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Kernel of matrix polynomials
The Next CEO of Stack OverflowLinear Algebra: Minimum Polynomial QuestionShowing diagonalisability using primary decompositionshow T is one one iff $ker(T)=0$Kernel of GCD between minimal polynomial and any polynomialSymmetric positive definite matrix, why $text{Im}(S^{-1}A) = (text{Ker}(A))^{bot_S}$?The kernel of nilpotent MatrixIf $P_1$ and $P_2$ are coprime, how do I show that $ker ( (P_1 times P_2 )(f) )=ker(P_1 (f)) oplus ker(P_2(f))$?Understanding sets of polynomials as subspacesChange of basis of the kernel of a rectangular matrixKernel of polynomial of matrix
$begingroup$
I am told that $f(A)=g(A)h(A), f(A)=0$ is a polynomial of an $n times n$ matrix $A$.
I am further told that $g(x)$ and $h(x)$ are coprime. I need to show that $mathbb{R}^n$ is the direct sum of $text{ker }g(A)$ and $text{ker }h(A)$.
My attempt at a solution:
Since $g(x)$ and $h(x)$ are coprime, this means that they have no roots or factors in common, so I can deduce that $text{ker }g(A) cap text{ker }h(A) ={0}$.
I now want to show that if a vector $v in mathbb{R}^n$ is not in $text{ker }g(A)$, then it is in $text{ker }h(A)$ and vice versa.
Since $f(A)=0$, we have $f(A)v=0 Rightarrow g(A)h(A)v=0$
Because of commutativity of polynomial multiplication, this also means that $h(A)g(A)v=0$.
If $v notin text{ker }g(A)$ then $g(A)vne 0$ so $h(A)v=0$. Is this true? Am I on the right track?
Any help would be greatly appreciated!
linear-algebra matrices polynomials
asked Mar 17 at 7:33
AndrewAndrew
357213
$endgroup$
add a comment |
$begingroup$
I am told that $f(A)=g(A)h(A), f(A)=0$ is a polynomial of an $n times n$ matrix $A$.
I am further told that $g(x)$ and $h(x)$ are coprime. I need to show that $mathbb{R}^n$ is the direct sum of $text{ker }g(A)$ and $text{ker }h(A)$.
My attempt at a solution:
Since $g(x)$ and $h(x)$ are coprime, this means that they have no roots or factors in common, so I can deduce that $text{ker }g(A) cap text{ker }h(A) ={0}$.
I now want to show that if a vector $v in mathbb{R}^n$ is not in $text{ker }g(A)$, then it is in $text{ker }h(A)$ and vice versa.
Since $f(A)=0$, we have $f(A)v=0 Rightarrow g(A)h(A)v=0$
Because of commutativity of polynomial multiplication, this also means that $h(A)g(A)v=0$.
If $v notin text{ker }g(A)$ then $g(A)vne 0$ so $h(A)v=0$. Is this true? Am I on the right track?
Any help would be greatly appreciated!
linear-algebra matrices polynomials
asked Mar 17 at 7:33
AndrewAndrew
357213
$endgroup$
1
$begingroup$
You target is wrong : the direct sum is not the disjoint union. You need to show that any $v$ can be written as $v_1 + v_2$ where $v_1 in ker g(A)$ and $v_2 in ker h(A)$. For this purpose you must use relatively primality.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 17 at 7:46
$begingroup$
Thanks. Do you have any hints on how to do this? All I can think of is showing that since $f(A)v=0 forall v in mathbb{R}^n$ then $f(A)(v_1+v_2)=0$ so $v_1+v_2$ must be in one of the kernels - but can't be in both.
$endgroup$
– Andrew
Mar 17 at 7:50
$begingroup$
@Andrew Hint: Bezout's identity.
$endgroup$
– M. Vinay
Mar 17 at 7:52
add a comment |
$begingroup$
I am told that $f(A)=g(A)h(A), f(A)=0$ is a polynomial of an $n times n$ matrix $A$.
I am further told that $g(x)$ and $h(x)$ are coprime. I need to show that $mathbb{R}^n$ is the direct sum of $text{ker }g(A)$ and $text{ker }h(A)$.
My attempt at a solution:
Since $g(x)$ and $h(x)$ are coprime, this means that they have no roots or factors in common, so I can deduce that $text{ker }g(A) cap text{ker }h(A) ={0}$.
I now want to show that if a vector $v in mathbb{R}^n$ is not in $text{ker }g(A)$, then it is in $text{ker }h(A)$ and vice versa.
Since $f(A)=0$, we have $f(A)v=0 Rightarrow g(A)h(A)v=0$
Because of commutativity of polynomial multiplication, this also means that $h(A)g(A)v=0$.
If $v notin text{ker }g(A)$ then $g(A)vne 0$ so $h(A)v=0$. Is this true? Am I on the right track?
