Matrices with all non-zero entries.
$begingroup$
I am reading a paper and it uses one of these facts, I would like to know if it has a simple proof:
Let $F$ be an infinite field and $nge2$ and integer. Then for any non-scalar matrices $A_1,A_2,...,A_k$ in $M_{n}(F)$, there exist some invertible matrix $Q in M_{n}(F)$ such that each matrix $QA_1Q^{-1}, QA_2Q^{-1},..., QA_{k}Q^{-1}$ have all non-zero entries.
I just don't know where to start at the first place, could have used diagonalizability but not all non-scalar matrices are diagonalizable. Maybe its too simple, please help.
Thanks in advance.
linear-algebra matrices ring-theory
asked Jan 15 at 6:35
RickRick
868
$endgroup$
add a comment |
$begingroup$
I am reading a paper and it uses one of these facts, I would like to know if it has a simple proof:
Let $F$ be an infinite field and $nge2$ and integer. Then for any non-scalar matrices $A_1,A_2,...,A_k$ in $M_{n}(F)$, there exist some invertible matrix $Q in M_{n}(F)$ such that each matrix $QA_1Q^{-1}, QA_2Q^{-1},..., QA_{k}Q^{-1}$ have all non-zero entries.
I just don't know where to start at the first place, could have used diagonalizability but not all non-scalar matrices are diagonalizable. Maybe its too simple, please help.
Thanks in advance.
linear-algebra matrices ring-theory
asked Jan 15 at 6:35
RickRick
868
$endgroup$
1
$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
Jan 15 at 6:50
3
$begingroup$
Non-scalar matrices mean that the matrix is not of the form $lambda I$ where $lambda in F$. $I$ is the identity matrix.
$endgroup$
– Rick
Jan 15 at 6:52
add a comment |
$begingroup$
I am reading a paper and it uses one of these facts, I would like to know if it has a simple proof:
Let $F$ be an infinite field and $nge2$ and integer. Then for any non-scalar matrices $A_1,A_2,...,A_k$ in $M_{n}(F)$, there exist some invertible matrix $Q in M_{n}(F)$ such that each matrix $QA_1Q^{-1}, QA_2Q^{-1},..., QA_{k}Q^{-1}$ have all non-zero entries.
I just don't know where to start at the first place, could have used diagonalizability but not all non-scalar matrices are diagonalizable. Maybe its too simple, please help.
Thanks in advance.
linear-algebra matrices ring-theory
asked Jan 15 at 6:35
RickRick
868
$endgroup$
I am reading a paper and it uses one of these facts, I would like to know if it has a simple proof:
Let $F$ be an infinite field and $nge2$ and integer. Then for any non-scalar matrices $A_1,A_2,...,A_k$ in $M_{n}(F)$, there exist some invertible matrix $Q in M_{n}(F)$ such that each matrix $QA_1Q^{-1}, QA_2Q^{-1},..., QA_{k}Q^{-1}$ have all non-zero entries.
I just don't know where to start at the first place, could have used diagonalizability but not all non-scalar matrices are diagonalizable. Maybe its too simple, please help.
Thanks in advance.
linear-algebra matrices ring-theory
linear-algebra matrices ring-theory
asked Jan 15 at 6:35
RickRick
868
asked Jan 15 at 6:35
RickRick
868
asked Jan 15 at 6:35
RickRick
868
asked Jan 15 at 6:35
RickRick
868
asked Jan 15 at 6:35
RickRick
868
868
1
$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
Jan 15 at 6:50
3
$begingroup$
Non-scalar matrices mean that the matrix is not of the form $lambda I$ where $lambda in F$. $I$ is the identity matrix.
$endgroup$
– Rick
Jan 15 at 6:52
add a comment |
1
$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
Jan 15 at 6:50
3
$begingroup$
Non-scalar matrices mean that the matrix is not of the form $lambda I$ where $lambda in F$. $I$ is the identity matrix.
$endgroup$
– Rick
Jan 15 at 6:52
1
1
$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
Jan 15 at 6:50
$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
Jan 15 at 6:50
3
3
$begingroup$
Non-scalar matrices mean that the matrix is not of the form $lambda I$ where $lambda in F$. $I$ is the identity matrix.
$endgroup$
– Rick
Jan 15 at 6:52
$begingroup$
Non-scalar matrices mean that the matrix is not of the form $lambda I$ where $lambda in F$. $I$ is the identity matrix.
