Do all elements belonging to kernel of a linear transformation represented by a skew symmetric matrix NOT belong to its range? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Help clearing doubt about expansion of $vec itimes(vec atimes vec i)$Do four dimensional vectors have a cross product property?$vecatimes(vecatimesvecR)-vecbtimes(vecbtimesvecR)$Making sense of a cross product of three vectorsShow matrix transformation, finding kernel & range of itIs taking sum inside cross product valid?A bit confused on usage of cross products.Cross Product PropertiesKernel and image of a matrix converting linear transformationThe proof of Triple Vector Products
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Do all elements belonging to kernel of a linear transformation represented by a skew symmetric matrix NOT belong to its range?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Help clearing doubt about expansion of $vec itimes(vec atimes vec i)$Do four dimensional vectors have a cross product property?$vecatimes(vecatimesvecR)-vecbtimes(vecbtimesvecR)$Making sense of a cross product of three vectorsShow matrix transformation, finding kernel & range of itIs taking sum inside cross product valid?A bit confused on usage of cross products.Cross Product PropertiesKernel and image of a matrix converting linear transformationThe proof of Triple Vector Products
$begingroup$
I have a linear transformation represented by a skew symmetric matrix $S(vec b(t))$ of rank $2$, which in my context, is the cross-product matrix of magnetic field $vec b(t)$. I have explained what a cross product matrix is towards the end.
I know that its kernel is the set of all vectors collinear to $vec b(t)$ (since cross product of a non-zero vector with another non-zero vector is zero if and only if the other vector is collinear to it). Now, this fact is used to imply that the linear transformation can never give a vector collinear to $vec b(t)$ as output. I do not understand why this implication is true.
Cross product matrix refers to the matrix $S(vec b(t))$ such that, $$ vec atimes vec b = S(vec b)a $$ where $a$ is a $3 times 1$ column matrix.
Note: Just in case this might be helpful to someone, in my case, I have torque, $$tau = S(vec b(t))m_coils$$ where $m_coils$ is the magnetic moment generated by current carrying coils. Now the above fact is used to imply torque can never be along magnetic field vector.
Edit: This can be trivially shown using the fact that $vec atimes vec b$ is perpendicular to $vec b$. But the implication made in the paper I am reading seems to have been made solely based on the fact that kernel is given by $vec b$. Now, I maybe wrong here in this interpretation or the paper itself might have a mistake but the wording there seems to be saying this.
matrices linear-transformations cross-product
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
$endgroup$
add a comment |
$begingroup$
I have a linear transformation represented by a skew symmetric matrix $S(vec b(t))$ of rank $2$, which in my context, is the cross-product matrix of magnetic field $vec b(t)$. I have explained what a cross product matrix is towards the end.
I know that its kernel is the set of all vectors collinear to $vec b(t)$ (since cross product of a non-zero vector with another non-zero vector is zero if and only if the other vector is collinear to it). Now, this fact is used to imply that the linear transformation can never give a vector collinear to $vec b(t)$ as output. I do not understand why this implication is true.
Cross product matrix refers to the matrix $S(vec b(t))$ such that, $$ vec atimes vec b = S(vec b)a $$ where $a$ is a $3 times 1$ column matrix.
Note: Just in case this might be helpful to someone, in my case, I have torque, $$tau = S(vec b(t))m_coils$$ where $m_coils$ is the magnetic moment generated by current carrying coils. Now the above fact is used to imply torque can never be along magnetic field vector.
Edit: This can be trivially shown using the fact that $vec atimes vec b$ is perpendicular to $vec b$. But the implication made in the paper I am reading seems to have been made solely based on the fact that kernel is given by $vec b$. Now, I maybe wrong here in this interpretation or the paper itself might have a mistake but the wording there seems to be saying this.
matrices linear-transformations cross-product
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
$endgroup$
$begingroup$
In your particular case, you just need the fact that $vecatimesvecb$ is orthogonal to $vecb$
$endgroup$
– Poon Levi
Apr 7 at 13:03
$begingroup$
@PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $vec b(t)$. I have edited my question to make this clear.
$endgroup$
– Niket Parikh
Apr 7 at 13:19
add a comment |
$begingroup$
I have a linear transformation represented by a skew symmetric matrix $S(vec b(t))$ of rank $2$, which in my context, is the cross-product matrix of magnetic field $vec b(t)$. I have explained what a cross product matrix is towards the end.
I know that its kernel is the set of all vectors collinear to $vec b(t)$ (since cross product of a non-zero vector with another non-zero vector is zero if and only if the other vector is collinear to it). Now, this fact is used to imply that the linear transformation can never give a vector collinear to $vec b(t)$ as output. I do not understand why this implication is true.
Cross product matrix refers to the matrix $S(vec b(t))$ such that, $$ vec atimes vec b = S(vec b)a $$ where $a$ is a $3 times 1$ column matrix.
Note: Just in case this might be helpful to someone, in my case, I have torque, $$tau = S(vec b(t))m_coils$$ where $m_coils$ is the magnetic moment generated by current carrying coils. Now the above fact is used to imply torque can never be along magnetic field vector.
