endgroup$
– jgon
Jan 30 at 19:00
$begingroup$
I'll try to clarify to see if I was just misunderstanding. From what I understand, the solution of a linear function in R^n is a hyperplane. If the constant term of the linear function is zero, then the hyperplane is a subspace of R^n since it preserves vector operations and passes through the origin. I think my teacher may have misused homogenous, as he used it to mean a linear function with zero as the constant term. So, if a linear function has a zero for the constant term, is it's solution necessarily a subspace (of R^n)? And the converse, if the hyperplane is a subspace, is the constant 0
$endgroup$
– Corsair64
Jan 30 at 19:13
$begingroup$
I'll try to clarify to see if I was just misunderstanding. From what I understand, the solution of a linear function in R^n is a hyperplane. If the constant term of the linear function is zero, then the hyperplane is a subspace of R^n since it preserves vector operations and passes through the origin. I think my teacher may have misused homogenous, as he used it to mean a linear function with zero as the constant term. So, if a linear function has a zero for the constant term, is it's solution necessarily a subspace (of R^n)? And the converse, if the hyperplane is a subspace, is the constant 0
$endgroup$
– Corsair64
Jan 30 at 19:13
1
1
$begingroup$
I see. It's actually the usage of linear functional that was what was confusing me about your question. I would recommend using "linear equation" instead of linear functional. Here is how I would suggest phrasing the result: "The solutions of a linear equation, $Avec{x}=vec{b}$, form a vector subspace of $Bbb{R}^n$ if and only if the linear equation is homogeneous (i.e. $vec{b}=0$)." I've used vector arrows over the vector variables just to be clear, though usually in mathematics we omit them.
$endgroup$
– jgon
Jan 30 at 19:17
$begingroup$
I see. It's actually the usage of linear functional that was what was confusing me about your question. I would recommend using "linear equation" instead of linear functional. Here is how I would suggest phrasing the result: "The solutions of a linear equation, $Avec{x}=vec{b}$, form a vector subspace of $Bbb{R}^n$ if and only if the linear equation is homogeneous (i.e. $vec{b}=0$)." I've used vector arrows over the vector variables just to be clear, though usually in mathematics we omit them.
$endgroup$
– jgon
Jan 30 at 19:17
1
1
$begingroup$
Lastly, if and only if statements don't have converses. If I say $P$ if and only if $Q$, then $P$ and $Q$ are equivalent. This means that they are either both false at the same time or both true at the same time. It is never the case that one is true, but the other isn't. This means that it is also true that $Q$ if and only if $P$ holds. Thus it is also true that a linear equation is homogeneous if and only if its solutions form a vector subspace.
$endgroup$
– jgon
Jan 30 at 19:20
$begingroup$
Lastly, if and only if statements don't have converses. If I say $P$ if and only if $Q$, then $P$ and $Q$ are equivalent. This means that they are either both false at the same time or both true at the same time. It is never the case that one is true, but the other isn't. This means that it is also true that $Q$ if and only if $P$ holds. Thus it is also true that a linear equation is homogeneous if and only if its solutions form a vector subspace.
$endgroup$
– jgon
Jan 30 at 19:20
add a comment |
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$begingroup$
Wow there's so much wrong with the statement you're reporting your teacher made it's hard to know where to begin with writing an answer here. Linear functionals are by definition linear, and thus homogeneous, something you might call a nonhomogeneous linear functional is probably what is more properly called an affine map. Also linear functionals are not sets, they're functions, although as such they can be identified with sets by identifying them with their graphs. If that's what you mean, then it's probably best to say so explicitly though.
$endgroup$
– jgon
Jan 30 at 18:59
$begingroup$
Also the "converse" of an if and only if statement is equivalent to the original statement, so if an if and only if statement is true, then so is its "converse." Converse in quotes because converses are really for one way implications rather than if and only ifs.
$endgroup$
– jgon
Jan 30 at 19:00
$begingroup$
I'll try to clarify to see if I was just misunderstanding. From what I understand, the solution of a linear function in R^n is a hyperplane. If the constant term of the linear function is zero, then the hyperplane is a subspace of R^n since it preserves vector operations and passes through the origin. I think my teacher may have misused homogenous, as he used it to mean a linear function with zero as the constant term. So, if a linear function has a zero for the constant term, is it's solution necessarily a subspace (of R^n)? And the converse, if the hyperplane is a subspace, is the constant 0
$endgroup$
– Corsair64
Jan 30 at 19:13
1
$begingroup$
I see. It's actually the usage of linear functional that was what was confusing me about your question. I would recommend using "linear equation" instead of linear functional. Here is how I would suggest phrasing the result: "The solutions of a linear equation, $Avec{x}=vec{b}$, form a vector subspace of $Bbb{R}^n$ if and only if the linear equation is homogeneous (i.e. $vec{b}=0$)." I've used vector arrows over the vector variables just to be clear, though usually in mathematics we omit them.
$endgroup$
– jgon
Jan 30 at 19:17
1
$begingroup$
Lastly, if and only if statements don't have converses. If I say $P$ if and only if $Q$, then $P$ and $Q$ are equivalent. This means that they are either both false at the same time or both true at the same time. It is never the case that one is true, but the other isn't. This means that it is also true that $Q$ if and only if $P$ holds. Thus it is also true that a linear equation is homogeneous if and only if its solutions form a vector subspace.
$endgroup$
– jgon
Jan 30 at 19:20