On linearly independent matrices












12












$begingroup$


I have been reading J.S. Milne's lecture notes on fields and Galois theory and came across the normal basis theorem (Thm 5.18 on page 66 of the notes). Trying to find my own proof, the following problem in linear algebra quickly arose:




Question: Let $F$ be a field. Given linearly independent matrices $A_1, dots, A_n in operatorname{GL}_n(F)$, does there necessarily exist some $bin F^n$ such that $A_1b, dots, A_nb$ are linearly independent over $F$?




This is clearly not true if the matrices are not linearly independent. Also, if they are not invertible, the claim is false in general: e.g. set $n=2$ and consider



$$A_1 = begin{pmatrix} 1 & 0 \ 0& 0end{pmatrix},quad A_2 = begin{pmatrix} 0 & 1 \ 0& 0end{pmatrix}$$



Going through a case-by-case analysis, I think I can prove the claim for $n=2$, so there seems to be some hope...



Any help would be appreciated. Thanks!






linear-algebra abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    (I'm rambling here) In characteristic zero, let $c_1, c_2 neq 0.$ Then $c_1 A_1 b + c_2 A_2 b = (c_1 A_1 + c_2 A_2) b.$ But $c_1 A_1 + c_2 A_2 neq 0,$ and so $c_1 A_1 b + c_2 A_2 b$ is non-zero as long as $b$ is in the row space of $c_1 A_1 + c_2 A_2.$ Now if we can find $c_1, c_2$ such that $c_1 A_1 + c_2 A_2$ is non-singular, then there necessarily exists such $b$??
    $endgroup$
    – user2468
    Jul 31 '12 at 18:19












  • $begingroup$
    You have $A_1 in GL_n(F)$ and $b in F^n$. What multiplication are you using for $A_1b$? (I'm a random noob trying to understand the problem)
    $endgroup$
    – xavierm02
    Jul 31 '12 at 18:48












  • $begingroup$
    @xavierm02: Just plain old matrix multiplication (identifying $F^n = F^{ntimes 1}$)
    $endgroup$
    – Sam
    Jul 31 '12 at 18:51








  • 1




    $begingroup$
    Proof for $n = 2$: since the problem is invariant under left multiplication by an invertible matrix, you may assume WLOG that $A_1 = I$. If $A_2 b$ is always a scalar multiple of $b = A_1 b$, then it is not hard to show that $A_2$ must be a scalar multiple of the identity, but this contradicts the assumption of linear independence.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '12 at 19:35








  • 2




    $begingroup$
    @EdGorcenski: it's just "the" definition of linear independence in an arbitrary vector space. Linear independence is a property of a set of vectors (and, of course, the set of all matrices is a vector space) .
    $endgroup$
    – Martin Argerami
    Jul 31 '12 at 20:48
















12












$begingroup$


I have been reading J.S. Milne's lecture notes on fields and Galois theory and came across the normal basis theorem (Thm 5.18 on page 66 of the notes). Trying to find my own proof, the following problem in linear algebra quickly arose:




Question: Let $F$ be a field. Given linearly independent matrices $A_1, dots, A_n in operatorname{GL}_n(F)$, does there necessarily exist some $bin F^n$ such that $A_1b, dots, A_nb$ are linearly independent over $F$?




This is clearly not true if the matrices are not linearly independent. Also, if they are not invertible, the claim is false in general: e.g. set $n=2$ and consider



$$A_1 = begin{pmatrix} 1 & 0 \ 0& 0end{pmatrix},quad A_2 = begin{pmatrix} 0 & 1 \ 0& 0end{pmatrix}$$



Going through a case-by-case analysis, I think I can prove the claim for $n=2$, so there seems to be some hope...