Any help would be greatly appreciated!
linear-algebra matrices polynomials
asked Mar 17 at 7:33
AndrewAndrew
357213
$endgroup$
I am told that $f(A)=g(A)h(A), f(A)=0$ is a polynomial of an $n times n$ matrix $A$.
I am further told that $g(x)$ and $h(x)$ are coprime. I need to show that $mathbb{R}^n$ is the direct sum of $text{ker }g(A)$ and $text{ker }h(A)$.
My attempt at a solution:
Since $g(x)$ and $h(x)$ are coprime, this means that they have no roots or factors in common, so I can deduce that $text{ker }g(A) cap text{ker }h(A) ={0}$.
I now want to show that if a vector $v in mathbb{R}^n$ is not in $text{ker }g(A)$, then it is in $text{ker }h(A)$ and vice versa.
Since $f(A)=0$, we have $f(A)v=0 Rightarrow g(A)h(A)v=0$
Because of commutativity of polynomial multiplication, this also means that $h(A)g(A)v=0$.
If $v notin text{ker }g(A)$ then $g(A)vne 0$ so $h(A)v=0$. Is this true? Am I on the right track?
Any help would be greatly appreciated!
linear-algebra matrices polynomials
linear-algebra matrices polynomials
asked Mar 17 at 7:33
AndrewAndrew
357213
asked Mar 17 at 7:33
AndrewAndrew
357213
asked Mar 17 at 7:33
AndrewAndrew
357213
asked Mar 17 at 7:33
AndrewAndrew
357213
asked Mar 17 at 7:33
AndrewAndrew
357213
357213
1
$begingroup$
You target is wrong : the direct sum is not the disjoint union. You need to show that any $v$ can be written as $v_1 + v_2$ where $v_1 in ker g(A)$ and $v_2 in ker h(A)$. For this purpose you must use relatively primality.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 17 at 7:46
$begingroup$
Thanks. Do you have any hints on how to do this? All I can think of is showing that since $f(A)v=0 forall v in mathbb{R}^n$ then $f(A)(v_1+v_2)=0$ so $v_1+v_2$ must be in one of the kernels - but can't be in both.
$endgroup$
– Andrew
Mar 17 at 7:50
$begingroup$
@Andrew Hint: Bezout's identity.
$endgroup$
– M. Vinay
Mar 17 at 7:52
add a comment |
1
$begingroup$
You target is wrong : the direct sum is not the disjoint union. You need to show that any $v$ can be written as $v_1 + v_2$ where $v_1 in ker g(A)$ and $v_2 in ker h(A)$. For this purpose you must use relatively primality.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 17 at 7:46
$begingroup$
Thanks. Do you have any hints on how to do this? All I can think of is showing that since $f(A)v=0 forall v in mathbb{R}^n$ then $f(A)(v_1+v_2)=0$ so $v_1+v_2$ must be in one of the kernels - but can't be in both.
$endgroup$
– Andrew
Mar 17 at 7:50
$begingroup$
@Andrew Hint: Bezout's identity.
$endgroup$
– M. Vinay
Mar 17 at 7:52
1
1
$begingroup$
You target is wrong : the direct sum is not the disjoint union. You need to show that any $v$ can be written as $v_1 + v_2$ where $v_1 in ker g(A)$ and $v_2 in ker h(A)$. For this purpose you must use relatively primality.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 17 at 7:46
$begingroup$
You target is wrong : the direct sum is not the disjoint union. You need to show that any $v$ can be written as $v_1 + v_2$ where $v_1 in ker g(A)$ and $v_2 in ker h(A)$. For this purpose you must use relatively primality.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 17 at 7:46
$begingroup$
Thanks. Do you have any hints on how to do this? All I can think of is showing that since $f(A)v=0 forall v in mathbb{R}^n$ then $f(A)(v_1+v_2)=0$ so $v_1+v_2$ must be in one of the kernels - but can't be in both.
$endgroup$
– Andrew
Mar 17 at 7:50
$begingroup$
Thanks. Do you have any hints on how to do this? All I can think of is showing that since $f(A)v=0 forall v in mathbb{R}^n$ then $f(A)(v_1+v_2)=0$ so $v_1+v_2$ must be in one of the kernels - but can't be in both.
$endgroup$
– Andrew
Mar 17 at 7:50
$begingroup$
@Andrew Hint: Bezout's identity.
$endgroup$
– M. Vinay
Mar 17 at 7:52
$begingroup$
@Andrew Hint: Bezout's identity.
$endgroup$
– M. Vinay
Mar 17 at 7:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use the fact that if $g,h$ are relatively prime polynomials then there exist polynomials $p(x),q(x)$ such that $p(x)g(x) + q(x)h(x) = 1$ for all $x$,(Bezout identity), so in particular, $p(A)g(A) + q(A)h(A) = I$.
