Proof that $ operatorname{rank }(A) le 4$












4












$begingroup$



Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $ operatorname{rank} (A) le 4$





I know that it is truth because if I have seven linearly independent vectors and I multiply a matrix with these vectors I must have at least $6$ zero rows because I have $Ax_{1}=Ax_{2}=...=Ax_{7}$. For example if I have $5$ zero rows then this vectors are not linearly independent.




However I don't know how to prove it in an elegant way.






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Rank nullity theorem. Think about $dim (mathrm {Ker}(A))$.
    $endgroup$
    – xbh
    Feb 1 at 16:42
















4












$begingroup$



Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $ operatorname{rank} (A) le 4$





I know that it is truth because if I have seven linearly independent vectors and I multiply a matrix with these vectors I must have at least $6$ zero rows because I have $Ax_{1}=Ax_{2}=...=Ax_{7}$. For example if I have $5$ zero rows then this vectors are not linearly independent.




However I don't know how to prove it in an elegant way.






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Rank nullity theorem. Think about $dim (mathrm {Ker}(A))$.
    $endgroup$
    – xbh
    Feb 1 at 16:42














4












4








4





$begingroup$



Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $ operatorname{rank} (A) le 4$





I know that it is truth because if I have seven linearly independent vectors and I multiply a matrix with these vectors I must have at least $6$ zero rows because I have $Ax_{1}=Ax_{2}=...=Ax_{7}$. For example if I have $5$ zero rows then this vectors are not linearly independent.




However I don't know how to prove it in an elegant way.






linear-algebra







share|cite|improve this question











$endgroup$





Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $ operatorname{rank} (A) le 4$





I know that it is truth because if I have seven linearly independent vectors and I multiply a matrix with these vectors I must have at least $6$ zero rows because I have $Ax_{1}=Ax_{2}=...=Ax_{7}$. For example if I have $5$ zero rows then this vectors are not linearly independent.




However I don't know how to prove it in an elegant way.






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 Feb 1 at 17:18









Bernard

124k741117




124k741117










asked Feb 1 at 16:40









MP3129MP3129

792211




792211








  • 1




    $begingroup$
    Rank nullity theorem. Think about $dim (mathrm {Ker}(A))$.
    $endgroup$
    – xbh
    Feb 1 at 16:42














  • 1




    $begingroup$
    Rank nullity theorem. Think about $dim (mathrm {Ker}(A))$.
    $endgroup$
    – xbh
    Feb 1 at 16:42








1




1




$begingroup$
Rank nullity theorem. Think about $dim (mathrm {Ker}(A))$.
$endgroup$
– xbh
Feb 1 at 16:42




$begingroup$
Rank nullity theorem. Think about $dim (mathrm {Ker}(A))$.
$endgroup$
– xbh
Feb 1 at 16:42










1 Answer
1






active

oldest

votes


















4












$begingroup$

You can find 6 linearly-independent vectors in the kernel of $A$ by considering
$y_1=x_2-x_1$, $y_2=x_3-x_1$,...,$y_6=x_7-x_1$. Indeed, you can check that $Ay_i=0$ for all $1geq i geq 6$.



So $dim Ker(A) geq 6$.
By the Rank-Nullity Theorem, $$rank(T) = 10 - dim Ker(T) leq 10-6=4$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
    $endgroup$
    – MP3129
    Feb 1 at 16:55








  • 1




    $begingroup$
    OK I added some more details.
    $endgroup$
    – Stefan Lafon
    Feb 1 at 17:04












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1 Answer
1






active

oldest

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4












$begingroup$

You can find 6 linearly-independent vectors in the kernel of $A$ by considering
$y_1=x_2-x_1$, $y_2=x_3-x_1$,...,$y_6=x_7-x_1$. Indeed, you can check that $Ay_i=0$ for all $1geq i geq 6$.



So $dim Ker(A) geq 6$.
By the Rank-Nullity Theorem, $$rank(T) = 10 - dim Ker(T) leq 10-6=4$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
    $endgroup$
    – MP3129
    Feb 1 at 16:55








  • 1




    $begingroup$
    OK I added some more details.
    $endgroup$
    – Stefan Lafon
    Feb 1 at 17:04
















4












$begingroup$

You can find 6 linearly-independent vectors in the kernel of $A$ by considering
$y_1=x_2-x_1$, $y_2=x_3-x_1$,...,$y_6=x_7-x_1$. Indeed, you can check that $Ay_i=0$ for all $1geq i geq 6$.



So $dim Ker(A) geq 6$.
By the Rank-Nullity Theorem, $$rank(T) = 10 - dim Ker(T) leq 10-6=4$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
    $endgroup$
    – MP3129
    Feb 1 at 16:55








  • 1




    $begingroup$
    OK I added some more details.
    $endgroup$
    – Stefan Lafon
    Feb 1 at 17:04














4












4








4





$begingroup$

You can find 6 linearly-independent vectors in the kernel of $A$ by considering
$y_1=x_2-x_1$, $y_2=x_3-x_1$,...,$y_6=x_7-x_1$. Indeed, you can check that $Ay_i=0$ for all $1geq i geq 6$.



So $dim Ker(A) geq 6$.
By the Rank-Nullity Theorem, $$rank(T) = 10 - dim Ker(T) leq 10-6=4$$






share|cite|improve this answer











$endgroup$



You can find 6 linearly-independent vectors in the kernel of $A$ by considering
$y_1=x_2-x_1$, $y_2=x_3-x_1$,...,$y_6=x_7-x_1$. Indeed, you can check that $Ay_i=0$ for all $1geq i geq 6$.



So $dim Ker(A) geq 6$.
By the Rank-Nullity Theorem, $$rank(T) = 10 - dim Ker(T) leq 10-6=4$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 1 at 17:03

























answered Feb 1 at 16:43









Stefan LafonStefan Lafon

3,005212




3,005212












  • $begingroup$
    Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
    $endgroup$
    – MP3129
    Feb 1 at 16:55








  • 1




    $begingroup$
    OK I added some more details.
    $endgroup$
    – Stefan Lafon
    Feb 1 at 17:04


















  • $begingroup$
    Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
    $endgroup$
    – MP3129
    Feb 1 at 16:55








  • 1




    $begingroup$
    OK I added some more details.
    $endgroup$
    – Stefan Lafon
    Feb 1 at 17:04
















$begingroup$
Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
$endgroup$
– MP3129
Feb 1 at 16:55






$begingroup$
Ok, you have right, but have you got an idea how to describe nicely? I think your words are insufficient
$endgroup$
– MP3129
Feb 1 at 16:55






1




1




$begingroup$
OK I added some more details.
$endgroup$
– Stefan Lafon
Feb 1 at 17:04




$begingroup$
OK I added some more details.
$endgroup$
– Stefan Lafon
Feb 1 at 17:04


















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