Given a set for generators of a subspace of $R^n$, is it always possible to extend it to form a base of...
up vote
1
down vote
favorite
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 at 1:08
mathnoob
67011
asked Nov 20 at 0:45
JorgeeFG
2661315
add a comment |
up vote
1
down vote
favorite
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 at 1:08
mathnoob
67011
asked Nov 20 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 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 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 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 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 1:08
mathnoob
67011
asked Nov 20 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 at 1:08
mathnoob
67011
asked Nov 20 at 0:45
JorgeeFG
2661315
edited Nov 20 at 1:08
mathnoob
67011
asked Nov 20 at 0:45
JorgeeFG
2661315
edited Nov 20 at 1:08
mathnoob
67011
edited Nov 20 at 1:08
mathnoob
67011
edited Nov 20 at 1:08
mathnoob
67011
67011
asked Nov 20 at 0:45
JorgeeFG
2661315
asked Nov 20 at 0:45
JorgeeFG
2661315
asked Nov 20 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 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 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 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 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
2 days ago
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 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 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 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 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
2 days ago
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 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 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 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 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 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 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 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 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
2 days ago
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
2 days ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005766%2fgiven-a-set-for-generators-of-a-subspace-of-rn-is-it-always-possible-to-exte%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 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 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 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 at 0:56
The “if possible” takes into account the possibility that the initial set of vectors is not linearly independent.
– amd
2 days ago