Any help would be appreciated. Thanks!






linear-algebra abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    (I'm rambling here) In characteristic zero, let $c_1, c_2 neq 0.$ Then $c_1 A_1 b + c_2 A_2 b = (c_1 A_1 + c_2 A_2) b.$ But $c_1 A_1 + c_2 A_2 neq 0,$ and so $c_1 A_1 b + c_2 A_2 b$ is non-zero as long as $b$ is in the row space of $c_1 A_1 + c_2 A_2.$ Now if we can find $c_1, c_2$ such that $c_1 A_1 + c_2 A_2$ is non-singular, then there necessarily exists such $b$??
    $endgroup$
    – user2468
    Jul 31 '12 at 18:19












  • $begingroup$
    You have $A_1 in GL_n(F)$ and $b in F^n$. What multiplication are you using for $A_1b$? (I'm a random noob trying to understand the problem)
    $endgroup$
    – xavierm02
    Jul 31 '12 at 18:48












  • $begingroup$
    @xavierm02: Just plain old matrix multiplication (identifying $F^n = F^{ntimes 1}$)
    $endgroup$
    – Sam
    Jul 31 '12 at 18:51








  • 1




    $begingroup$
    Proof for $n = 2$: since the problem is invariant under left multiplication by an invertible matrix, you may assume WLOG that $A_1 = I$. If $A_2 b$ is always a scalar multiple of $b = A_1 b$, then it is not hard to show that $A_2$ must be a scalar multiple of the identity, but this contradicts the assumption of linear independence.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '12 at 19:35








  • 2




    $begingroup$
    @EdGorcenski: it's just "the" definition of linear independence in an arbitrary vector space. Linear independence is a property of a set of vectors (and, of course, the set of all matrices is a vector space) .
    $endgroup$
    – Martin Argerami
    Jul 31 '12 at 20:48














12












12








12


2



$begingroup$


I have been reading J.S. Milne's lecture notes on fields and Galois theory and came across the normal basis theorem (Thm 5.18 on page 66 of the notes). Trying to find my own proof, the following problem in linear algebra quickly arose:




Question: Let $F$ be a field. Given linearly independent matrices $A_1, dots, A_n in operatorname{GL}_n(F)$, does there necessarily exist some $bin F^n$ such that $A_1b, dots, A_nb$ are linearly independent over $F$?




This is clearly not true if the matrices are not linearly independent. Also, if they are not invertible, the claim is false in general: e.g. set $n=2$ and consider



$$A_1 = begin{pmatrix} 1 & 0 \ 0& 0end{pmatrix},quad A_2 = begin{pmatrix} 0 & 1 \ 0& 0end{pmatrix}$$



Going through a case-by-case analysis, I think I can prove the claim for $n=2$, so there seems to be some hope...



Any help would be appreciated. Thanks!






linear-algebra abstract-algebra







share|cite|improve this question









$endgroup$




I have been reading J.S. Milne's lecture notes on fields and Galois theory and came across the normal basis theorem (Thm 5.18 on page 66 of the notes). Trying to find my own proof, the following problem in linear algebra quickly arose:




Question: Let $F$ be a field. Given linearly independent matrices $A_1, dots, A_n in operatorname{GL}_n(F)$, does there necessarily exist some $bin F^n$ such that $A_1b, dots, A_nb$ are linearly independent over $F$?




This is clearly not true if the matrices are not linearly independent. Also, if they are not invertible, the claim is false in general: e.g. set $n=2$ and consider



$$A_1 = begin{pmatrix} 1 & 0 \ 0& 0end{pmatrix},quad A_2 = begin{pmatrix} 0 & 1 \ 0& 0end{pmatrix}$$



Going through a case-by-case analysis, I think I can prove the claim for $n=2$, so there seems to be some hope...



Any help would be appreciated. Thanks!






linear-algebra abstract-algebra




linear-algebra abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 31 '12 at 18:01









SamSam

10.5k13376




10.5k13376












  • $begingroup$
    (I'm rambling here) In characteristic zero, let $c_1, c_2 neq 0.$ Then $c_1 A_1 b + c_2 A_2 b = (c_1 A_1 + c_2 A_2) b.$ But $c_1 A_1 + c_2 A_2 neq 0,$ and so $c_1 A_1 b + c_2 A_2 b$ is non-zero as long as $b$ is in the row space of $c_1 A_1 + c_2 A_2.$ Now if we can find $c_1, c_2$ such that $c_1 A_1 + c_2 A_2$ is non-singular, then there necessarily exists such $b$??
    $endgroup$
    – user2468
    Jul 31 '12 at 18:19












