Mixing
Vector Space as the set of solutions of matrix equation $AX=O$
$begingroup$
One of my professor's lecture notes on Vector Spaces start by the following lines:-
We have seen that if $det(A)$ = 0, then system $AX=O$ has infinite number of solutions. We shall now see that in this case, the set of solutions has a structure called vector space.
My doubt is in what sense do the set of an infinite number of the solution of equation $AX = O$ (given |A|=0) is actually a structure of Vector Space? How does the term Vector Space come into picture?
linear-algebra vector-spaces matrix-equations matrix-calculus
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
$endgroup$
add a comment |
$begingroup$
One of my professor's lecture notes on Vector Spaces start by the following lines:-
We have seen that if $det(A)$ = 0, then system $AX=O$ has infinite number of solutions. We shall now see that in this case, the set of solutions has a structure called vector space.
My doubt is in what sense do the set of an infinite number of the solution of equation $AX = O$ (given |A|=0) is actually a structure of Vector Space? How does the term Vector Space come into picture?
linear-algebra vector-spaces matrix-equations matrix-calculus
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
$endgroup$
add a comment |
$begingroup$
One of my professor's lecture notes on Vector Spaces start by the following lines:-
We have seen that if $det(A)$ = 0, then system $AX=O$ has infinite number of solutions. We shall now see that in this case, the set of solutions has a structure called vector space.
My doubt is in what sense do the set of an infinite number of the solution of equation $AX = O$ (given |A|=0) is actually a structure of Vector Space? How does the term Vector Space come into picture?
linear-algebra vector-spaces matrix-equations matrix-calculus
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
$endgroup$
One of my professor's lecture notes on Vector Spaces start by the following lines:-
We have seen that if $det(A)$ = 0, then system $AX=O$ has infinite number of solutions. We shall now see that in this case, the set of solutions has a structure called vector space.
My doubt is in what sense do the set of an infinite number of the solution of equation $AX = O$ (given |A|=0) is actually a structure of Vector Space? How does the term Vector Space come into picture?
linear-algebra vector-spaces matrix-equations matrix-calculus
linear-algebra vector-spaces matrix-equations matrix-calculus
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
edited Jan 30 at 19:57
Onkar Singh
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
asked Jan 29 at 18:56
Onkar SinghOnkar Singh
297
297
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
What your professor means is that the set of all $X$'s that satisfy $AX=0$ satisfies the properties of a vector space. You have to check all the ten axioms of a vector space. Here is only a few:
It has an identity under vector addition. Namely, $X=(0,0,cdots,0)^t$.
It's closed under addition because $A(X+Y)=AX+AY=0$ and if $X,Y$ are in the set, so is $X+Y$
It's closed under scalar multiplication because $A(lambda X)=lambda AX=0$.
And you can check all other axioms one by one. Another way to avoid checking many of the axioms, is that if you already have shown that $mathbb{F}^n$ is a vector space for any field $F$. Then, since our set is a subset of that, associativity and many other properties are inherited from the superset naturally. And we only need to show that $alpha X + beta Y$ is in the set which follows from what we proved about being closed under addition and scalar multiplication.
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
$endgroup$
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
1
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
add a comment |
$begingroup$
It's simply the fact that linear combinations of solutions are again a solution: if $AX=0$ and $AY=0$, then
$$
A(alpha X+beta Y)=alpha AX+beta AY=0.
$$
Thus, the set ${X: AX=0}$ is a vector space with the natural vector operations.
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
You must look into the actual definition of a vector space... In this this case, since the set of solutions of $AX=0$ is a subset of $mathbb{R}^n$, you just need to check that it is closed with respect to the sum of vectors and to the multiplication by scalars. Take $u,v in mathbb{R}^n$ such that $Au = Av = 0$ (basically any two solutions of $AX=0$) and $alpha in mathbb{R}$. If you prove that $u+v$ and $alpha u$ are also solutions of $AX=0$, you have your result. This is very straightforward to check:
$$ A(u+v)= Au + Av = 0 + 0 = 0$$
$$A(alpha u) = alpha (Au) = alpha cdot 0 = 0$$
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
$endgroup$
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
$begingroup$
What your professor means is that the set of all $X$'s that satisfy $AX=0$ satisfies the properties of a vector space. You have to check all the ten axioms of a vector space. Here is only a few:
It has an identity under vector addition. Namely, $X=(0,0,cdots,0)^t$.
