Rank of a simple matrix
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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 at 16:10
user376343
2,6432819
asked Nov 20 at 16:02
Fareed AF
39711
add a comment |
up vote
0
down vote
favorite
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 at 16:10
user376343
2,6432819
asked Nov 20 at 16:02
Fareed AF
39711
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 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 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 at 16:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 16:10
user376343
2,6432819
asked Nov 20 at 16:02
Fareed AF
39711
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 at 16:10
user376343
2,6432819
asked Nov 20 at 16:02
Fareed AF
39711
edited Nov 20 at 16:10
user376343
2,6432819
asked Nov 20 at 16:02
Fareed AF
39711
edited Nov 20 at 16:10
user376343
2,6432819
edited Nov 20 at 16:10
user376343
2,6432819
edited Nov 20 at 16:10
user376343
2,6432819
2,6432819
asked Nov 20 at 16:02
Fareed AF
39711
asked Nov 20 at 16:02
Fareed AF
39711
asked Nov 20 at 16:02
Fareed AF
39711
39711
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 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 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 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 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 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 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 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 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 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 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 at 16:15
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 at 16:15
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
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 at 16:19
bRost03
34319
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 at 16:19
bRost03
34319
add a comment |
up vote
2
down vote
accepted
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 at 16:19
bRost03
34319
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 at 16:19
bRost03
34319
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 at 16:19
bRost03
34319
edited Nov 20 at 17:15
answered Nov 20 at 16:19
bRost03
34319
answered Nov 20 at 16:19
bRost03
34319
answered Nov 20 at 16:19
bRost03
34319
34319
add a comment |
add a comment |
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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 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 at 16:11
It is $t$ fixed? Or you are using that coef. as functions?
– José Alejandro Aburto Araneda
Nov 20 at 16:15