  • $begingroup$
    You have $A_1 in GL_n(F)$ and $b in F^n$. What multiplication are you using for $A_1b$? (I'm a random noob trying to understand the problem)
    $endgroup$
    – xavierm02
    Jul 31 '12 at 18:48












  • $begingroup$
    @xavierm02: Just plain old matrix multiplication (identifying $F^n = F^{ntimes 1}$)
    $endgroup$
    – Sam
    Jul 31 '12 at 18:51








  • 1




    $begingroup$
    Proof for $n = 2$: since the problem is invariant under left multiplication by an invertible matrix, you may assume WLOG that $A_1 = I$. If $A_2 b$ is always a scalar multiple of $b = A_1 b$, then it is not hard to show that $A_2$ must be a scalar multiple of the identity, but this contradicts the assumption of linear independence.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '12 at 19:35








  • 2




    $begingroup$
    @EdGorcenski: it's just "the" definition of linear independence in an arbitrary vector space. Linear independence is a property of a set of vectors (and, of course, the set of all matrices is a vector space) .
    $endgroup$
    – Martin Argerami
    Jul 31 '12 at 20:48


















  • $begingroup$
    (I'm rambling here) In characteristic zero, let $c_1, c_2 neq 0.$ Then $c_1 A_1 b + c_2 A_2 b = (c_1 A_1 + c_2 A_2) b.$ But $c_1 A_1 + c_2 A_2 neq 0,$ and so $c_1 A_1 b + c_2 A_2 b$ is non-zero as long as $b$ is in the row space of $c_1 A_1 + c_2 A_2.$ Now if we can find $c_1, c_2$ such that $c_1 A_1 + c_2 A_2$ is non-singular, then there necessarily exists such $b$??
    $endgroup$
    – user2468
    Jul 31 '12 at 18:19












  • $begingroup$
    You have $A_1 in GL_n(F)$ and $b in F^n$. What multiplication are you using for $A_1b$? (I'm a random noob trying to understand the problem)
    $endgroup$
    – xavierm02
    Jul 31 '12 at 18:48












  • $begingroup$
    @xavierm02: Just plain old matrix multiplication (identifying $F^n = F^{ntimes 1}$)
    $endgroup$
    – Sam
    Jul 31 '12 at 18:51








  • 1




    $begingroup$
    Proof for $n = 2$: since the problem is invariant under left multiplication by an invertible matrix, you may assume WLOG that $A_1 = I$. If $A_2 b$ is always a scalar multiple of $b = A_1 b$, then it is not hard to show that $A_2$ must be a scalar multiple of the identity, but this contradicts the assumption of linear independence.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '12 at 19:35








  • 2




    $begingroup$
    @EdGorcenski: it's just "the" definition of linear independence in an arbitrary vector space. Linear independence is a property of a set of vectors (and, of course, the set of all matrices is a vector space) .
    $endgroup$
    – Martin Argerami
    Jul 31 '12 at 20:48
















$begingroup$
(I'm rambling here) In characteristic zero, let $c_1, c_2 neq 0.$ Then $c_1 A_1 b + c_2 A_2 b = (c_1 A_1 + c_2 A_2) b.$ But $c_1 A_1 + c_2 A_2 neq 0,$ and so $c_1 A_1 b + c_2 A_2 b$ is non-zero as long as $b$ is in the row space of $c_1 A_1 + c_2 A_2.$ Now if we can find $c_1, c_2$ such that $c_1 A_1 + c_2 A_2$ is non-singular, then there necessarily exists such $b$??
$endgroup$
– user2468
Jul 31 '12 at 18:19






$begingroup$
(I'm rambling here) In characteristic zero, let $c_1, c_2 neq 0.$ Then $c_1 A_1 b + c_2 A_2 b = (c_1 A_1 + c_2 A_2) b.$ But $c_1 A_1 + c_2 A_2 neq 0,$ and so $c_1 A_1 b + c_2 A_2 b$ is non-zero as long as $b$ is in the row space of $c_1 A_1 + c_2 A_2.$ Now if we can find $c_1, c_2$ such that $c_1 A_1 + c_2 A_2$ is non-singular, then there necessarily exists such $b$??
$endgroup$
– user2468
Jul 31 '12 at 18:19














