Mixing
Why is the dimension of the zero subspace 0 and not 1?
$begingroup$
A related question: If a single non-zero vector serves as a basis for a subspace, then is the dimension of that subspace 1 or 0?
I'm almost certain the answer to the above question is 1.
But my confusion lies with the fact that the dimension of the zero subspace (which consists only of the zero vector) is zero. There is one vector in this subspace (namely, the zero vector), so shouldn't the dimension be 1?
Is there an intuitive way to think about this, or is this just how it's been defined as this is the only way everything works?
On this similar post, a commenter said: "The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space."
I don't see see how this is true, unless the empty set is not counted when counting cardinality (i.e. a set of vectors with 5 vectors as well as the empty set would be counted as having 5 vectors and not 6). Again, is this just how we've defined things?
linear-algebra vector-spaces
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
$endgroup$
|
show 1 more comment
$begingroup$
A related question: If a single non-zero vector serves as a basis for a subspace, then is the dimension of that subspace 1 or 0?
I'm almost certain the answer to the above question is 1.
But my confusion lies with the fact that the dimension of the zero subspace (which consists only of the zero vector) is zero. There is one vector in this subspace (namely, the zero vector), so shouldn't the dimension be 1?
Is there an intuitive way to think about this, or is this just how it's been defined as this is the only way everything works?
On this similar post, a commenter said: "The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space."
I don't see see how this is true, unless the empty set is not counted when counting cardinality (i.e. a set of vectors with 5 vectors as well as the empty set would be counted as having 5 vectors and not 6). Again, is this just how we've defined things?
linear-algebra vector-spaces
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
$endgroup$
$begingroup$
The basis is the empty set. You should check that the zero vector is not a basis vector.
$endgroup$
– Randall
Jan 30 at 23:31
$begingroup$
The empty set has cardinality $0$ so that’s the dimension. Done.
$endgroup$
– Randall
Jan 30 at 23:32
5
$begingroup$
The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero.
$endgroup$
– MPW
Jan 30 at 23:33
$begingroup$
I thought this was a great question, as it makes you look closer at the definitions!
$endgroup$
– Metric
Jan 30 at 23:47
$begingroup$
Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere.
$endgroup$
– Klaus
Jan 30 at 23:50
|
show 1 more comment
$begingroup$
A related question: If a single non-zero vector serves as a basis for a subspace, then is the dimension of that subspace 1 or 0?
I'm almost certain the answer to the above question is 1.
But my confusion lies with the fact that the dimension of the zero subspace (which consists only of the zero vector) is zero. There is one vector in this subspace (namely, the zero vector), so shouldn't the dimension be 1?
Is there an intuitive way to think about this, or is this just how it's been defined as this is the only way everything works?
On this similar post, a commenter said: "The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space."
I don't see see how this is true, unless the empty set is not counted when counting cardinality (i.e. a set of vectors with 5 vectors as well as the empty set would be counted as having 5 vectors and not 6). Again, is this just how we've defined things?
linear-algebra vector-spaces
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
$endgroup$
A related question: If a single non-zero vector serves as a basis for a subspace, then is the dimension of that subspace 1 or 0?
I'm almost certain the answer to the above question is 1.
But my confusion lies with the fact that the dimension of the zero subspace (which consists only of the zero vector) is zero. There is one vector in this subspace (namely, the zero vector), so shouldn't the dimension be 1?
Is there an intuitive way to think about this, or is this just how it's been defined as this is the only way everything works?
On this similar post, a commenter said: "The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space."
I don't see see how this is true, unless the empty set is not counted when counting cardinality (i.e. a set of vectors with 5 vectors as well as the empty set would be counted as having 5 vectors and not 6). Again, is this just how we've defined things?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
asked Jan 30 at 23:25
James RonaldJames Ronald
3047
3047
$begingroup$
The basis is the empty set. You should check that the zero vector is not a basis vector.
$endgroup$
– Randall
Jan 30 at 23:31
$begingroup$
The empty set has cardinality $0$ so that’s the dimension. Done.
$endgroup$
– Randall
Jan 30 at 23:32
5
$begingroup$
The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero.
$endgroup$
– MPW
Jan 30 at 23:33
$begingroup$
I thought this was a great question, as it makes you look closer at the definitions!
$endgroup$
– Metric
Jan 30 at 23:47
$begingroup$
Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere.
$endgroup$
– Klaus
Jan 30 at 23:50
|
show 1 more comment
$begingroup$
The basis is the empty set. You should check that the zero vector is not a basis vector.
$endgroup$
– Randall
Jan 30 at 23:31
$begingroup$
The empty set has cardinality $0$ so that’s the dimension. Done.
$endgroup$
– Randall
Jan 30 at 23:32
5
$begingroup$
The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero.
$endgroup$
– MPW
Jan 30 at 23:33
$begingroup$
I thought this was a great question, as it makes you look closer at the definitions!
$endgroup$
– Metric
Jan 30 at 23:47
$begingroup$
Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere.
$endgroup$
– Klaus
Jan 30 at 23:50
$begingroup$
The basis is the empty set. You should check that the zero vector is not a basis vector.
$endgroup$
– Randall
Jan 30 at 23:31
$begingroup$
The basis is the empty set. You should check that the zero vector is not a basis vector.
$endgroup$
– Randall
Jan 30 at 23:31
$begingroup$
The empty set has cardinality $0$ so that’s the dimension. Done.
$endgroup$
– Randall
Jan 30 at 23:32
$begingroup$
The empty set has cardinality $0$ so that’s the dimension. Done.
$endgroup$
– Randall
Jan 30 at 23:32
5
5
$begingroup$
The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero.
$endgroup$
– MPW
Jan 30 at 23:33
$begingroup$
The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero.
