Proving $ker(F)neq{0}$












0












$begingroup$


I've been posed this question:



Suppose that $F:mathbb{R}^ntomathbb{R}^m$ is a linear transformation and that $m<n$. Use the Dimension Theorem to prove that $ker(F)neq{0}$. (Hint: $mathcal{R}(F)$ is a subspace of $mathbb{R}^m$ - what does that tell you about the rank of $F$?)



I'm stuck with how to go about proving this question. All I could gather from the hint is that the rank of $Fleq m$ and I'm not even sure that is correct.
Any help would be appreciated






linear-algebra







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    State the dimension theorem. See where the rank and the dimension of the kernel come in.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:24








  • 1




    $begingroup$
    so far, i have: dimension theorem: $rank(A) + dimNul(A) = n$. Alternatively, $rank(A) + dimKer(F) = n$. Since $rank(F)=rank(A) leq m$, $dimKer(F) = n-rank(A) geq 0$ since, $rank(F) leq m<n$
    $endgroup$
    – lohboys
    Jan 11 at 6:32








  • 1




    $begingroup$
    If $mbox{rank } A = mbox{rank } F$, then $mbox{rank } A < n$ as well, so $dim ker F = n - mbox{rank } A$ is?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:39








  • 1




    $begingroup$
    Yes, that is correct. Now since $dimker F>0,ker Fne{mathbf 0}$ as $dim{mathbf 0}=0$
    $endgroup$
    – Shubham Johri
    Jan 11 at 6:43








  • 1




    $begingroup$
    Exactly. It is positive. Now, that of course means that the kernel contains some vector other than zero, because the dimension of the zero subspace is zero. That completes the argument Write an answer below yourself.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:44


















0












$begingroup$


I've been posed this question:



Suppose that $F:mathbb{R}^ntomathbb{R}^m$ is a linear transformation and that $m<n$. Use the Dimension Theorem to prove that $ker(F)neq{0}$. (Hint: $mathcal{R}(F)$ is a subspace of $mathbb{R}^m$ - what does that tell you about the rank of $F$?)



I'm stuck with how to go about proving this question. All I could gather from the hint is that the rank of $Fleq m$ and I'm not even sure that is correct.
Any help would be appreciated






linear-algebra







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    State the dimension theorem. See where the rank and the dimension of the kernel come in.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:24








  • 1




    $begingroup$
    so far, i have: dimension theorem: $rank(A) + dimNul(A) = n$. Alternatively, $rank(A) + dimKer(F) = n$. Since $rank(F)=rank(A) leq m$, $dimKer(F) = n-rank(A) geq 0$ since, $rank(F) leq m<n$
    $endgroup$
    – lohboys
    Jan 11 at 6:32








  • 1




    $begingroup$
    If $mbox{rank } A = mbox{rank } F$, then $mbox{rank } A < n$ as well, so $dim ker F = n - mbox{rank } A$ is?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:39








  • 1




    $begingroup$
    Yes, that is correct. Now since $dimker F>0,ker Fne{mathbf 0}$ as $dim{mathbf 0}=0$
    $endgroup$
    – Shubham Johri
    Jan 11 at 6:43








  • 1




    $begingroup$
    Exactly. It is positive. Now, that of course means that the kernel contains some vector other than zero, because the dimension of the zero subspace is zero. That completes the argument Write an answer below yourself.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:44
















0












0








0





$begingroup$


I've been posed this question:



Suppose that $F:mathbb{R}^ntomathbb{R}^m$ is a linear transformation and that $m<n$. Use the Dimension Theorem to prove that $ker(F)neq{0}$. (Hint: $mathcal{R}(F)$ is a subspace of $mathbb{R}^m$ - what does that tell you about the rank of $F$?)



I'm stuck with how to go about proving this question. All I could gather from the hint is that the rank of $Fleq m$ and I'm not even sure that is correct.
Any help would be appreciated






linear-algebra







share|cite|improve this question











$endgroup$




I've been posed this question:



Suppose that $F:mathbb{R}^ntomathbb{R}^m$ is a linear transformation and that $m<n$. Use the Dimension Theorem to prove that $ker(F)neq{0}$. (Hint: $mathcal{R}(F)$ is a subspace of $mathbb{R}^m$ - what does that tell you about the rank of $F$?)



I'm stuck with how to go about proving this question. All I could gather from the hint is that the rank of $Fleq m$ and I'm not even sure that is correct.
Any help would be appreciated






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 6:48









user26857

39.3k124183




39.3k124183










asked Jan 11 at 6:22









lohboyslohboys

9518




9518








  • 5




    $begingroup$
    State the dimension theorem. See where the rank and the dimension of the kernel come in.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:24








  • 1




    $begingroup$
    so far, i have: dimension theorem: $rank(A) + dimNul(A) = n$. Alternatively, $rank(A) + dimKer(F) = n$. Since $rank(F)=rank(A) leq m$, $dimKer(F) = n-rank(A) geq 0$ since, $rank(F) leq m<n$
    $endgroup$
    – lohboys
    Jan 11 at 6:32








  • 1




    $begingroup$
    If $mbox{rank } A = mbox{rank } F$, then $mbox{rank } A < n$ as well, so $dim ker F = n - mbox{rank } A$ is?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:39








