Mixing
Given a set for generators of a subspace of $R^n$, is it always possible to extend it to form a base of...
In my heart I feel it is always possible, but the subject of an exercise makes me doubt.
It has two subspaces:
$S_1 = langle(1,-1,2,1),(2,-3,3,-1),(-1,2,-1,2)rangle$
$S_2 = langle(2,-3,1,4,1), (2,-3,3,-1,1),(1,0,-1,3,1),(1,0,-3,2,1)rangle$
It says to find base for $S_1$ and $S_2$, which I did.
then it says extend, if possible the set of generators to form a base of $mathbb{R}^4$ and $mathbb{R}^5$ respectively.
It is my understanding that when you have a n-tuple for a space of $mathbb{R}^n$ then you can always fill the missing generators. Am I right?
I understand it would not be possible for example to form a base of $mathbb{R}^4$ with vectors having only 3 components.
linear-algebra
edited Nov 20 '18 at 1:08
mathnoob
1,799422
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
add a comment |
In my heart I feel it is always possible, but the subject of an exercise makes me doubt.
It has two subspaces:
$S_1 = langle(1,-1,2,1),(2,-3,3,-1),(-1,2,-1,2)rangle$
$S_2 = langle(2,-3,1,4,1), (2,-3,3,-1,1),(1,0,-1,3,1),(1,0,-3,2,1)rangle$
It says to find base for $S_1$ and $S_2$, which I did.
then it says extend, if possible the set of generators to form a base of $mathbb{R}^4$ and $mathbb{R}^5$ respectively.
It is my understanding that when you have a n-tuple for a space of $mathbb{R}^n$ then you can always fill the missing generators. Am I right?
I understand it would not be possible for example to form a base of $mathbb{R}^4$ with vectors having only 3 components.
linear-algebra
edited Nov 20 '18 at 1:08
mathnoob
1,799422
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
1
A rather important theorem in linear algebra states that any linearly independent set in a linear space can be extended to a base of the space, so yes: you can always extend a basis of a subspace to a basis of the whole space
– DonAntonio
Nov 20 '18 at 0:48
1
You can think of vectors with 3 components in $mathbb{R}^4$ as having its 4th entry 0. Then you can extend to a basis for $mathbb{R}^4$.
– mathnoob
Nov 20 '18 at 0:52
Here is a small silly example: $(1,0,0),(0,1,0)$ forms a basis for a subspace of $Bbb R^3$. You can extend that basis by including an additional linearly independent vector (or vectors) to it, for example $(0,0,1)$, in order to make a basis for $Bbb R^3$.
– JMoravitz
Nov 20 '18 at 0:53
Of course in my silly example, and in your examples, there are infinitely many correct answers. I could have extended the basis by instead including $(1,1,1)$ or $(pi, e, sqrt{2})$ or any other valid choice so long as it was linearly independent. The question is effectively "can you find a vector from $Bbb R^n$ which is linearly independent to your list of basis vectors that you already have" and recognizing that if you were to look at the span of all of your vectors, your original basis and whatever vector(s) you choose to include, that it spans the entirety of $Bbb R^n$.
– JMoravitz
Nov 20 '18 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
Nov 21 '18 at 1:21
add a comment |
In my heart I feel it is always possible, but the subject of an exercise makes me doubt.
It has two subspaces:
$S_1 = langle(1,-1,2,1),(2,-3,3,-1),(-1,2,-1,2)rangle$
$S_2 = langle(2,-3,1,4,1), (2,-3,3,-1,1),(1,0,-1,3,1),(1,0,-3,2,1)rangle$
It says to find base for $S_1$ and $S_2$, which I did.
then it says extend, if possible the set of generators to form a base of $mathbb{R}^4$ and $mathbb{R}^5$ respectively.
It is my understanding that when you have a n-tuple for a space of $mathbb{R}^n$ then you can always fill the missing generators. Am I right?
