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1

$begingroup$

Let $A=(a_i,j)_ntimes n$ be an invertible matrix with the positive rational entry. Let $p(x)$ be a rational polynomial. Consider the following matrix
beginalign*
B=left(p(a_i,j)right)_ntimes n.
endalign*
Assume that $a_i,j$ is not a zero of $p(x)$ for any $1leq i,jleq n$.
Then I am wondering whether $B$ is invertible?

A more general question can be raised as follows:
Let $f(x)$ be a real function. If each $a_i,j$ is not a zero of $f(x)$, then when is the matrix $D=left(f(a_i,j)right)_ntimes n$ invertible?

Any help will be appreciated!:)

linear-algebra matrices polynomials determinant inverse

share|cite|improve this question

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

$endgroup$

add a comment | 

1

$begingroup$

Let $A=(a_i,j)_ntimes n$ be an invertible matrix with the positive rational entry. Let $p(x)$ be a rational polynomial. Consider the following matrix
beginalign*
B=left(p(a_i,j)right)_ntimes n.
endalign*
Assume that $a_i,j$ is not a zero of $p(x)$ for any $1leq i,jleq n$.
Then I am wondering whether $B$ is invertible?

A more general question can be raised as follows:
Let $f(x)$ be a real function. If each $a_i,j$ is not a zero of $f(x)$, then when is the matrix $D=left(f(a_i,j)right)_ntimes n$ invertible?

Any help will be appreciated!:)

linear-algebra matrices polynomials determinant inverse

share|cite|improve this question

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

$endgroup$

add a comment | 

$begingroup$

Let $A=(a_i,j)_ntimes n$ be an invertible matrix with the positive rational entry. Let $p(x)$ be a rational polynomial. Consider the following matrix
beginalign*
B=left(p(a_i,j)right)_ntimes n.
endalign*
Assume that $a_i,j$ is not a zero of $p(x)$ for any $1leq i,jleq n$.
Then I am wondering whether $B$ is invertible?

A more general question can be raised as follows:
Let $f(x)$ be a real function. If each $a_i,j$ is not a zero of $f(x)$, then when is the matrix $D=left(f(a_i,j)right)_ntimes n$ invertible?

Any help will be appreciated!:)

linear-algebra matrices polynomials determinant inverse

share|cite|improve this question

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

$endgroup$

share|cite|improve this question

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

share|cite|improve this question

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

asked Mar 11 at 10:16

VerMoriartyVerMoriarty

1538

2 Answers
2

active

oldest

votes

3

$begingroup$

No. Take $p(x)=(x-1)^2$ and let $A=left[beginsmallmatrix2&0\0&2endsmallmatrixright]$. Then$$beginvmatrixp(2)&p(0)\p(0)&p(2)endvmatrix=beginvmatrix1&1\1&1endvmatrix=0,$$but $det Aneq0$.

share|cite|improve this answer

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

add a comment | 

1

$begingroup$

I am not giving you a straight answer, but will say something that will help you see what all things can happen.

Actually we can easily see the other direction. All matrices are square and size $ntimes n$, for some fixed $n>1$.

Let $A$ be a fixed matrix with all entries different. For an arbitrary matrix $X$
we can find a polynomial $p(x)$ such that $x_i,j= p(a_i,j)$.
This is Lagrange Interpolation Theorem (Numerical mathematics, or Chinese Remainder Theorem).

So this covers $X$ singular as well as non-singular.

Now you should be able to see.

share|cite|improve this answer

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136

$endgroup$

add a comment | 

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3

$begingroup$

No. Take $p(x)=(x-1)^2$ and let $A=left[beginsmallmatrix2&0\0&2endsmallmatrixright]$. Then$$beginvmatrixp(2)&p(0)\p(0)&p(2)endvmatrix=beginvmatrix1&1\1&1endvmatrix=0,$$but $det Aneq0$.

share|cite|improve this answer

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

add a comment | 

3

$begingroup$

No. Take $p(x)=(x-1)^2$ and let $A=left[beginsmallmatrix2&0\0&2endsmallmatrixright]$. Then$$beginvmatrixp(2)&p(0)\p(0)&p(2)endvmatrix=beginvmatrix1&1\1&1endvmatrix=0,$$but $det Aneq0$.

share|cite|improve this answer

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

add a comment | 

$begingroup$

No. Take $p(x)=(x-1)^2$ and let $A=left[beginsmallmatrix2&0\0&2endsmallmatrixright]$. Then$$beginvmatrixp(2)&p(0)\p(0)&p(2)endvmatrix=beginvmatrix1&1\1&1endvmatrix=0,$$but $det Aneq0$.

share|cite|improve this answer

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

share|cite|improve this answer

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

answered Mar 11 at 10:21

José Carlos SantosJosé Carlos Santos

167k22132235

1

$begingroup$

I am not giving you a straight answer, but will say something that will help you see what all things can happen.

Actually we can easily see the other direction. All matrices are square and size $ntimes n$, for some fixed $n>1$.

Let $A$ be a fixed matrix with all entries different. For an arbitrary matrix $X$
we can find a polynomial $p(x)$ such that $x_i,j= p(a_i,j)$.
This is Lagrange Interpolation Theorem (Numerical mathematics, or Chinese Remainder Theorem).

So this covers $X$ singular as well as non-singular.

Now you should be able to see.

share|cite|improve this answer

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136

$endgroup$

add a comment | 

1

$begingroup$

I am not giving you a straight answer, but will say something that will help you see what all things can happen.

Actually we can easily see the other direction. All matrices are square and size $ntimes n$, for some fixed $n>1$.

Let $A$ be a fixed matrix with all entries different. For an arbitrary matrix $X$
we can find a polynomial $p(x)$ such that $x_i,j= p(a_i,j)$.
This is Lagrange Interpolation Theorem (Numerical mathematics, or Chinese Remainder Theorem).

So this covers $X$ singular as well as non-singular.

Now you should be able to see.

share|cite|improve this answer

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136

$endgroup$

add a comment | 

$begingroup$

I am not giving you a straight answer, but will say something that will help you see what all things can happen.

Actually we can easily see the other direction. All matrices are square and size $ntimes n$, for some fixed $n>1$.

Let $A$ be a fixed matrix with all entries different. For an arbitrary matrix $X$
we can find a polynomial $p(x)$ such that $x_i,j= p(a_i,j)$.
This is Lagrange Interpolation Theorem (Numerical mathematics, or Chinese Remainder Theorem).

So this covers $X$ singular as well as non-singular.

Now you should be able to see.

share|cite|improve this answer

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136

$endgroup$

share|cite|improve this answer

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136

answered Mar 11 at 10:32

P VanchinathanP Vanchinathan

15.5k12136