Representing the inverse of a matrix by a polynomial The Next CEO of Stack OverflowProperties of a matrix with the minimal polynomial $m_A(t) = t^3+2t^2+t+1$?Continuity of the inverse matrix functionWhat is the quickest way to find the characteristic polynomial of this matrix?Show that the minimal polynomial over $mathbbC$ of a real matrix has real coefficientsCharacteristic polynomial of a matrix polynomialDoes the inverse of a polynomial matrix have polynomial growth?Help with understanding the proof for: $AB$ and $BA$ have the same characteristic polynomial (for square complex matrices)Prove that matrix $A$ diagonalizable if $A^2=I$ using characteristic polynomialCounting Characteristic Polynomials?Determinant and trace of a matrix
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Representing the inverse of a matrix by a polynomial
The Next CEO of Stack OverflowProperties of a matrix with the minimal polynomial $m_A(t) = t^3+2t^2+t+1$?Continuity of the inverse matrix functionWhat is the quickest way to find the characteristic polynomial of this matrix?Show that the minimal polynomial over $mathbbC$ of a real matrix has real coefficientsCharacteristic polynomial of a matrix polynomialDoes the inverse of a polynomial matrix have polynomial growth?Help with understanding the proof for: $AB$ and $BA$ have the same characteristic polynomial (for square complex matrices)Prove that matrix $A$ diagonalizable if $A^2=I$ using characteristic polynomialCounting Characteristic Polynomials?Determinant and trace of a matrix
$begingroup$
I saw a question that asks in the first part to prove that if $A, B in M_n(mathbb C)$ such that $AB=BA$ and for any 2 polynomials $f$ and $g$, we have $f(A)g(B)=g(B)f(A)$. Thats not difficult.
In the second part the question wa to prove that $A^−1=p(A)$ for some polynomial with complex coefficients. So how can I use part 1 to prove part 2? Or are they not related to each other ?
I think in general they use the fact that the characteristic polynomial $c_A(t) = det(t I - A)$ of any square matrix $A$ satisfies $c_A(A) = 0$.
But I also didn't know how to prove from here.
matrices polynomials inverse
asked 2 days ago
Fareed AFFareed AF
62212
$endgroup$
|
show 1 more comment
$begingroup$
I saw a question that asks in the first part to prove that if $A, B in M_n(mathbb C)$ such that $AB=BA$ and for any 2 polynomials $f$ and $g$, we have $f(A)g(B)=g(B)f(A)$. Thats not difficult.
In the second part the question wa to prove that $A^−1=p(A)$ for some polynomial with complex coefficients. So how can I use part 1 to prove part 2? Or are they not related to each other ?
I think in general they use the fact that the characteristic polynomial $c_A(t) = det(t I - A)$ of any square matrix $A$ satisfies $c_A(A) = 0$.
But I also didn't know how to prove from here.
matrices polynomials inverse
asked 2 days ago
Fareed AFFareed AF
62212
$endgroup$
1
$begingroup$
The constant in the characteristic polynomial must be non zero if $A$ is invertible, so you can use the characteristic polynomial to find the required formula by just multiplying across by $A^-1$.
$endgroup$
– copper.hat
2 days ago
$begingroup$
I didn't understand what you mean, can you please explain it more?
$endgroup$
– Fareed AF
2 days ago
$begingroup$
What happens if you multiply the characteristic equation (with $A$) across by the inverse if $A$?
$endgroup$
– copper.hat
2 days ago
$begingroup$
The charactaristic equation is a determinant. So if I multiplied inside the determinant I will get $det(tA^-1-I)$. I didn't understand where you want me to multiply by $A^-1$ and do "multiply across by" have the same meaning as "multiply by"? Sorry maybe I am not that good in english...
$endgroup$
– Fareed AF
2 days ago
1
$begingroup$
I added an answer.
$endgroup$
– copper.hat
2 days ago
|
show 1 more comment
$begingroup$
I saw a question that asks in the first part to prove that if $A, B in M_n(mathbb C)$ such that $AB=BA$ and for any 2 polynomials $f$ and $g$, we have $f(A)g(B)=g(B)f(A)$. Thats not difficult.
In the second part the question wa to prove that $A^−1=p(A)$ for some polynomial with complex coefficients. So how can I use part 1 to prove part 2? Or are they not related to each other ?
I think in general they use the fact that the characteristic polynomial $c_A(t) = det(t I - A)$ of any square matrix $A$ satisfies $c_A(A) = 0$.
But I also didn't know how to prove from here.
matrices polynomials inverse
asked 2 days ago
Fareed AFFareed AF
62212
$endgroup$
I saw a question that asks in the first part to prove that if $A, B in M_n(mathbb C)$ such that $AB=BA$ and for any 2 polynomials $f$ and $g$, we have $f(A)g(B)=g(B)f(A)$. Thats not difficult.
