Vector spaces and span of linearly independent vectors [closed]
I have a question. Are any n linearly independent vectors in a vector space V with dimV=n a bases for V? why or why not? How can I be sure that they span the vector space V? I'm asking this because I wish to prove that for any given Matrix of size n x n over a field F, if it has n distinct eigenvalues then it is diagonalisable. So if I consider the linear transformation associated with the the matrix A, T: $F^n--->F^n$ then dim$F^n$=n. But i'm not sure how to proceed after that, may someone please help? The question I asked, I think should help me because that would mean that the number of distinct eigenvalues is n and so is dim$F^n$, but even if that was the case, then i'm not sure how to proceed, may someone please help?
linear-algebra matrices vector-spaces linear-transformations diagonalization
asked Nov 24 at 21:03
maths researcher
458
closed as off-topic by amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen Nov 25 at 8:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have a question. Are any n linearly independent vectors in a vector space V with dimV=n a bases for V? why or why not? How can I be sure that they span the vector space V? I'm asking this because I wish to prove that for any given Matrix of size n x n over a field F, if it has n distinct eigenvalues then it is diagonalisable. So if I consider the linear transformation associated with the the matrix A, T: $F^n--->F^n$ then dim$F^n$=n. But i'm not sure how to proceed after that, may someone please help? The question I asked, I think should help me because that would mean that the number of distinct eigenvalues is n and so is dim$F^n$, but even if that was the case, then i'm not sure how to proceed, may someone please help?
linear-algebra matrices vector-spaces linear-transformations diagonalization
asked Nov 24 at 21:03
maths researcher
458
closed as off-topic by amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen Nov 25 at 8:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have a question. Are any n linearly independent vectors in a vector space V with dimV=n a bases for V? why or why not? How can I be sure that they span the vector space V? I'm asking this because I wish to prove that for any given Matrix of size n x n over a field F, if it has n distinct eigenvalues then it is diagonalisable. So if I consider the linear transformation associated with the the matrix A, T: $F^n--->F^n$ then dim$F^n$=n. But i'm not sure how to proceed after that, may someone please help? The question I asked, I think should help me because that would mean that the number of distinct eigenvalues is n and so is dim$F^n$, but even if that was the case, then i'm not sure how to proceed, may someone please help?
linear-algebra matrices vector-spaces linear-transformations diagonalization
asked Nov 24 at 21:03
maths researcher
458
I have a question. Are any n linearly independent vectors in a vector space V with dimV=n a bases for V? why or why not? How can I be sure that they span the vector space V? I'm asking this because I wish to prove that for any given Matrix of size n x n over a field F, if it has n distinct eigenvalues then it is diagonalisable. So if I consider the linear transformation associated with the the matrix A, T: $F^n--->F^n$ then dim$F^n$=n. But i'm not sure how to proceed after that, may someone please help? The question I asked, I think should help me because that would mean that the number of distinct eigenvalues is n and so is dim$F^n$, but even if that was the case, then i'm not sure how to proceed, may someone please help?
linear-algebra matrices vector-spaces linear-transformations diagonalization
linear-algebra matrices vector-spaces linear-transformations diagonalization
asked Nov 24 at 21:03
maths researcher
458
asked Nov 24 at 21:03
maths researcher
458
edited Nov 24 at 21:10
asked Nov 24 at 21:03
maths researcher
458
asked Nov 24 at 21:03
maths researcher
458
asked Nov 24 at 21:03
maths researcher
458
458
closed as off-topic by amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen Nov 25 at 8:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen Nov 25 at 8:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, KReiser, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
That depends upon the facts that you can use. If you can use the fact that, in a $n$-dimensional space, any set with more that $n$ vectors is linearly dependent, that's easy. If a set $S$ has exactly $n$ elements and $S$ is linearly independent, suppose that $langle Srangle$ is nor the whole $V$. Take $win Vsetminuslangle Srangle$. Then no element of $Scup{w}$ can be written as a linear combination of the other elements of that set. Therefore, $Scup{w}$ is linearly independent. So, a contradiction is reached.
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
add a comment |
If $V_n$ is a set of $n$ linearly independent vectors of $V$ and $V$ is a n-dimensional vector space, then these $n$ vectors will span the space. This follows as elementarily as it sounds, as $n$ linearly independent vectors belonging to a space with $n$ dimensions means you have one elemental for every dimension of that space that a combination of them could build any element of the space.
answered Nov 24 at 21:10
Rebellos
14.2k31244
add a comment |
From the definition of a basis, a subset $B$ is a basis of space $V$ if its vector components are linearly independent and span $V$.
To check the span requirement, any vector in $V$ should be a linear combination of all vectors in $B$.
answered Nov 24 at 21:16
LeNoir
112
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
That depends upon the facts that you can use. If you can use the fact that, in a $n$-dimensional space, any set with more that $n$ vectors is linearly dependent, that's easy. If a set $S$ has exactly $n$ elements and $S$ is linearly independent, suppose that $langle Srangle$ is nor the whole $V$. Take $win Vsetminuslangle Srangle$. Then no element of $Scup{w}$ can be written as a linear combination of the other elements of that set. Therefore, $Scup{w}$ is linearly independent. So, a contradiction is reached.