$begingroup$
You have $A_1 in GL_n(F)$ and $b in F^n$. What multiplication are you using for $A_1b$? (I'm a random noob trying to understand the problem)
$endgroup$
– xavierm02
Jul 31 '12 at 18:48






$begingroup$
You have $A_1 in GL_n(F)$ and $b in F^n$. What multiplication are you using for $A_1b$? (I'm a random noob trying to understand the problem)
$endgroup$
– xavierm02
Jul 31 '12 at 18:48














$begingroup$
@xavierm02: Just plain old matrix multiplication (identifying $F^n = F^{ntimes 1}$)
$endgroup$
– Sam
Jul 31 '12 at 18:51






$begingroup$
@xavierm02: Just plain old matrix multiplication (identifying $F^n = F^{ntimes 1}$)
$endgroup$
– Sam
Jul 31 '12 at 18:51






1




1




$begingroup$
Proof for $n = 2$: since the problem is invariant under left multiplication by an invertible matrix, you may assume WLOG that $A_1 = I$. If $A_2 b$ is always a scalar multiple of $b = A_1 b$, then it is not hard to show that $A_2$ must be a scalar multiple of the identity, but this contradicts the assumption of linear independence.
$endgroup$
– Qiaochu Yuan
Jul 31 '12 at 19:35






$begingroup$
Proof for $n = 2$: since the problem is invariant under left multiplication by an invertible matrix, you may assume WLOG that $A_1 = I$. If $A_2 b$ is always a scalar multiple of $b = A_1 b$, then it is not hard to show that $A_2$ must be a scalar multiple of the identity, but this contradicts the assumption of linear independence.
$endgroup$
– Qiaochu Yuan
Jul 31 '12 at 19:35






2




2




$begingroup$
@EdGorcenski: it's just "the" definition of linear independence in an arbitrary vector space. Linear independence is a property of a set of vectors (and, of course, the set of all matrices is a vector space) .
$endgroup$
– Martin Argerami
Jul 31 '12 at 20:48




$begingroup$
@EdGorcenski: it's just "the" definition of linear independence in an arbitrary vector space. Linear independence is a property of a set of vectors (and, of course, the set of all matrices is a vector space) .
$endgroup$
– Martin Argerami
Jul 31 '12 at 20:48










1 Answer
1






active

oldest

votes


















10












$begingroup$

The answer is no.



Consider, for $F=mathbb{R}$, $n=3$,
$$
A_1=begin{bmatrix}1&0&0\0&1&0\ 1&2&3end{bmatrix},
A_2=begin{bmatrix}1&0&0\0&1&0\ 2&3&1end{bmatrix},
A_3=begin{bmatrix}1&0&0\0&1&0\ 3&1&2end{bmatrix}.
$$
These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal.



For any $b=begin{bmatrix}x\ y\ zend{bmatrix}inmathbb{R}^3$, the three vectors we obtain are
$$
A_1b=begin{bmatrix}x\ y\ x+2y+3zend{bmatrix},
A_2b=begin{bmatrix}x\ y\ 2x+3y+zend{bmatrix},
A_3b=begin{bmatrix}x\ y\ 3x+y+2zend{bmatrix}.
$$
And for any choice of $x,y,z$, these vectors are linearly dependent. To see this, we need to find coefficients, not all zero, such that $alpha,A_1b+beta,A_2b+gamma,A_3b=0$. The first two rows require $alpha+beta+gamma=0$. And the third row requires $(x+2y+3x)alpha+(2x+3y+z)beta+(2x+y+2x)gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
    $endgroup$
    – Gerry Myerson
    Aug 1 '12 at 3:46










  • $begingroup$
    Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
    $endgroup$
    – Martin Argerami
    Aug 1 '12 at 4:55










  • $begingroup$
    Nice. Thank you for the counterexample!
    $endgroup$
    – Sam
    Aug 1 '12 at 6:41











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









10












$begingroup$

The answer is no.