$endgroup$
– MPW
Jan 30 at 23:33
$begingroup$
I thought this was a great question, as it makes you look closer at the definitions!
$endgroup$
– Metric
Jan 30 at 23:47
$begingroup$
I thought this was a great question, as it makes you look closer at the definitions!
$endgroup$
– Metric
Jan 30 at 23:47
$begingroup$
Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere.
$endgroup$
– Klaus
Jan 30 at 23:50
$begingroup$
Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere.
$endgroup$
– Klaus
Jan 30 at 23:50
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $V = {0}$. Can you find a basis for $V$? Surely it can't be $V$, as it contains $0$ (and hence linearly dependent). Thus, all we have left to check is the empty set $emptyset$.
(1) Linearly independent: this follows vacuously, as you can't find a non-empty subset of $emptyset$.
(2) The empty set spans $V$. By definition of the span of a set $S$, it's the smallest subspace of $V$ that contains $S$. Clearly $emptyset$ is not a vector space (it's not a group as it doesn't contain an identity element), and so the smallest subspace of $V$ that contains $emptyset$ is $V$.
Thus the empty set $emptyset$ is a basis for $V$, so the dimension of $V$ is the size of $emptyset$, which is $0$.
answered Jan 30 at 23:41
MetricMetric
1,23659
$endgroup$
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
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%2f3094268%2fwhy-is-the-dimension-of-the-zero-subspace-0-and-not-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $V = {0}$. Can you find a basis for $V$? Surely it can't be $V$, as it contains $0$ (and hence linearly dependent). Thus, all we have left to check is the empty set $emptyset$.
(1) Linearly independent: this follows vacuously, as you can't find a non-empty subset of $emptyset$.
(2) The empty set spans $V$. By definition of the span of a set $S$, it's the smallest subspace of $V$ that contains $S$. Clearly $emptyset$ is not a vector space (it's not a group as it doesn't contain an identity element), and so the smallest subspace of $V$ that contains $emptyset$ is $V$.
Thus the empty set $emptyset$ is a basis for $V$, so the dimension of $V$ is the size of $emptyset$, which is $0$.
answered Jan 30 at 23:41
MetricMetric
1,23659
$endgroup$
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
add a comment |
$begingroup$
Let $V = {0}$. Can you find a basis for $V$? Surely it can't be $V$, as it contains $0$ (and hence linearly dependent). Thus, all we have left to check is the empty set $emptyset$.
(1) Linearly independent: this follows vacuously, as you can't find a non-empty subset of $emptyset$.
(2) The empty set spans $V$. By definition of the span of a set $S$, it's the smallest subspace of $V$ that contains $S$. Clearly $emptyset$ is not a vector space (it's not a group as it doesn't contain an identity element), and so the smallest subspace of $V$ that contains $emptyset$ is $V$.
Thus the empty set $emptyset$ is a basis for $V$, so the dimension of $V$ is the size of $emptyset$, which is $0$.
answered Jan 30 at 23:41
MetricMetric
1,23659
$endgroup$
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
add a comment |
$begingroup$
Let $V = {0}$. Can you find a basis for $V$? Surely it can't be $V$, as it contains $0$ (and hence linearly dependent). Thus, all we have left to check is the empty set $emptyset$.
(1) Linearly independent: this follows vacuously, as you can't find a non-empty subset of $emptyset$.
(2) The empty set spans $V$. By definition of the span of a set $S$, it's the smallest subspace of $V$ that contains $S$. Clearly $emptyset$ is not a vector space (it's not a group as it doesn't contain an identity element), and so the smallest subspace of $V$ that contains $emptyset$ is $V$.
Thus the empty set $emptyset$ is a basis for $V$, so the dimension of $V$ is the size of $emptyset$, which is $0$.
answered Jan 30 at 23:41
MetricMetric
1,23659
$endgroup$
Let $V = {0}$. Can you find a basis for $V$? Surely it can't be $V$, as it contains $0$ (and hence linearly dependent). Thus, all we have left to check is the empty set $emptyset$.
(1) Linearly independent: this follows vacuously, as you can't find a non-empty subset of $emptyset$.
(2) The empty set spans $V$. By definition of the span of a set $S$, it's the smallest subspace of $V$ that contains $S$. Clearly $emptyset$ is not a vector space (it's not a group as it doesn't contain an identity element), and so the smallest subspace of $V$ that contains $emptyset$ is $V$.
Thus the empty set $emptyset$ is a basis for $V$, so the dimension of $V$ is the size of $emptyset$, which is $0$.
answered Jan 30 at 23:41
MetricMetric
1,23659
edited Jan 31 at 1:35
answered Jan 30 at 23:41
MetricMetric
1,23659
answered Jan 30 at 23:41
MetricMetric
1,23659
answered Jan 30 at 23:41
MetricMetric
1,23659
1,23659
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
add a comment |
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
$begingroup$
I think this is nice and complete.
$endgroup$
– Randall
Jan 31 at 0:27
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.
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%2f3094268%2fwhy-is-the-dimension-of-the-zero-subspace-0-and-not-1%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
$begingroup$
The basis is the empty set. You should check that the zero vector is not a basis vector.
$endgroup$
– Randall
Jan 30 at 23:31
$begingroup$
The empty set has cardinality $0$ so that’s the dimension. Done.
$endgroup$
– Randall
Jan 30 at 23:32
5
$begingroup$
The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero.
$endgroup$
– MPW
Jan 30 at 23:33
$begingroup$
I thought this was a great question, as it makes you look closer at the definitions!
$endgroup$
– Metric
Jan 30 at 23:47
$begingroup$
Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere.
$endgroup$
– Klaus
Jan 30 at 23:50