When asked to find the basis of the row space of a matrix, what's the point of reducing the matrix? Row...
$begingroup$
If matrix A is row equivalent to matrix B, then row(A) = row(B). This is because the row space of A is just the span of the row vectors of A. The rows of B are a linear combination of the rows of A, so the rows of B lie within the row space of A. And vice versa. Therefore row(A) = row(B)
So why is it that whenever I see a problem asking me to find a basis of the row space of a matrix, the matrix is reduced? Can't you just take the rows of the matrix as they are, and say that those row vectors make up a basis? What's the point of row reducing first?
Side question, whenever you're asked for a basis, would it not be valid to just give the unit vectors of that dimension?
Any help is appreciated.
linear-algebra matrices vector-spaces
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
$endgroup$
add a comment |
$begingroup$
If matrix A is row equivalent to matrix B, then row(A) = row(B). This is because the row space of A is just the span of the row vectors of A. The rows of B are a linear combination of the rows of A, so the rows of B lie within the row space of A. And vice versa. Therefore row(A) = row(B)
So why is it that whenever I see a problem asking me to find a basis of the row space of a matrix, the matrix is reduced? Can't you just take the rows of the matrix as they are, and say that those row vectors make up a basis? What's the point of row reducing first?
Side question, whenever you're asked for a basis, would it not be valid to just give the unit vectors of that dimension?
Any help is appreciated.
linear-algebra matrices vector-spaces
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
$endgroup$
add a comment |
$begingroup$
If matrix A is row equivalent to matrix B, then row(A) = row(B). This is because the row space of A is just the span of the row vectors of A. The rows of B are a linear combination of the rows of A, so the rows of B lie within the row space of A. And vice versa. Therefore row(A) = row(B)
So why is it that whenever I see a problem asking me to find a basis of the row space of a matrix, the matrix is reduced? Can't you just take the rows of the matrix as they are, and say that those row vectors make up a basis? What's the point of row reducing first?
Side question, whenever you're asked for a basis, would it not be valid to just give the unit vectors of that dimension?
Any help is appreciated.
linear-algebra matrices vector-spaces
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
$endgroup$
If matrix A is row equivalent to matrix B, then row(A) = row(B). This is because the row space of A is just the span of the row vectors of A. The rows of B are a linear combination of the rows of A, so the rows of B lie within the row space of A. And vice versa. Therefore row(A) = row(B)
So why is it that whenever I see a problem asking me to find a basis of the row space of a matrix, the matrix is reduced? Can't you just take the rows of the matrix as they are, and say that those row vectors make up a basis? What's the point of row reducing first?
Side question, whenever you're asked for a basis, would it not be valid to just give the unit vectors of that dimension?
Any help is appreciated.
linear-algebra matrices vector-spaces
linear-algebra matrices vector-spaces
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
edited Dec 4 '18 at 16:33
James Ronald
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
asked Dec 4 '18 at 16:27
James RonaldJames Ronald
957
957
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Yes, the rows of $A$ span the row space of $A$. So what? A basis is an independent spanning set, and the rows of $A$ need not be independent.
But if $B$ is an echelon form for $A$ then the rows of $B$ span the row space of $A$, as you point out; hence the non-zero rows of $B$ span the row space of $A$, and also the non-zero rows of $B$ are independent.
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
add a comment |
$begingroup$
What if the matrix has $5$ rows but the row space only has $3$ dimensions?
Given an arbitrary $5times 5$ matrix,
how can you be sure that the matrix's row space really has $5$ dimensions, and therefore has five vectors in its basis?
NOTE: When I wrote "has $5$ dimensions, and therefore has five vectors in its basis", I assumed you were aware that a set of five vectors in a five-dimensional vector space does not necessarily span the entire five-dimensional space, but might span a subspace of fewer dimensions,
and also that the number of vectors in any basis of that subspace is exactly equal to the number of dimensions of the subspace.
Regarding the side question, of course you can name a basis of a subspace using unit vectors as the basis.
If you are dealing with a proper subspace of a vector space
(i.e. the subspace is not equal to the whole vector space), however,
there is no guarantee that any of the vectors $(1,0,0,0,0),$ $(0,1,0,0,0),$
$(0,0,1,0,0),$ etc. will be in the subspace.
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
$endgroup$
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
2
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, the rows of $A$ span the row space of $A$. So what? A basis is an independent spanning set, and the rows of $A$ need not be independent.
But if $B$ is an echelon form for $A$ then the rows of $B$ span the row space of $A$, as you point out; hence the non-zero rows of $B$ span the row space of $A$, and also the non-zero rows of $B$ are independent.
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
add a comment |
$begingroup$
Yes, the rows of $A$ span the row space of $A$. So what? A basis is an independent spanning set, and the rows of $A$ need not be independent.
But if $B$ is an echelon form for $A$ then the rows of $B$ span the row space of $A$, as you point out; hence the non-zero rows of $B$ span the row space of $A$, and also the non-zero rows of $B$ are independent.
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
add a comment |
$begingroup$
Yes, the rows of $A$ span the row space of $A$. So what? A basis is an independent spanning set, and the rows of $A$ need not be independent.
But if $B$ is an echelon form for $A$ then the rows of $B$ span the row space of $A$, as you point out; hence the non-zero rows of $B$ span the row space of $A$, and also the non-zero rows of $B$ are independent.