I understand it would not be possible for example to form a base of $mathbb{R}^4$ with vectors having only 3 components.
linear-algebra
edited Nov 20 '18 at 1:08
mathnoob
1,799422
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
In my heart I feel it is always possible, but the subject of an exercise makes me doubt.
It has two subspaces:
$S_1 = langle(1,-1,2,1),(2,-3,3,-1),(-1,2,-1,2)rangle$
$S_2 = langle(2,-3,1,4,1), (2,-3,3,-1,1),(1,0,-1,3,1),(1,0,-3,2,1)rangle$
It says to find base for $S_1$ and $S_2$, which I did.
then it says extend, if possible the set of generators to form a base of $mathbb{R}^4$ and $mathbb{R}^5$ respectively.
It is my understanding that when you have a n-tuple for a space of $mathbb{R}^n$ then you can always fill the missing generators. Am I right?
I understand it would not be possible for example to form a base of $mathbb{R}^4$ with vectors having only 3 components.
linear-algebra
linear-algebra
edited Nov 20 '18 at 1:08
mathnoob
1,799422
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
edited Nov 20 '18 at 1:08
mathnoob
1,799422
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
edited Nov 20 '18 at 1:08
mathnoob
1,799422
edited Nov 20 '18 at 1:08
mathnoob
1,799422
edited Nov 20 '18 at 1:08
mathnoob
1,799422
1,799422
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
asked Nov 20 '18 at 0:45
JorgeeFG
2661315
2661315
1
A rather important theorem in linear algebra states that any linearly independent set in a linear space can be extended to a base of the space, so yes: you can always extend a basis of a subspace to a basis of the whole space
– DonAntonio
Nov 20 '18 at 0:48
1
You can think of vectors with 3 components in $mathbb{R}^4$ as having its 4th entry 0. Then you can extend to a basis for $mathbb{R}^4$.
– mathnoob
Nov 20 '18 at 0:52
Here is a small silly example: $(1,0,0),(0,1,0)$ forms a basis for a subspace of $Bbb R^3$. You can extend that basis by including an additional linearly independent vector (or vectors) to it, for example $(0,0,1)$, in order to make a basis for $Bbb R^3$.
– JMoravitz
Nov 20 '18 at 0:53
Of course in my silly example, and in your examples, there are infinitely many correct answers. I could have extended the basis by instead including $(1,1,1)$ or $(pi, e, sqrt{2})$ or any other valid choice so long as it was linearly independent. The question is effectively "can you find a vector from $Bbb R^n$ which is linearly independent to your list of basis vectors that you already have" and recognizing that if you were to look at the span of all of your vectors, your original basis and whatever vector(s) you choose to include, that it spans the entirety of $Bbb R^n$.
– JMoravitz
Nov 20 '18 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
Nov 21 '18 at 1:21
add a comment |
1
A rather important theorem in linear algebra states that any linearly independent set in a linear space can be extended to a base of the space, so yes: you can always extend a basis of a subspace to a basis of the whole space
– DonAntonio
Nov 20 '18 at 0:48
1
You can think of vectors with 3 components in $mathbb{R}^4$ as having its 4th entry 0. Then you can extend to a basis for $mathbb{R}^4$.
– mathnoob
Nov 20 '18 at 0:52
Here is a small silly example: $(1,0,0),(0,1,0)$ forms a basis for a subspace of $Bbb R^3$. You can extend that basis by including an additional linearly independent vector (or vectors) to it, for example $(0,0,1)$, in order to make a basis for $Bbb R^3$.
– JMoravitz
Nov 20 '18 at 0:53
Of course in my silly example, and in your examples, there are infinitely many correct answers. I could have extended the basis by instead including $(1,1,1)$ or $(pi, e, sqrt{2})$ or any other valid choice so long as it was linearly independent. The question is effectively "can you find a vector from $Bbb R^n$ which is linearly independent to your list of basis vectors that you already have" and recognizing that if you were to look at the span of all of your vectors, your original basis and whatever vector(s) you choose to include, that it spans the entirety of $Bbb R^n$.
