Rank of a simple matrix












1














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







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  • 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














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







share|cite|improve this question




















  • 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








1







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 16:10









user376343

2,8582823




2,8582823










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














  • 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










1 Answer
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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 $1$.






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    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 $1$.






    share|cite|improve this answer




























      2














      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 $1$.






      share|cite|improve this answer


























        2












        2








        2






        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 $1$.






        share|cite|improve this answer














        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 $1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 20 '18 at 17:15

























        answered Nov 20 '18 at 16:19









        bRost03

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