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
Yes, the rows of $A$ span the row space of $A$. So what? A basis is an independent spanning set, and the rows of $A$ need not be independent.
But if $B$ is an echelon form for $A$ then the rows of $B$ span the row space of $A$, as you point out; hence the non-zero rows of $B$ span the row space of $A$, and also the non-zero rows of $B$ are independent.
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
answered Dec 4 '18 at 16:40
David C. UllrichDavid C. Ullrich
59.3k43893
59.3k43893
add a comment |
add a comment |
$begingroup$
What if the matrix has $5$ rows but the row space only has $3$ dimensions?
Given an arbitrary $5times 5$ matrix,
how can you be sure that the matrix's row space really has $5$ dimensions, and therefore has five vectors in its basis?
NOTE: When I wrote "has $5$ dimensions, and therefore has five vectors in its basis", I assumed you were aware that a set of five vectors in a five-dimensional vector space does not necessarily span the entire five-dimensional space, but might span a subspace of fewer dimensions,
and also that the number of vectors in any basis of that subspace is exactly equal to the number of dimensions of the subspace.
Regarding the side question, of course you can name a basis of a subspace using unit vectors as the basis.
If you are dealing with a proper subspace of a vector space
(i.e. the subspace is not equal to the whole vector space), however,
there is no guarantee that any of the vectors $(1,0,0,0,0),$ $(0,1,0,0,0),$
$(0,0,1,0,0),$ etc. will be in the subspace.
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
$endgroup$
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
2
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
$begingroup$
What if the matrix has $5$ rows but the row space only has $3$ dimensions?
Given an arbitrary $5times 5$ matrix,
how can you be sure that the matrix's row space really has $5$ dimensions, and therefore has five vectors in its basis?
NOTE: When I wrote "has $5$ dimensions, and therefore has five vectors in its basis", I assumed you were aware that a set of five vectors in a five-dimensional vector space does not necessarily span the entire five-dimensional space, but might span a subspace of fewer dimensions,
and also that the number of vectors in any basis of that subspace is exactly equal to the number of dimensions of the subspace.
Regarding the side question, of course you can name a basis of a subspace using unit vectors as the basis.
If you are dealing with a proper subspace of a vector space
(i.e. the subspace is not equal to the whole vector space), however,
there is no guarantee that any of the vectors $(1,0,0,0,0),$ $(0,1,0,0,0),$
$(0,0,1,0,0),$ etc. will be in the subspace.
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
$endgroup$
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
2
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
$begingroup$
What if the matrix has $5$ rows but the row space only has $3$ dimensions?
Given an arbitrary $5times 5$ matrix,
how can you be sure that the matrix's row space really has $5$ dimensions, and therefore has five vectors in its basis?
NOTE: When I wrote "has $5$ dimensions, and therefore has five vectors in its basis", I assumed you were aware that a set of five vectors in a five-dimensional vector space does not necessarily span the entire five-dimensional space, but might span a subspace of fewer dimensions,
and also that the number of vectors in any basis of that subspace is exactly equal to the number of dimensions of the subspace.
Regarding the side question, of course you can name a basis of a subspace using unit vectors as the basis.
If you are dealing with a proper subspace of a vector space
(i.e. the subspace is not equal to the whole vector space), however,
there is no guarantee that any of the vectors $(1,0,0,0,0),$ $(0,1,0,0,0),$
$(0,0,1,0,0),$ etc. will be in the subspace.
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
$endgroup$
What if the matrix has $5$ rows but the row space only has $3$ dimensions?
Given an arbitrary $5times 5$ matrix,
how can you be sure that the matrix's row space really has $5$ dimensions, and therefore has five vectors in its basis?
NOTE: When I wrote "has $5$ dimensions, and therefore has five vectors in its basis", I assumed you were aware that a set of five vectors in a five-dimensional vector space does not necessarily span the entire five-dimensional space, but might span a subspace of fewer dimensions,
and also that the number of vectors in any basis of that subspace is exactly equal to the number of dimensions of the subspace.
Regarding the side question, of course you can name a basis of a subspace using unit vectors as the basis.
If you are dealing with a proper subspace of a vector space
(i.e. the subspace is not equal to the whole vector space), however,
there is no guarantee that any of the vectors $(1,0,0,0,0),$ $(0,1,0,0,0),$
$(0,0,1,0,0),$ etc. will be in the subspace.
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
edited Dec 4 '18 at 18:56
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
answered Dec 4 '18 at 16:32
David KDavid K
52.8k340115
52.8k340115
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
2
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
2
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
$begingroup$
Thank you for the response. With your example, it's likely that 2 of the 5 row vectors in the 5x5 matrix are linearly dependent on the other row vectors. I'm guessing that it'll always turn out this way: there will always be the exactly correct amount of vectors required in the row space, nothing more or less. So using this fact, could I not apply what I described originally (just taking the row vectors as-is)?
$endgroup$
– James Ronald
Dec 4 '18 at 16:36
2
2
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
@JamesRonald It’s not just likely, but it’s certain that two of the rows can be written as linear combinations of the other three. And, no, you can’t take the row vectors “as is” because a basis must consist of linearly independent vectors. You would do well to review those basic definitions.
$endgroup$
– amd
Dec 4 '18 at 17:25
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
$begingroup$
The closest you can get to obtaining a basis by your "take the rows as they are" strategy would be to take a linearly independent subset of the rows that suffices to span all the other rows. But it's not obvious how to find such a subset.
$endgroup$
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
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