When asked to find the basis of the row space of a matrix, what's the point of reducing the matrix? Row...
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 Ronald
977
add a comment |
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 Ronald
977
add a comment |
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 Ronald
977
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 Ronald
977
asked Dec 4 '18 at 16:27
James Ronald
977
edited Dec 4 '18 at 16:33
asked Dec 4 '18 at 16:27
James Ronald
977
asked Dec 4 '18 at 16:27
James Ronald
977
asked Dec 4 '18 at 16:27
James Ronald
977
977
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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. Ullrich
59k43892
add a comment |
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 K
52.7k340115
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)?
– James Ronald
Dec 4 '18 at 16:36
2
@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.
– amd
Dec 4 '18 at 17:25
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.
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3025783%2fwhen-asked-to-find-the-basis-of-the-row-space-of-a-matrix-whats-the-point-of-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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. Ullrich
59k43892
add a comment |
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. Ullrich
59k43892
add a comment |
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. Ullrich
59k43892
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. Ullrich
59k43892
answered Dec 4 '18 at 16:40
David C. Ullrich
59k43892
answered Dec 4 '18 at 16:40
David C. Ullrich
59k43892
answered Dec 4 '18 at 16:40
David C. Ullrich
59k43892
59k43892
add a comment |
add a comment |
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 K
52.7k340115
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)?
– James Ronald
Dec 4 '18 at 16:36
2
@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.
– amd
Dec 4 '18 at 17:25
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.
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
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 K
52.7k340115
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)?
– James Ronald
Dec 4 '18 at 16:36
2
@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.
– amd
Dec 4 '18 at 17:25
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.
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
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 K
52.7k340115
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 K
52.7k340115
edited Dec 4 '18 at 18:56
answered Dec 4 '18 at 16:32
David K
52.7k340115
answered Dec 4 '18 at 16:32
David K
52.7k340115
answered Dec 4 '18 at 16:32
David K
52.7k340115
52.7k340115
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)?
– James Ronald
Dec 4 '18 at 16:36
2
@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.
– amd
Dec 4 '18 at 17:25
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.
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
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)?
– James Ronald
Dec 4 '18 at 16:36
2
@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.
– amd
Dec 4 '18 at 17:25
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.
– Andreas Blass
Dec 4 '18 at 18:05
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)?
– James Ronald
Dec 4 '18 at 16:36
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)?
– James Ronald
Dec 4 '18 at 16:36
2
2
@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.
– amd
Dec 4 '18 at 17:25
@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.
– amd
Dec 4 '18 at 17:25
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.
– Andreas Blass
Dec 4 '18 at 18:05
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.
– Andreas Blass
Dec 4 '18 at 18:05
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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%2f3025783%2fwhen-asked-to-find-the-basis-of-the-row-space-of-a-matrix-whats-the-point-of-r%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
