Mixing
Rank of a simple matrix
While solving an exercise I arrived to this simple matrix and I want to get its rank
It is
$$begin{pmatrix}2cos 2t\
cos t
end{pmatrix}$$
It is written in the solution that the rank is 1, they took 2 cases if $cos t=0$ which is easy and the other case when $cos t ne 0$, and this second case is the one I didn't understand.
Help please by showing steps and by giving the precise definition of a rank, it is the number of nonzero rows in a matrix?
linear-algebra matrices matrix-rank
edited Nov 20 '18 at 16:10
user376343
2,8582823
asked Nov 20 '18 at 16:02
Fareed AF
44211
add a comment |
While solving an exercise I arrived to this simple matrix and I want to get its rank
It is
$$begin{pmatrix}2cos 2t\
cos t
end{pmatrix}$$
It is written in the solution that the rank is 1, they took 2 cases if $cos t=0$ which is easy and the other case when $cos t ne 0$, and this second case is the one I didn't understand.
Help please by showing steps and by giving the precise definition of a rank, it is the number of nonzero rows in a matrix?
linear-algebra matrices matrix-rank
edited Nov 20 '18 at 16:10
user376343
2,8582823
asked Nov 20 '18 at 16:02
Fareed AF
44211
1
The rank cannot be larger than $1$ because the matrix has only one column. The rank cannot be $0$ because $cos 2t$ and $cos t$ are never simultaneously $0.$ See lso en.wikipedia.org/wiki/Rank_(linear_algebra)
– user376343
Nov 20 '18 at 16:09
1
The rank of an $mtimes n$ matrix can never be larger than $min{m,n}$ because the rank is the number of linearly independent rows (or linearly independent columns, these are always the same). So in this case the rank must be either $0$ or $1$.
– MPW
Nov 20 '18 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 '18 at 16:15
add a comment |
While solving an exercise I arrived to this simple matrix and I want to get its rank
It is
$$begin{pmatrix}2cos 2t\
cos t
end{pmatrix}$$
It is written in the solution that the rank is 1, they took 2 cases if $cos t=0$ which is easy and the other case when $cos t ne 0$, and this second case is the one I didn't understand.
Help please by showing steps and by giving the precise definition of a rank, it is the number of nonzero rows in a matrix?
linear-algebra matrices matrix-rank
edited Nov 20 '18 at 16:10
user376343
2,8582823
asked Nov 20 '18 at 16:02
Fareed AF
44211
While solving an exercise I arrived to this simple matrix and I want to get its rank
It is
$$begin{pmatrix}2cos 2t\
cos t
end{pmatrix}$$
It is written in the solution that the rank is 1, they took 2 cases if $cos t=0$ which is easy and the other case when $cos t ne 0$, and this second case is the one I didn't understand.
Help please by showing steps and by giving the precise definition of a rank, it is the number of nonzero rows in a matrix?
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
edited Nov 20 '18 at 16:10
user376343
2,8582823
asked Nov 20 '18 at 16:02
Fareed AF
44211
edited Nov 20 '18 at 16:10
user376343
2,8582823
asked Nov 20 '18 at 16:02
Fareed AF
44211
edited Nov 20 '18 at 16:10
user376343
2,8582823
edited Nov 20 '18 at 16:10
user376343
2,8582823
edited Nov 20 '18 at 16:10
user376343
2,8582823
2,8582823
asked Nov 20 '18 at 16:02
Fareed AF
44211
asked Nov 20 '18 at 16:02
Fareed AF
44211
asked Nov 20 '18 at 16:02
Fareed AF
44211
44211
1
The rank cannot be larger than $1$ because the matrix has only one column. The rank cannot be $0$ because $cos 2t$ and $cos t$ are never simultaneously $0.$ See lso en.wikipedia.org/wiki/Rank_(linear_algebra)
– user376343
Nov 20 '18 at 16:09
1
The rank of an $mtimes n$ matrix can never be larger than $min{m,n}$ because the rank is the number of linearly independent rows (or linearly independent columns, these are always the same). So in this case the rank must be either $0$ or $1$.
– MPW
Nov 20 '18 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 '18 at 16:15
add a comment |
1
The rank cannot be larger than $1$ because the matrix has only one column. The rank cannot be $0$ because $cos 2t$ and $cos t$ are never simultaneously $0.$ See lso en.wikipedia.org/wiki/Rank_(linear_algebra)
– user376343
Nov 20 '18 at 16:09
1
The rank of an $mtimes n$ matrix can never be larger than $min{m,n}$ because the rank is the number of linearly independent rows (or linearly independent columns, these are always the same). So in this case the rank must be either $0$ or $1$.
– MPW
Nov 20 '18 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 '18 at 16:15
1
1
The rank cannot be larger than $1$ because the matrix has only one column. The rank cannot be $0$ because $cos 2t$ and $cos t$ are never simultaneously $0.$ See lso en.wikipedia.org/wiki/Rank_(linear_algebra)
– user376343
Nov 20 '18 at 16:09
The rank cannot be larger than $1$ because the matrix has only one column. The rank cannot be $0$ because $cos 2t$ and $cos t$ are never simultaneously $0.$ See lso en.wikipedia.org/wiki/Rank_(linear_algebra)
– user376343
Nov 20 '18 at 16:09
1
1
The rank of an $mtimes n$ matrix can never be larger than $min{m,n}$ because the rank is the number of linearly independent rows (or linearly independent columns, these are always the same). So in this case the rank must be either $0$ or $1$.