  • 1




    $begingroup$
    Yes, that is correct. Now since $dimker F>0,ker Fne{mathbf 0}$ as $dim{mathbf 0}=0$
    $endgroup$
    – Shubham Johri
    Jan 11 at 6:43








  • 1




    $begingroup$
    Exactly. It is positive. Now, that of course means that the kernel contains some vector other than zero, because the dimension of the zero subspace is zero. That completes the argument Write an answer below yourself.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:44
















  • 5




    $begingroup$
    State the dimension theorem. See where the rank and the dimension of the kernel come in.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:24








  • 1




    $begingroup$
    so far, i have: dimension theorem: $rank(A) + dimNul(A) = n$. Alternatively, $rank(A) + dimKer(F) = n$. Since $rank(F)=rank(A) leq m$, $dimKer(F) = n-rank(A) geq 0$ since, $rank(F) leq m<n$
    $endgroup$
    – lohboys
    Jan 11 at 6:32








  • 1




    $begingroup$
    If $mbox{rank } A = mbox{rank } F$, then $mbox{rank } A < n$ as well, so $dim ker F = n - mbox{rank } A$ is?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:39








  • 1




    $begingroup$
    Yes, that is correct. Now since $dimker F>0,ker Fne{mathbf 0}$ as $dim{mathbf 0}=0$
    $endgroup$
    – Shubham Johri
    Jan 11 at 6:43








  • 1




    $begingroup$
    Exactly. It is positive. Now, that of course means that the kernel contains some vector other than zero, because the dimension of the zero subspace is zero. That completes the argument Write an answer below yourself.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 11 at 6:44










5




5




$begingroup$
State the dimension theorem. See where the rank and the dimension of the kernel come in.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 6:24






$begingroup$
State the dimension theorem. See where the rank and the dimension of the kernel come in.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 6:24






1




1




$begingroup$
so far, i have: dimension theorem: $rank(A) + dimNul(A) = n$. Alternatively, $rank(A) + dimKer(F) = n$. Since $rank(F)=rank(A) leq m$, $dimKer(F) = n-rank(A) geq 0$ since, $rank(F) leq m<n$
$endgroup$
– lohboys
Jan 11 at 6:32






$begingroup$
so far, i have: dimension theorem: $rank(A) + dimNul(A) = n$. Alternatively, $rank(A) + dimKer(F) = n$. Since $rank(F)=rank(A) leq m$, $dimKer(F) = n-rank(A) geq 0$ since, $rank(F) leq m<n$
$endgroup$
– lohboys
Jan 11 at 6:32






1




1




$begingroup$
If $mbox{rank } A = mbox{rank } F$, then $mbox{rank } A < n$ as well, so $dim ker F = n - mbox{rank } A$ is?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 6:39






$begingroup$
If $mbox{rank } A = mbox{rank } F$, then $mbox{rank } A < n$ as well, so $dim ker F = n - mbox{rank } A$ is?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 6:39






1




1




$begingroup$
Yes, that is correct. Now since $dimker F>0,ker Fne{mathbf 0}$ as $dim{mathbf 0}=0$
$endgroup$
– Shubham Johri
Jan 11 at 6:43






$begingroup$
Yes, that is correct. Now since $dimker F>0,ker Fne{mathbf 0}$ as $dim{mathbf 0}=0$
$endgroup$
– Shubham Johri
Jan 11 at 6:43






1




1




$begingroup$
Exactly. It is positive. Now, that of course means that the kernel contains some vector other than zero, because the dimension of the zero subspace is zero. That completes the argument Write an answer below yourself.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 6:44






$begingroup$
Exactly. It is positive. Now, that of course means that the kernel contains some vector other than zero, because the dimension of the zero subspace is zero. That completes the argument Write an answer below yourself.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 11 at 6:44












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$begingroup$

You know that since F is linear $dim(Ran(F))+dim(Ker(F))=n$ so, like you said, $dim(Ran(F))le m$. For equality to be true, it is necessary that $dim(Ker(F))>0$ so $Ker(F)neq{0}$.






share|cite|improve this answer









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    $begingroup$

    You know that since F is linear $dim(Ran(F))+dim(Ker(F))=n$ so, like you said, $dim(Ran(F))le m$. For equality to be true, it is necessary that $dim(Ker(F))>0$ so $Ker(F)neq{0}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You know that since F is linear $dim(Ran(F))+dim(Ker(F))=n$ so, like you said, $dim(Ran(F))le m$. For equality to be true, it is necessary that $dim(Ker(F))>0$ so $Ker(F)neq{0}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You know that since F is linear $dim(Ran(F))+dim(Ker(F))=n$ so, like you said, $dim(Ran(F))le m$. For equality to be true, it is necessary that $dim(Ker(F))>0$ so $Ker(F)neq{0}$.






        share|cite|improve this answer









        $endgroup$



        You know that since F is linear $dim(Ran(F))+dim(Ker(F))=n$ so, like you said, $dim(Ran(F))le m$. For equality to be true, it is necessary that $dim(Ker(F))>0$ so $Ker(F)neq{0}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 10:36









        Antonio BAntonio B

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