It's closed under addition because $A(X+Y)=AX+AY=0$ and if $X,Y$ are in the set, so is $X+Y$
It's closed under scalar multiplication because $A(lambda X)=lambda AX=0$.
And you can check all other axioms one by one. Another way to avoid checking many of the axioms, is that if you already have shown that $mathbb{F}^n$ is a vector space for any field $F$. Then, since our set is a subset of that, associativity and many other properties are inherited from the superset naturally. And we only need to show that $alpha X + beta Y$ is in the set which follows from what we proved about being closed under addition and scalar multiplication.
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
$endgroup$
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
1
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
add a comment |
$begingroup$
What your professor means is that the set of all $X$'s that satisfy $AX=0$ satisfies the properties of a vector space. You have to check all the ten axioms of a vector space. Here is only a few:
It has an identity under vector addition. Namely, $X=(0,0,cdots,0)^t$.
It's closed under addition because $A(X+Y)=AX+AY=0$ and if $X,Y$ are in the set, so is $X+Y$
It's closed under scalar multiplication because $A(lambda X)=lambda AX=0$.
And you can check all other axioms one by one. Another way to avoid checking many of the axioms, is that if you already have shown that $mathbb{F}^n$ is a vector space for any field $F$. Then, since our set is a subset of that, associativity and many other properties are inherited from the superset naturally. And we only need to show that $alpha X + beta Y$ is in the set which follows from what we proved about being closed under addition and scalar multiplication.
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
$endgroup$
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
1
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
add a comment |
$begingroup$
What your professor means is that the set of all $X$'s that satisfy $AX=0$ satisfies the properties of a vector space. You have to check all the ten axioms of a vector space. Here is only a few:
It has an identity under vector addition. Namely, $X=(0,0,cdots,0)^t$.
It's closed under addition because $A(X+Y)=AX+AY=0$ and if $X,Y$ are in the set, so is $X+Y$
It's closed under scalar multiplication because $A(lambda X)=lambda AX=0$.
And you can check all other axioms one by one. Another way to avoid checking many of the axioms, is that if you already have shown that $mathbb{F}^n$ is a vector space for any field $F$. Then, since our set is a subset of that, associativity and many other properties are inherited from the superset naturally. And we only need to show that $alpha X + beta Y$ is in the set which follows from what we proved about being closed under addition and scalar multiplication.
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
$endgroup$
What your professor means is that the set of all $X$'s that satisfy $AX=0$ satisfies the properties of a vector space. You have to check all the ten axioms of a vector space. Here is only a few:
It has an identity under vector addition. Namely, $X=(0,0,cdots,0)^t$.
It's closed under addition because $A(X+Y)=AX+AY=0$ and if $X,Y$ are in the set, so is $X+Y$
It's closed under scalar multiplication because $A(lambda X)=lambda AX=0$.
And you can check all other axioms one by one. Another way to avoid checking many of the axioms, is that if you already have shown that $mathbb{F}^n$ is a vector space for any field $F$. Then, since our set is a subset of that, associativity and many other properties are inherited from the superset naturally. And we only need to show that $alpha X + beta Y$ is in the set which follows from what we proved about being closed under addition and scalar multiplication.
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
answered Jan 29 at 19:00
stressed outstressed out
6,5131939
6,5131939
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
1
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
add a comment |
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
1
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
$begingroup$
Seeing the examples of vector spaces, I realized that even polynomials can be a vector space. So, is it that the true notion of 'vector' is being lost here! I mean a polynomial can have no directional sense. Everyone talks about $R^n$, but how to bring that in here in case of a polynomial example?
$endgroup$
– Onkar Singh
Jan 29 at 19:11
1
1
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
@OnkarSingh You are correct. In mathematics, we deal with abstract objects. The fact that a vector space can have objects that are not geometric vectors should not bother you. Even though, it's unfortunate. Some authors use the term 'linear space' instead of 'vector space' because the objects are not necessarily vectors from the kind of geometry we're used to. Even the set of continuous real functions with pointwise addition and scalar multiplication form a vector space and therefore, one may call a function a vector! The terminology is unfortunate, but the idea of abstraction proves useful.