Consider, for $F=mathbb{R}$, $n=3$,
$$
A_1=begin{bmatrix}1&0&0\0&1&0\ 1&2&3end{bmatrix},
A_2=begin{bmatrix}1&0&0\0&1&0\ 2&3&1end{bmatrix},
A_3=begin{bmatrix}1&0&0\0&1&0\ 3&1&2end{bmatrix}.
$$
These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal.



For any $b=begin{bmatrix}x\ y\ zend{bmatrix}inmathbb{R}^3$, the three vectors we obtain are
$$
A_1b=begin{bmatrix}x\ y\ x+2y+3zend{bmatrix},
A_2b=begin{bmatrix}x\ y\ 2x+3y+zend{bmatrix},
A_3b=begin{bmatrix}x\ y\ 3x+y+2zend{bmatrix}.
$$
And for any choice of $x,y,z$, these vectors are linearly dependent. To see this, we need to find coefficients, not all zero, such that $alpha,A_1b+beta,A_2b+gamma,A_3b=0$. The first two rows require $alpha+beta+gamma=0$. And the third row requires $(x+2y+3x)alpha+(2x+3y+z)beta+(2x+y+2x)gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
    $endgroup$
    – Gerry Myerson
    Aug 1 '12 at 3:46










  • $begingroup$
    Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
    $endgroup$
    – Martin Argerami
    Aug 1 '12 at 4:55










  • $begingroup$
    Nice. Thank you for the counterexample!
    $endgroup$
    – Sam
    Aug 1 '12 at 6:41
















10












$begingroup$

The answer is no.



Consider, for $F=mathbb{R}$, $n=3$,
$$
A_1=begin{bmatrix}1&0&0\0&1&0\ 1&2&3end{bmatrix},
A_2=begin{bmatrix}1&0&0\0&1&0\ 2&3&1end{bmatrix},
A_3=begin{bmatrix}1&0&0\0&1&0\ 3&1&2end{bmatrix}.
$$
These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal.



For any $b=begin{bmatrix}x\ y\ zend{bmatrix}inmathbb{R}^3$, the three vectors we obtain are
$$
A_1b=begin{bmatrix}x\ y\ x+2y+3zend{bmatrix},
A_2b=begin{bmatrix}x\ y\ 2x+3y+zend{bmatrix},
A_3b=begin{bmatrix}x\ y\ 3x+y+2zend{bmatrix}.
$$
And for any choice of $x,y,z$, these vectors are linearly dependent. To see this, we need to find coefficients, not all zero, such that $alpha,A_1b+beta,A_2b+gamma,A_3b=0$. The first two rows require $alpha+beta+gamma=0$. And the third row requires $(x+2y+3x)alpha+(2x+3y+z)beta+(2x+y+2x)gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
    $endgroup$
    – Gerry Myerson
    Aug 1 '12 at 3:46










  • $begingroup$
    Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
    $endgroup$
    – Martin Argerami
    Aug 1 '12 at 4:55










  • $begingroup$
    Nice. Thank you for the counterexample!
    $endgroup$
    – Sam
    Aug 1 '12 at 6:41














10












10








10





$begingroup$

The answer is no.



Consider, for $F=mathbb{R}$, $n=3$,
$$
A_1=begin{bmatrix}1&0&0\0&1&0\ 1&2&3end{bmatrix},
A_2=begin{bmatrix}1&0&0\0&1&0\ 2&3&1end{bmatrix},
A_3=begin{bmatrix}1&0&0\0&1&0\ 3&1&2end{bmatrix}.
$$
These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal.



