Algorithm to return the characteristic polynomial formula using the characteristic polynomial itself as a...
$begingroup$
We consider an $n×n$ matrix $A$. The characteristic polynomial of $A$, denoted by $pA(t)$, is the polynomial defined by
${p_{A}(t)=det left(tI-Aright)}$
This is the 'algorithm' to have a characteristic polynomial.
Suppose we want to compute the characteristic polynomial of the matrix
${A=begin{pmatrix}2&1\-1&0end{pmatrix}}$
We now compute the determinant of
${ tI-A={begin{pmatrix}t-2&-1\1&t-0end{pmatrix}}}$
which is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$ , the characteristic polynomial of A.
Well, suppose this situation:
I take ${displaystyle t^{2}-2t+1,!}$ and I want to return the algorithm (${p_{A}(t)=det left(tI-Aright)}$ that generated (and that I know before) that characteristic polynomial using the characteristic polynomial itself as a 'vehicle' to do it.
How to do it?
polynomials algorithms characteristic-functions
asked Jan 11 at 14:45
JeffJeff
1
$endgroup$
|
show 8 more comments
$begingroup$
We consider an $n×n$ matrix $A$. The characteristic polynomial of $A$, denoted by $pA(t)$, is the polynomial defined by
${p_{A}(t)=det left(tI-Aright)}$
This is the 'algorithm' to have a characteristic polynomial.
Suppose we want to compute the characteristic polynomial of the matrix
${A=begin{pmatrix}2&1\-1&0end{pmatrix}}$
We now compute the determinant of
${ tI-A={begin{pmatrix}t-2&-1\1&t-0end{pmatrix}}}$
which is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$ , the characteristic polynomial of A.
Well, suppose this situation:
I take ${displaystyle t^{2}-2t+1,!}$ and I want to return the algorithm (${p_{A}(t)=det left(tI-Aright)}$ that generated (and that I know before) that characteristic polynomial using the characteristic polynomial itself as a 'vehicle' to do it.
How to do it?
polynomials algorithms characteristic-functions
asked Jan 11 at 14:45
JeffJeff
1
$endgroup$
1
$begingroup$
It's a little unclear what you're asking. Are you asking what matrix, given a characteristic polynomial, has eigenvalues equal to the roots of the polynomial?
$endgroup$
– Adrian Keister
Jan 11 at 14:52
$begingroup$
No I ask what matrix or what I need to use to represent or what algorithm to generate(not an algorithm to use to solve) I need if we know the characteristic polynomial (we know it, in this case is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$, if we know also that polynomial is defined by ${p_{A}(t)=det left(tI-Aright)}$. I want to make a link btween that characteristic polynomial ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$ generated by ${p_{A}(t)=det left(tI-Aright)}$ and the ${p_{A}(t)=det left(tI-Aright)}$ formula itself that we used to generate my characteristic polynomial
$endgroup$
– Jeff
Jan 11 at 14:59
1
$begingroup$
Sorry, I can't follow that comment at all. Any polynomial can be a characteristic polynomial of some matrix.
$endgroup$
– Adrian Keister
Jan 11 at 15:04
$begingroup$
In my situation ${p_{A}(t)=det left(tI-Aright)}$ generates the characteristic polynomial of A, denoted by pA(t). In my situation the characteristic polynomial is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$. Is the ${p_{A}(t)=det left(tI-Aright)}$ a polynomial ?
$endgroup$
– Jeff
Jan 11 at 15:08
$begingroup$
If $A$ is a finite-dimensional matrix, then $det(tI-A)$ is always a polynomial, yes.
$endgroup$
– Adrian Keister
Jan 11 at 15:09
|
show 8 more comments
$begingroup$
We consider an $n×n$ matrix $A$. The characteristic polynomial of $A$, denoted by $pA(t)$, is the polynomial defined by
${p_{A}(t)=det left(tI-Aright)}$
This is the 'algorithm' to have a characteristic polynomial.
Suppose we want to compute the characteristic polynomial of the matrix
${A=begin{pmatrix}2&1\-1&0end{pmatrix}}$
We now compute the determinant of
${ tI-A={begin{pmatrix}t-2&-1\1&t-0end{pmatrix}}}$
which is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$ , the characteristic polynomial of A.
Well, suppose this situation:
I take ${displaystyle t^{2}-2t+1,!}$ and I want to return the algorithm (${p_{A}(t)=det left(tI-Aright)}$ that generated (and that I know before) that characteristic polynomial using the characteristic polynomial itself as a 'vehicle' to do it.
