Matrices with all non-zero entries.












5












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    What does "for any non-scalar matrices" mean?
    $endgroup$
    – Theo Bendit
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago


















5












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    What does "for any non-scalar matrices" mean?
    $endgroup$
    – Theo Bendit
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago
















5












5








5


3



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









RickRick

656




656












  • $begingroup$
    What does "for any non-scalar matrices" mean?
    $endgroup$
    – Theo Bendit
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago




















  • $begingroup$
    What does "for any non-scalar matrices" mean?
    $endgroup$
    – Theo Bendit
    2 hours ago






  • 2




    $begingroup$
    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
    2 hours ago


















$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
2 hours ago




$begingroup$
What does "for any non-scalar matrices" mean?
$endgroup$
– Theo Bendit
2 hours ago




2




2




$begingroup$
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
2 hours ago






$begingroup$
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
2 hours ago












1 Answer
1






active

oldest

votes


















6












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the answer.
    $endgroup$
    – Rick
    32 mins ago











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the answer.
    $endgroup$
    – Rick
    32 mins ago
















6












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the answer.
    $endgroup$
    – Rick
    32 mins ago














6












6








6





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 2 hours ago









Eric WofseyEric Wofsey

181k12208336




181k12208336












  • $begingroup$
    thanks for the answer.
    $endgroup$
    – Rick
    32 mins ago


















  • $begingroup$
    thanks for the answer.
    $endgroup$
    – Rick
    32 mins ago
















$begingroup$
thanks for the answer.
$endgroup$
– Rick
32 mins ago




$begingroup$
thanks for the answer.
$endgroup$
– Rick
32 mins ago


















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