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
add a comment |
That depends upon the facts that you can use. If you can use the fact that, in a $n$-dimensional space, any set with more that $n$ vectors is linearly dependent, that's easy. If a set $S$ has exactly $n$ elements and $S$ is linearly independent, suppose that $langle Srangle$ is nor the whole $V$. Take $win Vsetminuslangle Srangle$. Then no element of $Scup{w}$ can be written as a linear combination of the other elements of that set. Therefore, $Scup{w}$ is linearly independent. So, a contradiction is reached.
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
add a comment |
That depends upon the facts that you can use. If you can use the fact that, in a $n$-dimensional space, any set with more that $n$ vectors is linearly dependent, that's easy. If a set $S$ has exactly $n$ elements and $S$ is linearly independent, suppose that $langle Srangle$ is nor the whole $V$. Take $win Vsetminuslangle Srangle$. Then no element of $Scup{w}$ can be written as a linear combination of the other elements of that set. Therefore, $Scup{w}$ is linearly independent. So, a contradiction is reached.
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
That depends upon the facts that you can use. If you can use the fact that, in a $n$-dimensional space, any set with more that $n$ vectors is linearly dependent, that's easy. If a set $S$ has exactly $n$ elements and $S$ is linearly independent, suppose that $langle Srangle$ is nor the whole $V$. Take $win Vsetminuslangle Srangle$. Then no element of $Scup{w}$ can be written as a linear combination of the other elements of that set. Therefore, $Scup{w}$ is linearly independent. So, a contradiction is reached.
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
answered Nov 24 at 21:12
José Carlos Santos
148k22117218
148k22117218
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
add a comment |
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Oh, so I can prove it in the following way: Suppose the span of the n linearly independent vectors does not equal to the n-dimensional vector space, that means I can add a number of vectors in the set of the linearly independent vectors, but I have reached a contradiction because any collection of m>n vectors is linearly dependent, am I correct?
– maths researcher
Nov 24 at 21:21
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Yes, that would be correct.
– José Carlos Santos
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
Thank you very much, that was very helpful!
– maths researcher
Nov 24 at 21:22
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
@mathsresearcher If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Nov 24 at 21:26
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
I just did, thanks again.
– maths researcher
Nov 24 at 21:27
add a comment |
If $V_n$ is a set of $n$ linearly independent vectors of $V$ and $V$ is a n-dimensional vector space, then these $n$ vectors will span the space. This follows as elementarily as it sounds, as $n$ linearly independent vectors belonging to a space with $n$ dimensions means you have one elemental for every dimension of that space that a combination of them could build any element of the space.
answered Nov 24 at 21:10
Rebellos
14.2k31244
add a comment |
If $V_n$ is a set of $n$ linearly independent vectors of $V$ and $V$ is a n-dimensional vector space, then these $n$ vectors will span the space. This follows as elementarily as it sounds, as $n$ linearly independent vectors belonging to a space with $n$ dimensions means you have one elemental for every dimension of that space that a combination of them could build any element of the space.
answered Nov 24 at 21:10
Rebellos
14.2k31244
add a comment |
If $V_n$ is a set of $n$ linearly independent vectors of $V$ and $V$ is a n-dimensional vector space, then these $n$ vectors will span the space. This follows as elementarily as it sounds, as $n$ linearly independent vectors belonging to a space with $n$ dimensions means you have one elemental for every dimension of that space that a combination of them could build any element of the space.
answered Nov 24 at 21:10
Rebellos
14.2k31244
If $V_n$ is a set of $n$ linearly independent vectors of $V$ and $V$ is a n-dimensional vector space, then these $n$ vectors will span the space. This follows as elementarily as it sounds, as $n$ linearly independent vectors belonging to a space with $n$ dimensions means you have one elemental for every dimension of that space that a combination of them could build any element of the space.
answered Nov 24 at 21:10
Rebellos
14.2k31244
answered Nov 24 at 21:10
Rebellos
14.2k31244
answered Nov 24 at 21:10
Rebellos
14.2k31244
answered Nov 24 at 21:10
Rebellos
14.2k31244
14.2k31244
add a comment |
add a comment |
From the definition of a basis, a subset $B$ is a basis of space $V$ if its vector components are linearly independent and span $V$.
To check the span requirement, any vector in $V$ should be a linear combination of all vectors in $B$.
answered Nov 24 at 21:16
LeNoir
112
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
add a comment |
From the definition of a basis, a subset $B$ is a basis of space $V$ if its vector components are linearly independent and span $V$.
To check the span requirement, any vector in $V$ should be a linear combination of all vectors in $B$.
answered Nov 24 at 21:16
LeNoir
112
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
add a comment |
From the definition of a basis, a subset $B$ is a basis of space $V$ if its vector components are linearly independent and span $V$.
To check the span requirement, any vector in $V$ should be a linear combination of all vectors in $B$.
answered Nov 24 at 21:16
LeNoir
112
From the definition of a basis, a subset $B$ is a basis of space $V$ if its vector components are linearly independent and span $V$.
To check the span requirement, any vector in $V$ should be a linear combination of all vectors in $B$.
answered Nov 24 at 21:16
LeNoir
112
answered Nov 24 at 21:16
LeNoir
112
answered Nov 24 at 21:16
LeNoir
112
answered Nov 24 at 21:16
LeNoir
112
112
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
add a comment |
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
But how can I prove that any any set with linearly independent vectors of size n span the vector space with dimension n?
– maths researcher
Nov 24 at 21:17
add a comment |