$endgroup$
– Rick
Jan 15 at 6:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each $(i,j,ell)$, note that the $(i,j)$ entry of $QA_ell Q^{-1}cdot det(Q)$ is some polynomial $p_{ijell}$ in the entries in $Q$ (we multiply by $det(Q)$ since the entries of $Q^{-1}$ are rational functions in the entries of $Q$ with denominator $det(Q)$). None of these polynomials are identically zero, since each $A_{ell}$ is not scalar and so $Q$ can be chosen to make any individual one of its entries nonzero. In more detail, the fact that $A_{ell}$ is not scalar implies there is a vector $vin F^n$ such that $A_{ell}v$ is linearly independent from $v$. We can thus choose a basis with $v$ and $A_{ell}v$ as the $i$th and $j$th basis vectors, and in this basis the $(i,j)$ entry of $A_{ell}$ will be nonzero.
The result then follows from the following well-known theorem, applied to the polynomial $p$ in the entries of $Q$ which is the product of all the $p_{ijell}$ and $det(Q)$.
Theorem: Let $F$ be an infinite field and let $pin F[x_1,dots,x_m]$ be nonzero. Then there exist $a_1,dots,a_min F$ such that $p(a_1,dots,a_m)neq 0$.
Proof: We use induction on $m$; the base case $m=0$ is trivial. Now suppose $m>0$ and the result is known for $m-1$. Think of $p$ as a polynomial in the variables $x_1,dots,x_{m-1}$ with coefficients in $F[x_m]$. At least one of its coefficients is nonzero, and that coefficient has only finitely many roots. Since $F$ is infinite, there is some $a_min F$ which is not a root of that coefficient, and so the polynomial $p(x_1,dots,x_{m-1},a_m)in F[x_1,dots,x_{m-1}]$ is nonzero. We can then find the required $a_1,dots,a_{m-1}$ by the induction hypothesis.
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
$endgroup$
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
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%2f3074138%2fmatrices-with-all-non-zero-entries%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 each $(i,j,ell)$, note that the $(i,j)$ entry of $QA_ell Q^{-1}cdot det(Q)$ is some polynomial $p_{ijell}$ in the entries in $Q$ (we multiply by $det(Q)$ since the entries of $Q^{-1}$ are rational functions in the entries of $Q$ with denominator $det(Q)$). None of these polynomials are identically zero, since each $A_{ell}$ is not scalar and so $Q$ can be chosen to make any individual one of its entries nonzero. In more detail, the fact that $A_{ell}$ is not scalar implies there is a vector $vin F^n$ such that $A_{ell}v$ is linearly independent from $v$. We can thus choose a basis with $v$ and $A_{ell}v$ as the $i$th and $j$th basis vectors, and in this basis the $(i,j)$ entry of $A_{ell}$ will be nonzero.
The result then follows from the following well-known theorem, applied to the polynomial $p$ in the entries of $Q$ which is the product of all the $p_{ijell}$ and $det(Q)$.
Theorem: Let $F$ be an infinite field and let $pin F[x_1,dots,x_m]$ be nonzero. Then there exist $a_1,dots,a_min F$ such that $p(a_1,dots,a_m)neq 0$.
Proof: We use induction on $m$; the base case $m=0$ is trivial. Now suppose $m>0$ and the result is known for $m-1$. Think of $p$ as a polynomial in the variables $x_1,dots,x_{m-1}$ with coefficients in $F[x_m]$. At least one of its coefficients is nonzero, and that coefficient has only finitely many roots. Since $F$ is infinite, there is some $a_min F$ which is not a root of that coefficient, and so the polynomial $p(x_1,dots,x_{m-1},a_m)in F[x_1,dots,x_{m-1}]$ is nonzero. We can then find the required $a_1,dots,a_{m-1}$ by the induction hypothesis.
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
$endgroup$
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
add a comment |
$begingroup$
For each $(i,j,ell)$, note that the $(i,j)$ entry of $QA_ell Q^{-1}cdot det(Q)$ is some polynomial $p_{ijell}$ in the entries in $Q$ (we multiply by $det(Q)$ since the entries of $Q^{-1}$ are rational functions in the entries of $Q$ with denominator $det(Q)$). None of these polynomials are identically zero, since each $A_{ell}$ is not scalar and so $Q$ can be chosen to make any individual one of its entries nonzero. In more detail, the fact that $A_{ell}$ is not scalar implies there is a vector $vin F^n$ such that $A_{ell}v$ is linearly independent from $v$. We can thus choose a basis with $v$ and $A_{ell}v$ as the $i$th and $j$th basis vectors, and in this basis the $(i,j)$ entry of $A_{ell}$ will be nonzero.
The result then follows from the following well-known theorem, applied to the polynomial $p$ in the entries of $Q$ which is the product of all the $p_{ijell}$ and $det(Q)$.
Theorem: Let $F$ be an infinite field and let $pin F[x_1,dots,x_m]$ be nonzero. Then there exist $a_1,dots,a_min F$ such that $p(a_1,dots,a_m)neq 0$.