In the second part the question wa to prove that $A^−1=p(A)$ for some polynomial with complex coefficients. So how can I use part 1 to prove part 2? Or are they not related to each other ?
I think in general they use the fact that the characteristic polynomial $c_A(t) = det(t I - A)$ of any square matrix $A$ satisfies $c_A(A) = 0$.
But I also didn't know how to prove from here.
matrices polynomials inverse
matrices polynomials inverse
asked 2 days ago
Fareed AFFareed AF
62212
asked 2 days ago
Fareed AFFareed AF
62212
asked 2 days ago
Fareed AFFareed AF
62212
asked 2 days ago
Fareed AFFareed AF
62212
asked 2 days ago
Fareed AFFareed AF
62212
62212
1
$begingroup$
The constant in the characteristic polynomial must be non zero if $A$ is invertible, so you can use the characteristic polynomial to find the required formula by just multiplying across by $A^-1$.
$endgroup$
– copper.hat
2 days ago
$begingroup$
I didn't understand what you mean, can you please explain it more?
$endgroup$
– Fareed AF
2 days ago
$begingroup$
What happens if you multiply the characteristic equation (with $A$) across by the inverse if $A$?
$endgroup$
– copper.hat
2 days ago
$begingroup$
The charactaristic equation is a determinant. So if I multiplied inside the determinant I will get $det(tA^-1-I)$. I didn't understand where you want me to multiply by $A^-1$ and do "multiply across by" have the same meaning as "multiply by"? Sorry maybe I am not that good in english...
$endgroup$
– Fareed AF
2 days ago
1
$begingroup$
I added an answer.
$endgroup$
– copper.hat
2 days ago
|
show 1 more comment
1
$begingroup$
The constant in the characteristic polynomial must be non zero if $A$ is invertible, so you can use the characteristic polynomial to find the required formula by just multiplying across by $A^-1$.
$endgroup$
– copper.hat
2 days ago
$begingroup$
I didn't understand what you mean, can you please explain it more?
$endgroup$
– Fareed AF
2 days ago
$begingroup$
What happens if you multiply the characteristic equation (with $A$) across by the inverse if $A$?
$endgroup$
– copper.hat
2 days ago
$begingroup$
The charactaristic equation is a determinant. So if I multiplied inside the determinant I will get $det(tA^-1-I)$. I didn't understand where you want me to multiply by $A^-1$ and do "multiply across by" have the same meaning as "multiply by"? Sorry maybe I am not that good in english...
$endgroup$
– Fareed AF
2 days ago
1
$begingroup$
I added an answer.
$endgroup$
– copper.hat
2 days ago
1
1
$begingroup$
The constant in the characteristic polynomial must be non zero if $A$ is invertible, so you can use the characteristic polynomial to find the required formula by just multiplying across by $A^-1$.
$endgroup$
– copper.hat
2 days ago
$begingroup$
The constant in the characteristic polynomial must be non zero if $A$ is invertible, so you can use the characteristic polynomial to find the required formula by just multiplying across by $A^-1$.
$endgroup$
– copper.hat
2 days ago
$begingroup$
I didn't understand what you mean, can you please explain it more?
$endgroup$
– Fareed AF
2 days ago
$begingroup$
I didn't understand what you mean, can you please explain it more?
$endgroup$
– Fareed AF
2 days ago
$begingroup$
What happens if you multiply the characteristic equation (with $A$) across by the inverse if $A$?
$endgroup$
– copper.hat
2 days ago
$begingroup$
What happens if you multiply the characteristic equation (with $A$) across by the inverse if $A$?
$endgroup$
– copper.hat
2 days ago
$begingroup$
The charactaristic equation is a determinant. So if I multiplied inside the determinant I will get $det(tA^-1-I)$. I didn't understand where you want me to multiply by $A^-1$ and do "multiply across by" have the same meaning as "multiply by"? Sorry maybe I am not that good in english...
$endgroup$
– Fareed AF
2 days ago
$begingroup$
The charactaristic equation is a determinant. So if I multiplied inside the determinant I will get $det(tA^-1-I)$. I didn't understand where you want me to multiply by $A^-1$ and do "multiply across by" have the same meaning as "multiply by"? Sorry maybe I am not that good in english...
$endgroup$
– Fareed AF
2 days ago
1
1
$begingroup$
I added an answer.
$endgroup$
– copper.hat
2 days ago
$begingroup$
I added an answer.