$endgroup$
– stressed out
Jan 29 at 19:13
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
Ok!, so is this the notion of ABSTRACT algebra, that's what 'abstract' word stands for!
$endgroup$
– Onkar Singh
Jan 29 at 19:17
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
$begingroup$
@OnkarSingh Exactly! Mathematicians study things abstractly. That's what makes us different from engineers or physicists.
$endgroup$
– stressed out
Jan 29 at 19:19
add a comment |
$begingroup$
It's simply the fact that linear combinations of solutions are again a solution: if $AX=0$ and $AY=0$, then
$$
A(alpha X+beta Y)=alpha AX+beta AY=0.
$$
Thus, the set ${X: AX=0}$ is a vector space with the natural vector operations.
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
It's simply the fact that linear combinations of solutions are again a solution: if $AX=0$ and $AY=0$, then
$$
A(alpha X+beta Y)=alpha AX+beta AY=0.
$$
Thus, the set ${X: AX=0}$ is a vector space with the natural vector operations.
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
It's simply the fact that linear combinations of solutions are again a solution: if $AX=0$ and $AY=0$, then
$$
A(alpha X+beta Y)=alpha AX+beta AY=0.
$$
Thus, the set ${X: AX=0}$ is a vector space with the natural vector operations.
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
It's simply the fact that linear combinations of solutions are again a solution: if $AX=0$ and $AY=0$, then
$$
A(alpha X+beta Y)=alpha AX+beta AY=0.
$$
Thus, the set ${X: AX=0}$ is a vector space with the natural vector operations.
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 29 at 19:00
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
$begingroup$
You must look into the actual definition of a vector space... In this this case, since the set of solutions of $AX=0$ is a subset of $mathbb{R}^n$, you just need to check that it is closed with respect to the sum of vectors and to the multiplication by scalars. Take $u,v in mathbb{R}^n$ such that $Au = Av = 0$ (basically any two solutions of $AX=0$) and $alpha in mathbb{R}$. If you prove that $u+v$ and $alpha u$ are also solutions of $AX=0$, you have your result. This is very straightforward to check:
$$ A(u+v)= Au + Av = 0 + 0 = 0$$
$$A(alpha u) = alpha (Au) = alpha cdot 0 = 0$$
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
You must look into the actual definition of a vector space... In this this case, since the set of solutions of $AX=0$ is a subset of $mathbb{R}^n$, you just need to check that it is closed with respect to the sum of vectors and to the multiplication by scalars. Take $u,v in mathbb{R}^n$ such that $Au = Av = 0$ (basically any two solutions of $AX=0$) and $alpha in mathbb{R}$. If you prove that $u+v$ and $alpha u$ are also solutions of $AX=0$, you have your result. This is very straightforward to check:
$$ A(u+v)= Au + Av = 0 + 0 = 0$$
$$A(alpha u) = alpha (Au) = alpha cdot 0 = 0$$
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
You must look into the actual definition of a vector space... In this this case, since the set of solutions of $AX=0$ is a subset of $mathbb{R}^n$, you just need to check that it is closed with respect to the sum of vectors and to the multiplication by scalars. Take $u,v in mathbb{R}^n$ such that $Au = Av = 0$ (basically any two solutions of $AX=0$) and $alpha in mathbb{R}$. If you prove that $u+v$ and $alpha u$ are also solutions of $AX=0$, you have your result. This is very straightforward to check:
$$ A(u+v)= Au + Av = 0 + 0 = 0$$
$$A(alpha u) = alpha (Au) = alpha cdot 0 = 0$$
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
$endgroup$
You must look into the actual definition of a vector space... In this this case, since the set of solutions of $AX=0$ is a subset of $mathbb{R}^n$, you just need to check that it is closed with respect to the sum of vectors and to the multiplication by scalars. Take $u,v in mathbb{R}^n$ such that $Au = Av = 0$ (basically any two solutions of $AX=0$) and $alpha in mathbb{R}$. If you prove that $u+v$ and $alpha u$ are also solutions of $AX=0$, you have your result. This is very straightforward to check:
$$ A(u+v)= Au + Av = 0 + 0 = 0$$
$$A(alpha u) = alpha (Au) = alpha cdot 0 = 0$$
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
answered Jan 29 at 19:06
PierreCarrePierreCarre
1,665212
1,665212
add a comment |
add a comment |
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