Edit: This can be trivially shown using the fact that $vec atimes vec b$ is perpendicular to $vec b$. But the implication made in the paper I am reading seems to have been made solely based on the fact that kernel is given by $vec b$. Now, I maybe wrong here in this interpretation or the paper itself might have a mistake but the wording there seems to be saying this.
matrices linear-transformations cross-product
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
$endgroup$
I have a linear transformation represented by a skew symmetric matrix $S(vec b(t))$ of rank $2$, which in my context, is the cross-product matrix of magnetic field $vec b(t)$. I have explained what a cross product matrix is towards the end.
I know that its kernel is the set of all vectors collinear to $vec b(t)$ (since cross product of a non-zero vector with another non-zero vector is zero if and only if the other vector is collinear to it). Now, this fact is used to imply that the linear transformation can never give a vector collinear to $vec b(t)$ as output. I do not understand why this implication is true.
Cross product matrix refers to the matrix $S(vec b(t))$ such that, $$ vec atimes vec b = S(vec b)a $$ where $a$ is a $3 times 1$ column matrix.
Note: Just in case this might be helpful to someone, in my case, I have torque, $$tau = S(vec b(t))m_coils$$ where $m_coils$ is the magnetic moment generated by current carrying coils. Now the above fact is used to imply torque can never be along magnetic field vector.
Edit: This can be trivially shown using the fact that $vec atimes vec b$ is perpendicular to $vec b$. But the implication made in the paper I am reading seems to have been made solely based on the fact that kernel is given by $vec b$. Now, I maybe wrong here in this interpretation or the paper itself might have a mistake but the wording there seems to be saying this.
matrices linear-transformations cross-product
matrices linear-transformations cross-product
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
edited Apr 7 at 13:26
Niket Parikh
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
asked Apr 7 at 8:49
Niket ParikhNiket Parikh
397
397
$begingroup$
In your particular case, you just need the fact that $vecatimesvecb$ is orthogonal to $vecb$
$endgroup$
– Poon Levi
Apr 7 at 13:03
$begingroup$
@PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $vec b(t)$. I have edited my question to make this clear.
$endgroup$
– Niket Parikh
Apr 7 at 13:19
add a comment |
$begingroup$
In your particular case, you just need the fact that $vecatimesvecb$ is orthogonal to $vecb$
$endgroup$
– Poon Levi
Apr 7 at 13:03
$begingroup$
@PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $vec b(t)$. I have edited my question to make this clear.
$endgroup$
– Niket Parikh
Apr 7 at 13:19
$begingroup$
In your particular case, you just need the fact that $vecatimesvecb$ is orthogonal to $vecb$
$endgroup$
– Poon Levi
Apr 7 at 13:03
$begingroup$
In your particular case, you just need the fact that $vecatimesvecb$ is orthogonal to $vecb$
$endgroup$
– Poon Levi
Apr 7 at 13:03
$begingroup$
@PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $vec b(t)$. I have edited my question to make this clear.
$endgroup$
– Niket Parikh
Apr 7 at 13:19
$begingroup$
@PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $vec b(t)$. I have edited my question to make this clear.
$endgroup$
– Niket Parikh
Apr 7 at 13:19
add a comment |
1 Answer
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$begingroup$
Inspired by the $3times 3$ case (which reduces to a cross product), we may consider the quantity $u^T M v$, where $u$ is in the kernel of $M$, $v$ is an arbitrary column vector and $M$ is skew-symmetric. But we have
$$u^T M v=-(Mu)^T v=0$$
Hence a non-zero real vector cannot belong to both the kernel and range of $M$.
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Inspired by the $3times 3$ case (which reduces to a cross product), we may consider the quantity $u^T M v$, where $u$ is in the kernel of $M$, $v$ is an arbitrary column vector and $M$ is skew-symmetric. But we have
$$u^T M v=-(Mu)^T v=0$$
Hence a non-zero real vector cannot belong to both the kernel and range of $M$.
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
$endgroup$
add a comment |
$begingroup$
Inspired by the $3times 3$ case (which reduces to a cross product), we may consider the quantity $u^T M v$, where $u$ is in the kernel of $M$, $v$ is an arbitrary column vector and $M$ is skew-symmetric. But we have
$$u^T M v=-(Mu)^T v=0$$
Hence a non-zero real vector cannot belong to both the kernel and range of $M$.
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
$endgroup$
add a comment |
$begingroup$
Inspired by the $3times 3$ case (which reduces to a cross product), we may consider the quantity $u^T M v$, where $u$ is in the kernel of $M$, $v$ is an arbitrary column vector and $M$ is skew-symmetric. But we have
$$u^T M v=-(Mu)^T v=0$$
Hence a non-zero real vector cannot belong to both the kernel and range of $M$.
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
$endgroup$
Inspired by the $3times 3$ case (which reduces to a cross product), we may consider the quantity $u^T M v$, where $u$ is in the kernel of $M$, $v$ is an arbitrary column vector and $M$ is skew-symmetric. But we have
$$u^T M v=-(Mu)^T v=0$$
Hence a non-zero real vector cannot belong to both the kernel and range of $M$.
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
edited Apr 8 at 9:25
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
answered Apr 8 at 8:04
Poon LeviPoon Levi
73639
73639
add a comment |
add a comment |
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$begingroup$
In your particular case, you just need the fact that $vecatimesvecb$ is orthogonal to $vecb$
$endgroup$
– Poon Levi
Apr 7 at 13:03
$begingroup$
@PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $vec b(t)$. I have edited my question to make this clear.
$endgroup$
– Niket Parikh
Apr 7 at 13:19