For any $b=begin{bmatrix}x\ y\ zend{bmatrix}inmathbb{R}^3$, the three vectors we obtain are
$$
A_1b=begin{bmatrix}x\ y\ x+2y+3zend{bmatrix},
A_2b=begin{bmatrix}x\ y\ 2x+3y+zend{bmatrix},
A_3b=begin{bmatrix}x\ y\ 3x+y+2zend{bmatrix}.
$$
And for any choice of $x,y,z$, these vectors are linearly dependent. To see this, we need to find coefficients, not all zero, such that $alpha,A_1b+beta,A_2b+gamma,A_3b=0$. The first two rows require $alpha+beta+gamma=0$. And the third row requires $(x+2y+3x)alpha+(2x+3y+z)beta+(2x+y+2x)gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions.






share|cite|improve this answer











$endgroup$



The answer is no.



Consider, for $F=mathbb{R}$, $n=3$,
$$
A_1=begin{bmatrix}1&0&0\0&1&0\ 1&2&3end{bmatrix},
A_2=begin{bmatrix}1&0&0\0&1&0\ 2&3&1end{bmatrix},
A_3=begin{bmatrix}1&0&0\0&1&0\ 3&1&2end{bmatrix}.
$$
These three matrices are linearly independent, because the last three rows are. They are invertible, since the are triangular with nonzero diagonal.



For any $b=begin{bmatrix}x\ y\ zend{bmatrix}inmathbb{R}^3$, the three vectors we obtain are
$$
A_1b=begin{bmatrix}x\ y\ x+2y+3zend{bmatrix},
A_2b=begin{bmatrix}x\ y\ 2x+3y+zend{bmatrix},
A_3b=begin{bmatrix}x\ y\ 3x+y+2zend{bmatrix}.
$$
And for any choice of $x,y,z$, these vectors are linearly dependent. To see this, we need to find coefficients, not all zero, such that $alpha,A_1b+beta,A_2b+gamma,A_3b=0$. The first two rows require $alpha+beta+gamma=0$. And the third row requires $(x+2y+3x)alpha+(2x+3y+z)beta+(2x+y+2x)gamma=0$. As it is a homogeneous system of two equations in three variables, it will always have nonzero solutions.







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edited Dec 8 '18 at 11:23

























answered Aug 1 '12 at 2:38









Martin ArgeramiMartin Argerami

125k1177177




125k1177177








  • 2




    $begingroup$
    A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
    $endgroup$
    – Gerry Myerson
    Aug 1 '12 at 3:46










  • $begingroup$
    Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
    $endgroup$
    – Martin Argerami
    Aug 1 '12 at 4:55










  • $begingroup$
    Nice. Thank you for the counterexample!
    $endgroup$
    – Sam
    Aug 1 '12 at 6:41














  • 2




    $begingroup$
    A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
    $endgroup$
    – Gerry Myerson
    Aug 1 '12 at 3:46










  • $begingroup$
    Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
    $endgroup$
    – Martin Argerami
    Aug 1 '12 at 4:55










  • $begingroup$
    Nice. Thank you for the counterexample!
    $endgroup$
    – Sam
    Aug 1 '12 at 6:41








2




2




$begingroup$
A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
$endgroup$
– Gerry Myerson
Aug 1 '12 at 3:46




$begingroup$
A tiny bit simpler: $$pmatrix{1&0&0cr0&1&0cr1&0&1cr},pmatrix{1&0&0cr0&1&0cr0&1&1cr},pmatrix{1&0&0cr0&1&0cr0&0&1cr}$$ Then $yA_1b-xA_2b+(x-y)A_3b=0$ gives a non-trivial linear dependence unless $x=y=0$, in which case $A_1b-A_2b=0$.
$endgroup$
– Gerry Myerson
Aug 1 '12 at 3:46












$begingroup$
Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
$endgroup$
– Martin Argerami
Aug 1 '12 at 4:55




$begingroup$
Nice. I didn't really think about optimizing the choice (which I like to do). I just noticed that one needs only 3 coordinates to make the matrices linearly independent, and the other 6 would likely fix a lot of things if equal.
$endgroup$
– Martin Argerami
Aug 1 '12 at 4:55












$begingroup$
Nice. Thank you for the counterexample!
$endgroup$
– Sam
Aug 1 '12 at 6:41




$begingroup$
Nice. Thank you for the counterexample!
$endgroup$
– Sam
Aug 1 '12 at 6:41


















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