How to do it?
polynomials algorithms characteristic-functions
asked Jan 11 at 14:45
JeffJeff
1
$endgroup$
We consider an $n×n$ matrix $A$. The characteristic polynomial of $A$, denoted by $pA(t)$, is the polynomial defined by
${p_{A}(t)=det left(tI-Aright)}$
This is the 'algorithm' to have a characteristic polynomial.
Suppose we want to compute the characteristic polynomial of the matrix
${A=begin{pmatrix}2&1\-1&0end{pmatrix}}$
We now compute the determinant of
${ tI-A={begin{pmatrix}t-2&-1\1&t-0end{pmatrix}}}$
which is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$ , the characteristic polynomial of A.
Well, suppose this situation:
I take ${displaystyle t^{2}-2t+1,!}$ and I want to return the algorithm (${p_{A}(t)=det left(tI-Aright)}$ that generated (and that I know before) that characteristic polynomial using the characteristic polynomial itself as a 'vehicle' to do it.
How to do it?
polynomials algorithms characteristic-functions
polynomials algorithms characteristic-functions
asked Jan 11 at 14:45
JeffJeff
1
asked Jan 11 at 14:45
JeffJeff
1
asked Jan 11 at 14:45
JeffJeff
1
asked Jan 11 at 14:45
JeffJeff
1
asked Jan 11 at 14:45
JeffJeff
1
1
1
$begingroup$
It's a little unclear what you're asking. Are you asking what matrix, given a characteristic polynomial, has eigenvalues equal to the roots of the polynomial?
$endgroup$
– Adrian Keister
Jan 11 at 14:52
$begingroup$
No I ask what matrix or what I need to use to represent or what algorithm to generate(not an algorithm to use to solve) I need if we know the characteristic polynomial (we know it, in this case is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$, if we know also that polynomial is defined by ${p_{A}(t)=det left(tI-Aright)}$. I want to make a link btween that characteristic polynomial ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$ generated by ${p_{A}(t)=det left(tI-Aright)}$ and the ${p_{A}(t)=det left(tI-Aright)}$ formula itself that we used to generate my characteristic polynomial
$endgroup$
– Jeff
Jan 11 at 14:59
1
$begingroup$
Sorry, I can't follow that comment at all. Any polynomial can be a characteristic polynomial of some matrix.
$endgroup$
– Adrian Keister
Jan 11 at 15:04
$begingroup$
In my situation ${p_{A}(t)=det left(tI-Aright)}$ generates the characteristic polynomial of A, denoted by pA(t). In my situation the characteristic polynomial is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$. Is the ${p_{A}(t)=det left(tI-Aright)}$ a polynomial ?
$endgroup$
– Jeff
Jan 11 at 15:08
$begingroup$
If $A$ is a finite-dimensional matrix, then $det(tI-A)$ is always a polynomial, yes.
$endgroup$
– Adrian Keister
Jan 11 at 15:09
|
show 8 more comments
1
$begingroup$
It's a little unclear what you're asking. Are you asking what matrix, given a characteristic polynomial, has eigenvalues equal to the roots of the polynomial?
$endgroup$
– Adrian Keister
Jan 11 at 14:52
$begingroup$
No I ask what matrix or what I need to use to represent or what algorithm to generate(not an algorithm to use to solve) I need if we know the characteristic polynomial (we know it, in this case is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$, if we know also that polynomial is defined by ${p_{A}(t)=det left(tI-Aright)}$. I want to make a link btween that characteristic polynomial ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$ generated by ${p_{A}(t)=det left(tI-Aright)}$ and the ${p_{A}(t)=det left(tI-Aright)}$ formula itself that we used to generate my characteristic polynomial
$endgroup$
– Jeff
Jan 11 at 14:59
1
$begingroup$
Sorry, I can't follow that comment at all. Any polynomial can be a characteristic polynomial of some matrix.
$endgroup$
– Adrian Keister
Jan 11 at 15:04
$begingroup$
In my situation ${p_{A}(t)=det left(tI-Aright)}$ generates the characteristic polynomial of A, denoted by pA(t). In my situation the characteristic polynomial is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$. Is the ${p_{A}(t)=det left(tI-Aright)}$ a polynomial ?
$endgroup$
– Jeff
Jan 11 at 15:08
$begingroup$
If $A$ is a finite-dimensional matrix, then $det(tI-A)$ is always a polynomial, yes.
$endgroup$
– Adrian Keister
Jan 11 at 15:09
1
1
$begingroup$
It's a little unclear what you're asking. Are you asking what matrix, given a characteristic polynomial, has eigenvalues equal to the roots of the polynomial?