– JMoravitz
Nov 20 '18 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
Nov 21 '18 at 1:21
1
1
A rather important theorem in linear algebra states that any linearly independent set in a linear space can be extended to a base of the space, so yes: you can always extend a basis of a subspace to a basis of the whole space
– DonAntonio
Nov 20 '18 at 0:48
A rather important theorem in linear algebra states that any linearly independent set in a linear space can be extended to a base of the space, so yes: you can always extend a basis of a subspace to a basis of the whole space
– DonAntonio
Nov 20 '18 at 0:48
1
1
You can think of vectors with 3 components in $mathbb{R}^4$ as having its 4th entry 0. Then you can extend to a basis for $mathbb{R}^4$.
– mathnoob
Nov 20 '18 at 0:52
You can think of vectors with 3 components in $mathbb{R}^4$ as having its 4th entry 0. Then you can extend to a basis for $mathbb{R}^4$.
– mathnoob
Nov 20 '18 at 0:52
Here is a small silly example: $(1,0,0),(0,1,0)$ forms a basis for a subspace of $Bbb R^3$. You can extend that basis by including an additional linearly independent vector (or vectors) to it, for example $(0,0,1)$, in order to make a basis for $Bbb R^3$.
– JMoravitz
Nov 20 '18 at 0:53
Here is a small silly example: $(1,0,0),(0,1,0)$ forms a basis for a subspace of $Bbb R^3$. You can extend that basis by including an additional linearly independent vector (or vectors) to it, for example $(0,0,1)$, in order to make a basis for $Bbb R^3$.
– JMoravitz
Nov 20 '18 at 0:53
Of course in my silly example, and in your examples, there are infinitely many correct answers. I could have extended the basis by instead including $(1,1,1)$ or $(pi, e, sqrt{2})$ or any other valid choice so long as it was linearly independent. The question is effectively "can you find a vector from $Bbb R^n$ which is linearly independent to your list of basis vectors that you already have" and recognizing that if you were to look at the span of all of your vectors, your original basis and whatever vector(s) you choose to include, that it spans the entirety of $Bbb R^n$.
– JMoravitz
Nov 20 '18 at 0:56
Of course in my silly example, and in your examples, there are infinitely many correct answers. I could have extended the basis by instead including $(1,1,1)$ or $(pi, e, sqrt{2})$ or any other valid choice so long as it was linearly independent. The question is effectively "can you find a vector from $Bbb R^n$ which is linearly independent to your list of basis vectors that you already have" and recognizing that if you were to look at the span of all of your vectors, your original basis and whatever vector(s) you choose to include, that it spans the entirety of $Bbb R^n$.
– JMoravitz
Nov 20 '18 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
Nov 21 '18 at 1:21
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
Nov 21 '18 at 1:21
add a comment |
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1
A rather important theorem in linear algebra states that any linearly independent set in a linear space can be extended to a base of the space, so yes: you can always extend a basis of a subspace to a basis of the whole space
– DonAntonio
Nov 20 '18 at 0:48
1
You can think of vectors with 3 components in $mathbb{R}^4$ as having its 4th entry 0. Then you can extend to a basis for $mathbb{R}^4$.
– mathnoob
Nov 20 '18 at 0:52
Here is a small silly example: $(1,0,0),(0,1,0)$ forms a basis for a subspace of $Bbb R^3$. You can extend that basis by including an additional linearly independent vector (or vectors) to it, for example $(0,0,1)$, in order to make a basis for $Bbb R^3$.
– JMoravitz
Nov 20 '18 at 0:53
Of course in my silly example, and in your examples, there are infinitely many correct answers. I could have extended the basis by instead including $(1,1,1)$ or $(pi, e, sqrt{2})$ or any other valid choice so long as it was linearly independent. The question is effectively "can you find a vector from $Bbb R^n$ which is linearly independent to your list of basis vectors that you already have" and recognizing that if you were to look at the span of all of your vectors, your original basis and whatever vector(s) you choose to include, that it spans the entirety of $Bbb R^n$.
– JMoravitz
Nov 20 '18 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
Nov 21 '18 at 1:21