Proof: We use induction on $m$; the base case $m=0$ is trivial. Now suppose $m>0$ and the result is known for $m-1$. Think of $p$ as a polynomial in the variables $x_1,dots,x_{m-1}$ with coefficients in $F[x_m]$. At least one of its coefficients is nonzero, and that coefficient has only finitely many roots. Since $F$ is infinite, there is some $a_min F$ which is not a root of that coefficient, and so the polynomial $p(x_1,dots,x_{m-1},a_m)in F[x_1,dots,x_{m-1}]$ is nonzero. We can then find the required $a_1,dots,a_{m-1}$ by the induction hypothesis.
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
$endgroup$
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
add a comment |
$begingroup$
For each $(i,j,ell)$, note that the $(i,j)$ entry of $QA_ell Q^{-1}cdot det(Q)$ is some polynomial $p_{ijell}$ in the entries in $Q$ (we multiply by $det(Q)$ since the entries of $Q^{-1}$ are rational functions in the entries of $Q$ with denominator $det(Q)$). None of these polynomials are identically zero, since each $A_{ell}$ is not scalar and so $Q$ can be chosen to make any individual one of its entries nonzero. In more detail, the fact that $A_{ell}$ is not scalar implies there is a vector $vin F^n$ such that $A_{ell}v$ is linearly independent from $v$. We can thus choose a basis with $v$ and $A_{ell}v$ as the $i$th and $j$th basis vectors, and in this basis the $(i,j)$ entry of $A_{ell}$ will be nonzero.
The result then follows from the following well-known theorem, applied to the polynomial $p$ in the entries of $Q$ which is the product of all the $p_{ijell}$ and $det(Q)$.
Theorem: Let $F$ be an infinite field and let $pin F[x_1,dots,x_m]$ be nonzero. Then there exist $a_1,dots,a_min F$ such that $p(a_1,dots,a_m)neq 0$.
Proof: We use induction on $m$; the base case $m=0$ is trivial. Now suppose $m>0$ and the result is known for $m-1$. Think of $p$ as a polynomial in the variables $x_1,dots,x_{m-1}$ with coefficients in $F[x_m]$. At least one of its coefficients is nonzero, and that coefficient has only finitely many roots. Since $F$ is infinite, there is some $a_min F$ which is not a root of that coefficient, and so the polynomial $p(x_1,dots,x_{m-1},a_m)in F[x_1,dots,x_{m-1}]$ is nonzero. We can then find the required $a_1,dots,a_{m-1}$ by the induction hypothesis.
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
$endgroup$
For each $(i,j,ell)$, note that the $(i,j)$ entry of $QA_ell Q^{-1}cdot det(Q)$ is some polynomial $p_{ijell}$ in the entries in $Q$ (we multiply by $det(Q)$ since the entries of $Q^{-1}$ are rational functions in the entries of $Q$ with denominator $det(Q)$). None of these polynomials are identically zero, since each $A_{ell}$ is not scalar and so $Q$ can be chosen to make any individual one of its entries nonzero. In more detail, the fact that $A_{ell}$ is not scalar implies there is a vector $vin F^n$ such that $A_{ell}v$ is linearly independent from $v$. We can thus choose a basis with $v$ and $A_{ell}v$ as the $i$th and $j$th basis vectors, and in this basis the $(i,j)$ entry of $A_{ell}$ will be nonzero.
The result then follows from the following well-known theorem, applied to the polynomial $p$ in the entries of $Q$ which is the product of all the $p_{ijell}$ and $det(Q)$.
Theorem: Let $F$ be an infinite field and let $pin F[x_1,dots,x_m]$ be nonzero. Then there exist $a_1,dots,a_min F$ such that $p(a_1,dots,a_m)neq 0$.
Proof: We use induction on $m$; the base case $m=0$ is trivial. Now suppose $m>0$ and the result is known for $m-1$. Think of $p$ as a polynomial in the variables $x_1,dots,x_{m-1}$ with coefficients in $F[x_m]$. At least one of its coefficients is nonzero, and that coefficient has only finitely many roots. Since $F$ is infinite, there is some $a_min F$ which is not a root of that coefficient, and so the polynomial $p(x_1,dots,x_{m-1},a_m)in F[x_1,dots,x_{m-1}]$ is nonzero. We can then find the required $a_1,dots,a_{m-1}$ by the induction hypothesis.
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
edited Jan 15 at 7:03
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
answered Jan 15 at 6:51
Eric WofseyEric Wofsey
185k13213339
185k13213339
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
add a comment |
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
$begingroup$
thanks for the answer.
$endgroup$
– Rick
Jan 15 at 9:16
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%2f3074138%2fmatrices-with-all-non-zero-entries%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
1
$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
Jan 15 at 6:50
3
$begingroup$
Non-scalar matrices mean that the matrix is not of the form $lambda I$ where $lambda in F$. $I$ is the identity matrix.
$endgroup$
– Rick
Jan 15 at 6:52