$endgroup$
– Adrian Keister
Jan 11 at 14:52
$begingroup$
It's a little unclear what you're asking. Are you asking what matrix, given a characteristic polynomial, has eigenvalues equal to the roots of the polynomial?
$endgroup$
– Adrian Keister
Jan 11 at 14:52
$begingroup$
No I ask what matrix or what I need to use to represent or what algorithm to generate(not an algorithm to use to solve) I need if we know the characteristic polynomial (we know it, in this case is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$, if we know also that polynomial is defined by ${p_{A}(t)=det left(tI-Aright)}$. I want to make a link btween that characteristic polynomial ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$ generated by ${p_{A}(t)=det left(tI-Aright)}$ and the ${p_{A}(t)=det left(tI-Aright)}$ formula itself that we used to generate my characteristic polynomial
$endgroup$
– Jeff
Jan 11 at 14:59
$begingroup$
No I ask what matrix or what I need to use to represent or what algorithm to generate(not an algorithm to use to solve) I need if we know the characteristic polynomial (we know it, in this case is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$, if we know also that polynomial is defined by ${p_{A}(t)=det left(tI-Aright)}$. I want to make a link btween that characteristic polynomial ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$ generated by ${p_{A}(t)=det left(tI-Aright)}$ and the ${p_{A}(t)=det left(tI-Aright)}$ formula itself that we used to generate my characteristic polynomial
$endgroup$
– Jeff
Jan 11 at 14:59
1
1
$begingroup$
Sorry, I can't follow that comment at all. Any polynomial can be a characteristic polynomial of some matrix.
$endgroup$
– Adrian Keister
Jan 11 at 15:04
$begingroup$
Sorry, I can't follow that comment at all. Any polynomial can be a characteristic polynomial of some matrix.
$endgroup$
– Adrian Keister
Jan 11 at 15:04
$begingroup$
In my situation ${p_{A}(t)=det left(tI-Aright)}$ generates the characteristic polynomial of A, denoted by pA(t). In my situation the characteristic polynomial is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$. Is the ${p_{A}(t)=det left(tI-Aright)}$ a polynomial ?
$endgroup$
– Jeff
Jan 11 at 15:08
$begingroup$
In my situation ${p_{A}(t)=det left(tI-Aright)}$ generates the characteristic polynomial of A, denoted by pA(t). In my situation the characteristic polynomial is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$. Is the ${p_{A}(t)=det left(tI-Aright)}$ a polynomial ?
$endgroup$
– Jeff
Jan 11 at 15:08
$begingroup$
If $A$ is a finite-dimensional matrix, then $det(tI-A)$ is always a polynomial, yes.
$endgroup$
– Adrian Keister
Jan 11 at 15:09
$begingroup$
If $A$ is a finite-dimensional matrix, then $det(tI-A)$ is always a polynomial, yes.
$endgroup$
– Adrian Keister
Jan 11 at 15:09
|
show 8 more comments
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1
$begingroup$
It's a little unclear what you're asking. Are you asking what matrix, given a characteristic polynomial, has eigenvalues equal to the roots of the polynomial?
$endgroup$
– Adrian Keister
Jan 11 at 14:52
$begingroup$
No I ask what matrix or what I need to use to represent or what algorithm to generate(not an algorithm to use to solve) I need if we know the characteristic polynomial (we know it, in this case is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$, if we know also that polynomial is defined by ${p_{A}(t)=det left(tI-Aright)}$. I want to make a link btween that characteristic polynomial ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!}$ generated by ${p_{A}(t)=det left(tI-Aright)}$ and the ${p_{A}(t)=det left(tI-Aright)}$ formula itself that we used to generate my characteristic polynomial
$endgroup$
– Jeff
Jan 11 at 14:59
1
$begingroup$
Sorry, I can't follow that comment at all. Any polynomial can be a characteristic polynomial of some matrix.
$endgroup$
– Adrian Keister
Jan 11 at 15:04
$begingroup$
In my situation ${p_{A}(t)=det left(tI-Aright)}$ generates the characteristic polynomial of A, denoted by pA(t). In my situation the characteristic polynomial is ${displaystyle (t-2)t-1(-1)=t^{2}-2t+1,!,}$. Is the ${p_{A}(t)=det left(tI-Aright)}$ a polynomial ?
$endgroup$
– Jeff
Jan 11 at 15:08
$begingroup$
If $A$ is a finite-dimensional matrix, then $det(tI-A)$ is always a polynomial, yes.
$endgroup$
– Adrian Keister
Jan 11 at 15:09