– MPW
Nov 20 '18 at 16:11
The rank of an $mtimes n$ matrix can never be larger than $min{m,n}$ because the rank is the number of linearly independent rows (or linearly independent columns, these are always the same). So in this case the rank must be either $0$ or $1$.
– MPW
Nov 20 '18 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 '18 at 16:15
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 '18 at 16:15
add a comment |
1 Answer
1
active
oldest
votes
If $cos(t)=0$ then $t=pi n-frac{pi }{2}$ with $nin mathbb{Z}$ so $2cos(2t)=2cosleft(2left(pi n-frac{pi }{2}right)right)=2cosleft(-piright)=-2$ and your matrix is $left(-2,0right)$.
If $cos(t)neq0$ then your matrix is not $(0,0)$
The rank is the dimension of the space spanned by the columns. Since your matrix has only a single column, it has rank $0$ if it is the $0$ matrix (i.e. it spans a zero dimensional space because it can only generate a single point, the origin) and $1$ otherwise since it generates a line, a one-dimensional object.
Since we've shown the matrix is never the $0$ matrix, it has rank $1$.
answered Nov 20 '18 at 16:19
bRost03
34819
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%2f3006501%2frank-of-a-simple-matrix%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
If $cos(t)=0$ then $t=pi n-frac{pi }{2}$ with $nin mathbb{Z}$ so $2cos(2t)=2cosleft(2left(pi n-frac{pi }{2}right)right)=2cosleft(-piright)=-2$ and your matrix is $left(-2,0right)$.
If $cos(t)neq0$ then your matrix is not $(0,0)$
The rank is the dimension of the space spanned by the columns. Since your matrix has only a single column, it has rank $0$ if it is the $0$ matrix (i.e. it spans a zero dimensional space because it can only generate a single point, the origin) and $1$ otherwise since it generates a line, a one-dimensional object.
Since we've shown the matrix is never the $0$ matrix, it has rank $1$.
answered Nov 20 '18 at 16:19
bRost03
34819
add a comment |
If $cos(t)=0$ then $t=pi n-frac{pi }{2}$ with $nin mathbb{Z}$ so $2cos(2t)=2cosleft(2left(pi n-frac{pi }{2}right)right)=2cosleft(-piright)=-2$ and your matrix is $left(-2,0right)$.
If $cos(t)neq0$ then your matrix is not $(0,0)$
The rank is the dimension of the space spanned by the columns. Since your matrix has only a single column, it has rank $0$ if it is the $0$ matrix (i.e. it spans a zero dimensional space because it can only generate a single point, the origin) and $1$ otherwise since it generates a line, a one-dimensional object.
Since we've shown the matrix is never the $0$ matrix, it has rank $1$.
answered Nov 20 '18 at 16:19
bRost03
34819
add a comment |
If $cos(t)=0$ then $t=pi n-frac{pi }{2}$ with $nin mathbb{Z}$ so $2cos(2t)=2cosleft(2left(pi n-frac{pi }{2}right)right)=2cosleft(-piright)=-2$ and your matrix is $left(-2,0right)$.
If $cos(t)neq0$ then your matrix is not $(0,0)$
The rank is the dimension of the space spanned by the columns. Since your matrix has only a single column, it has rank $0$ if it is the $0$ matrix (i.e. it spans a zero dimensional space because it can only generate a single point, the origin) and $1$ otherwise since it generates a line, a one-dimensional object.
Since we've shown the matrix is never the $0$ matrix, it has rank $1$.
answered Nov 20 '18 at 16:19
bRost03
34819
If $cos(t)=0$ then $t=pi n-frac{pi }{2}$ with $nin mathbb{Z}$ so $2cos(2t)=2cosleft(2left(pi n-frac{pi }{2}right)right)=2cosleft(-piright)=-2$ and your matrix is $left(-2,0right)$.
If $cos(t)neq0$ then your matrix is not $(0,0)$
The rank is the dimension of the space spanned by the columns. Since your matrix has only a single column, it has rank $0$ if it is the $0$ matrix (i.e. it spans a zero dimensional space because it can only generate a single point, the origin) and $1$ otherwise since it generates a line, a one-dimensional object.
Since we've shown the matrix is never the $0$ matrix, it has rank $1$.
answered Nov 20 '18 at 16:19
bRost03
34819
edited Nov 20 '18 at 17:15
answered Nov 20 '18 at 16:19
bRost03
34819
answered Nov 20 '18 at 16:19
bRost03
34819
answered Nov 20 '18 at 16:19
bRost03
34819
34819
add a comment |
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%2f3006501%2frank-of-a-simple-matrix%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
The rank cannot be larger than $1$ because the matrix has only one column. The rank cannot be $0$ because $cos 2t$ and $cos t$ are never simultaneously $0.$ See lso en.wikipedia.org/wiki/Rank_(linear_algebra)
– user376343
Nov 20 '18 at 16:09
1
The rank of an $mtimes n$ matrix can never be larger than $min{m,n}$ because the rank is the number of linearly independent rows (or linearly independent columns, these are always the same). So in this case the rank must be either $0$ or $1$.
– MPW
Nov 20 '18 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 '18 at 16:15