Left inverse of a matrix $3 times 2$ in $mathbb{F}_7[x]$
$begingroup$
Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?
I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$
begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}
matrices polynomials finite-fields inverse vector-fields
edited Dec 4 '18 at 3:43
Batominovski
1
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
$endgroup$
add a comment |
$begingroup$
Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?
I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$
begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}
matrices polynomials finite-fields inverse vector-fields
edited Dec 4 '18 at 3:43
Batominovski
1
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
$endgroup$
$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50
add a comment |
$begingroup$
Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?
I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$
begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}
matrices polynomials finite-fields inverse vector-fields
edited Dec 4 '18 at 3:43
Batominovski
1
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
$endgroup$
Do you know a method to calculate inverse matrix in $mathbb{F}_7[x]$?
I want to calculate left inverse the following matrix of $3 times 2$ in $mathbb{F}_7[x]$
begin{bmatrix}
x^2+1 & x-1 \
3(x-3)(x+3) & x-2 \
5(x-1)(x+1) & 2x-3
end{bmatrix}
matrices polynomials finite-fields inverse vector-fields
matrices polynomials finite-fields inverse vector-fields
edited Dec 4 '18 at 3:43
Batominovski
1
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
edited Dec 4 '18 at 3:43
Batominovski
1
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
edited Dec 4 '18 at 3:43
Batominovski
1
edited Dec 4 '18 at 3:43
Batominovski
1
edited Dec 4 '18 at 3:43
Batominovski
1
1
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
asked Dec 4 '18 at 2:27
Tom RyddleTom Ryddle
828
828
$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50
add a comment |
$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50
$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50
$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For this specific case, I would make an Ansatz that a left inverse of
$$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
is of the form
$$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.
For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
$$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
$$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
making
$$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$
Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
$$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
$$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025045%2fleft-inverse-of-a-matrix-3-times-2-in-mathbbf-7x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For this specific case, I would make an Ansatz that a left inverse of
$$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
is of the form
$$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.
For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
$$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
$$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
making
$$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$
Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
$$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
$$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
For this specific case, I would make an Ansatz that a left inverse of
$$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
is of the form
$$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.
For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
$$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
$$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
making
$$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$
Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
$$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
$$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
For this specific case, I would make an Ansatz that a left inverse of
$$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
is of the form
$$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.
For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
$$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
$$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
making
$$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$
Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
$$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
$$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
$endgroup$
For this specific case, I would make an Ansatz that a left inverse of
$$X:=begin{bmatrix}x^2+1&x-1\3x^2-27&x-2\5x^2-5&2x-3end{bmatrix}=begin{bmatrix}vert&vert\X_1&X_2\vert&vertend{bmatrix}$$
is of the form
$$L=begin{bmatrix}b_{i,j}x+c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}=Bx+C,,$$
where $B:=begin{bmatrix}b_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ and $C:=begin{bmatrix}c_{i,j}end{bmatrix}_{iin{1,2},jin{1,2,3}}$ for some $b_{i,j},c_{i,j}inmathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $mathbb{F}_7[x]$.
For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $begin{bmatrix}1&-2&1end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that
$$C_1=begin{bmatrix}1&-2&1end{bmatrix},.$$
The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $begin{bmatrix}1&3&-2end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain
$$C_2=begin{bmatrix}-1&-3&2end{bmatrix},,$$
making
$$C=begin{bmatrix}1&-2&1\-1&-3&2end{bmatrix},.$$
Note that $$CX=begin{bmatrix}1&x\0&1end{bmatrix},.$$ Therefore, we need that $$BX=begin{bmatrix}0&-1\0&0end{bmatrix},.$$
The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=begin{bmatrix}1&3&-2end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=begin{bmatrix}0&0&0end{bmatrix}$. Consequently,
$$B=begin{bmatrix}1&3&-2\0&0&0end{bmatrix}$$ and so
$$L=Bx+C=begin{bmatrix}x+1&3x-2&-2x+1\-1&-3&2end{bmatrix},.$$
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
edited Dec 4 '18 at 13:18
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
answered Dec 4 '18 at 3:35
BatominovskiBatominovski
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025045%2fleft-inverse-of-a-matrix-3-times-2-in-mathbbf-7x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Have you checked the invariant factors of the matrix? IIRC you need both of them to be equal to one for the one sided inverse to exist. I also suspect that from the Smith normal form you can cook up a recipe for finding one, but I don't hve the time to think it through now.
$endgroup$
– Jyrki Lahtonen
Dec 4 '18 at 4:50