$endgroup$
– copper.hat
2 days ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The characteristic polynomial is $chi_A(s) = det (sI-a)$ and the Cayley Hamilton theorem
tells us that $chi_A(A) = 0$. Since $A$ is invertible, we have
$c_0=chi_A(0) neq 0$ (the parameter $0$ means the scalar zero)
and so we have
$chi_A(A) = 0 = A^n+c_n-1A^n-1+cdots+ c_1A +c_0 I = 0$. Now multiply through by $A^-1$ to get
$A^n-1+c_n-2A^n-1+cdots+ c_1 I +c_0 A^-1 = 0$, from which we get
$A^-1 = - 1 over c_0 (A^n-1+c_n-2A^n-1+cdots+ c_1 I)$.
answered 2 days ago
copper.hatcopper.hat
128k559161
$endgroup$
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
add a comment |
$begingroup$
Here's a quick and dirty way, not using determinants.
Because $M_n$ itself has dimension $n^2$, we can find constants $c_1,dots,c_n^2+1$, not all zero, such that $c_1 A + dots + c_n^2+1 A^n^2+1=0$.
Suppose now $A$ is invertible. Let $j_0$ the smallest index satisfying $c_j_0ne 0$. Then $j_0<n^2+1$ because $A$ is invertible.
Multiply the linear combination by $frac1c_j_0A^-(j_0+1)$ to obtain
$$ boxed A^-1 + fracc_j_0+1c_j_0 A + dots + fracc_n^2+1c_j_0A^n^2+1-j_0 =0,$$
completing the proof.
answered 2 days ago
FnacoolFnacool
5,281611
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The characteristic polynomial is $chi_A(s) = det (sI-a)$ and the Cayley Hamilton theorem
tells us that $chi_A(A) = 0$. Since $A$ is invertible, we have
$c_0=chi_A(0) neq 0$ (the parameter $0$ means the scalar zero)
and so we have
$chi_A(A) = 0 = A^n+c_n-1A^n-1+cdots+ c_1A +c_0 I = 0$. Now multiply through by $A^-1$ to get
$A^n-1+c_n-2A^n-1+cdots+ c_1 I +c_0 A^-1 = 0$, from which we get
$A^-1 = - 1 over c_0 (A^n-1+c_n-2A^n-1+cdots+ c_1 I)$.
answered 2 days ago
copper.hatcopper.hat
128k559161
$endgroup$
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
add a comment |
$begingroup$
The characteristic polynomial is $chi_A(s) = det (sI-a)$ and the Cayley Hamilton theorem
tells us that $chi_A(A) = 0$. Since $A$ is invertible, we have
$c_0=chi_A(0) neq 0$ (the parameter $0$ means the scalar zero)
and so we have
$chi_A(A) = 0 = A^n+c_n-1A^n-1+cdots+ c_1A +c_0 I = 0$. Now multiply through by $A^-1$ to get
$A^n-1+c_n-2A^n-1+cdots+ c_1 I +c_0 A^-1 = 0$, from which we get
$A^-1 = - 1 over c_0 (A^n-1+c_n-2A^n-1+cdots+ c_1 I)$.
answered 2 days ago
copper.hatcopper.hat
128k559161
$endgroup$
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
add a comment |
$begingroup$
The characteristic polynomial is $chi_A(s) = det (sI-a)$ and the Cayley Hamilton theorem
tells us that $chi_A(A) = 0$. Since $A$ is invertible, we have
$c_0=chi_A(0) neq 0$ (the parameter $0$ means the scalar zero)
and so we have
$chi_A(A) = 0 = A^n+c_n-1A^n-1+cdots+ c_1A +c_0 I = 0$. Now multiply through by $A^-1$ to get
$A^n-1+c_n-2A^n-1+cdots+ c_1 I +c_0 A^-1 = 0$, from which we get
$A^-1 = - 1 over c_0 (A^n-1+c_n-2A^n-1+cdots+ c_1 I)$.
answered 2 days ago
copper.hatcopper.hat
128k559161
$endgroup$
The characteristic polynomial is $chi_A(s) = det (sI-a)$ and the Cayley Hamilton theorem
tells us that $chi_A(A) = 0$. Since $A$ is invertible, we have
$c_0=chi_A(0) neq 0$ (the parameter $0$ means the scalar zero)
and so we have
$chi_A(A) = 0 = A^n+c_n-1A^n-1+cdots+ c_1A +c_0 I = 0$. Now multiply through by $A^-1$ to get
$A^n-1+c_n-2A^n-1+cdots+ c_1 I +c_0 A^-1 = 0$, from which we get
$A^-1 = - 1 over c_0 (A^n-1+c_n-2A^n-1+cdots+ c_1 I)$.
answered 2 days ago
copper.hatcopper.hat
128k559161
edited 2 days ago
answered 2 days ago
copper.hatcopper.hat
128k559161
answered 2 days ago
copper.hatcopper.hat
128k559161
answered 2 days ago
copper.hatcopper.hat
128k559161
128k559161
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
add a comment |
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
$begingroup$
Thank you, I appreciate that
$endgroup$
– Fareed AF
2 days ago
add a comment |
$begingroup$
Here's a quick and dirty way, not using determinants.