Now, with this in mind, let $v in Bbb R^n$. Then, we may write $v = p(A)g(A)v + q(A)h(A)v$.
Now, it is clear(by commutativity of the polynomial operators in $A$) that :
$$
h(A)[p(A)g(A)v] = p(A)[g(A)h(A)v] = 0 \
g(A)[q(A)h(A)v] = q(A)[g(A)h(A)v] = 0
$$
Therefore $v$ is a sum of two elements lying in the kernels of $h(A)$ and $g(A)$ respectively. Adding the disjointness of these subspaces gives the required conclusion.
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
add a comment |
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$begingroup$
Use the fact that if $g,h$ are relatively prime polynomials then there exist polynomials $p(x),q(x)$ such that $p(x)g(x) + q(x)h(x) = 1$ for all $x$,(Bezout identity), so in particular, $p(A)g(A) + q(A)h(A) = I$.
Now, with this in mind, let $v in Bbb R^n$. Then, we may write $v = p(A)g(A)v + q(A)h(A)v$.
Now, it is clear(by commutativity of the polynomial operators in $A$) that :
$$
h(A)[p(A)g(A)v] = p(A)[g(A)h(A)v] = 0 \
g(A)[q(A)h(A)v] = q(A)[g(A)h(A)v] = 0
$$
Therefore $v$ is a sum of two elements lying in the kernels of $h(A)$ and $g(A)$ respectively. Adding the disjointness of these subspaces gives the required conclusion.
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
add a comment |
$begingroup$
Use the fact that if $g,h$ are relatively prime polynomials then there exist polynomials $p(x),q(x)$ such that $p(x)g(x) + q(x)h(x) = 1$ for all $x$,(Bezout identity), so in particular, $p(A)g(A) + q(A)h(A) = I$.
Now, with this in mind, let $v in Bbb R^n$. Then, we may write $v = p(A)g(A)v + q(A)h(A)v$.
Now, it is clear(by commutativity of the polynomial operators in $A$) that :
$$
h(A)[p(A)g(A)v] = p(A)[g(A)h(A)v] = 0 \
g(A)[q(A)h(A)v] = q(A)[g(A)h(A)v] = 0
$$
Therefore $v$ is a sum of two elements lying in the kernels of $h(A)$ and $g(A)$ respectively. Adding the disjointness of these subspaces gives the required conclusion.
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
add a comment |
$begingroup$
Use the fact that if $g,h$ are relatively prime polynomials then there exist polynomials $p(x),q(x)$ such that $p(x)g(x) + q(x)h(x) = 1$ for all $x$,(Bezout identity), so in particular, $p(A)g(A) + q(A)h(A) = I$.
Now, with this in mind, let $v in Bbb R^n$. Then, we may write $v = p(A)g(A)v + q(A)h(A)v$.
Now, it is clear(by commutativity of the polynomial operators in $A$) that :
$$
h(A)[p(A)g(A)v] = p(A)[g(A)h(A)v] = 0 \
g(A)[q(A)h(A)v] = q(A)[g(A)h(A)v] = 0
$$
Therefore $v$ is a sum of two elements lying in the kernels of $h(A)$ and $g(A)$ respectively. Adding the disjointness of these subspaces gives the required conclusion.
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
Use the fact that if $g,h$ are relatively prime polynomials then there exist polynomials $p(x),q(x)$ such that $p(x)g(x) + q(x)h(x) = 1$ for all $x$,(Bezout identity), so in particular, $p(A)g(A) + q(A)h(A) = I$.
Now, with this in mind, let $v in Bbb R^n$. Then, we may write $v = p(A)g(A)v + q(A)h(A)v$.
Now, it is clear(by commutativity of the polynomial operators in $A$) that :
$$
h(A)[p(A)g(A)v] = p(A)[g(A)h(A)v] = 0 \
g(A)[q(A)h(A)v] = q(A)[g(A)h(A)v] = 0
$$
Therefore $v$ is a sum of two elements lying in the kernels of $h(A)$ and $g(A)$ respectively. Adding the disjointness of these subspaces gives the required conclusion.
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
answered Mar 17 at 8:12
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
40.1k33577
add a comment |
add a comment |
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1
$begingroup$
You target is wrong : the direct sum is not the disjoint union. You need to show that any $v$ can be written as $v_1 + v_2$ where $v_1 in ker g(A)$ and $v_2 in ker h(A)$. For this purpose you must use relatively primality.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 17 at 7:46
$begingroup$
Thanks. Do you have any hints on how to do this? All I can think of is showing that since $f(A)v=0 forall v in mathbb{R}^n$ then $f(A)(v_1+v_2)=0$ so $v_1+v_2$ must be in one of the kernels - but can't be in both.
$endgroup$
– Andrew
Mar 17 at 7:50
$begingroup$
@Andrew Hint: Bezout's identity.
$endgroup$
– M. Vinay
Mar 17 at 7:52