Because $M_n$ itself has dimension $n^2$, we can find constants $c_1,dots,c_n^2+1$, not all zero, such that $c_1 A + dots + c_n^2+1 A^n^2+1=0$.
Suppose now $A$ is invertible. Let $j_0$ the smallest index satisfying $c_j_0ne 0$. Then $j_0<n^2+1$ because $A$ is invertible.
Multiply the linear combination by $frac1c_j_0A^-(j_0+1)$ to obtain
$$ boxed A^-1 + fracc_j_0+1c_j_0 A + dots + fracc_n^2+1c_j_0A^n^2+1-j_0 =0,$$
completing the proof.
answered 2 days ago
FnacoolFnacool
5,281611
$endgroup$
add a comment |
$begingroup$
Here's a quick and dirty way, not using determinants.
Because $M_n$ itself has dimension $n^2$, we can find constants $c_1,dots,c_n^2+1$, not all zero, such that $c_1 A + dots + c_n^2+1 A^n^2+1=0$.
Suppose now $A$ is invertible. Let $j_0$ the smallest index satisfying $c_j_0ne 0$. Then $j_0<n^2+1$ because $A$ is invertible.
Multiply the linear combination by $frac1c_j_0A^-(j_0+1)$ to obtain
$$ boxed A^-1 + fracc_j_0+1c_j_0 A + dots + fracc_n^2+1c_j_0A^n^2+1-j_0 =0,$$
completing the proof.
answered 2 days ago
FnacoolFnacool
5,281611
$endgroup$
add a comment |
$begingroup$
Here's a quick and dirty way, not using determinants.
Because $M_n$ itself has dimension $n^2$, we can find constants $c_1,dots,c_n^2+1$, not all zero, such that $c_1 A + dots + c_n^2+1 A^n^2+1=0$.
Suppose now $A$ is invertible. Let $j_0$ the smallest index satisfying $c_j_0ne 0$. Then $j_0<n^2+1$ because $A$ is invertible.
Multiply the linear combination by $frac1c_j_0A^-(j_0+1)$ to obtain
$$ boxed A^-1 + fracc_j_0+1c_j_0 A + dots + fracc_n^2+1c_j_0A^n^2+1-j_0 =0,$$
completing the proof.
answered 2 days ago
FnacoolFnacool
5,281611
$endgroup$
Here's a quick and dirty way, not using determinants.
Because $M_n$ itself has dimension $n^2$, we can find constants $c_1,dots,c_n^2+1$, not all zero, such that $c_1 A + dots + c_n^2+1 A^n^2+1=0$.
Suppose now $A$ is invertible. Let $j_0$ the smallest index satisfying $c_j_0ne 0$. Then $j_0<n^2+1$ because $A$ is invertible.
Multiply the linear combination by $frac1c_j_0A^-(j_0+1)$ to obtain
$$ boxed A^-1 + fracc_j_0+1c_j_0 A + dots + fracc_n^2+1c_j_0A^n^2+1-j_0 =0,$$
completing the proof.
answered 2 days ago
FnacoolFnacool
5,281611
answered 2 days ago
FnacoolFnacool
5,281611
answered 2 days ago
FnacoolFnacool
5,281611
answered 2 days ago
FnacoolFnacool
5,281611
5,281611
add a comment |
add a comment |
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1
$begingroup$
The constant in the characteristic polynomial must be non zero if $A$ is invertible, so you can use the characteristic polynomial to find the required formula by just multiplying across by $A^-1$.
$endgroup$
– copper.hat
2 days ago
$begingroup$
I didn't understand what you mean, can you please explain it more?
$endgroup$
– Fareed AF
2 days ago
$begingroup$
What happens if you multiply the characteristic equation (with $A$) across by the inverse if $A$?
$endgroup$
– copper.hat
2 days ago
$begingroup$
The charactaristic equation is a determinant. So if I multiplied inside the determinant I will get $det(tA^-1-I)$. I didn't understand where you want me to multiply by $A^-1$ and do "multiply across by" have the same meaning as "multiply by"? Sorry maybe I am not that good in english...
$endgroup$
– Fareed AF
2 days ago
1
$begingroup$
I added an answer.
$endgroup$
– copper.hat
2 days ago