Linear Transformations between 2 non-standard basis of Polynomials

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If
$$
A = beginpmatrix 1 & -1 & 2 \
-2 & 1 &-1 \ 1 & 2 & 3 endpmatrix
$$
is the matrix representation of a
linear transformation $T : P_3(x) to P_3(x)$ with respect to
bases $1-x,x(1-x),x(1+x)$ and $1,1+x,1+x^2$. Find T.

While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.

$T[1-x] = 1(1) -2(1+x) +1(1+x^2)$

$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$

$T[x(1+x)] = 2(1) -1(1+x) +3(1+x^2)$

Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$

Is this fine ?

linear-algebra matrices linear-transformations

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edited Jul 21 at 16:06

mvw

30.4k22250

asked Jul 21 at 15:41

Sarkar

52

add a comment | 

up vote
0
down vote

favorite

If
$$
A = beginpmatrix 1 & -1 & 2 \
-2 & 1 &-1 \ 1 & 2 & 3 endpmatrix
$$
is the matrix representation of a
linear transformation $T : P_3(x) to P_3(x)$ with respect to
bases $1-x,x(1-x),x(1+x)$ and $1,1+x,1+x^2$. Find T.

While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.

$T[1-x] = 1(1) -2(1+x) +1(1+x^2)$

$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$

$T[x(1+x)] = 2(1) -1(1+x) +3(1+x^2)$

Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$

Is this fine ?

linear-algebra matrices linear-transformations

share|cite|improve this question

edited Jul 21 at 16:06

mvw

30.4k22250

asked Jul 21 at 15:41

Sarkar

52

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If
$$
A = beginpmatrix 1 & -1 & 2 \
-2 & 1 &-1 \ 1 & 2 & 3 endpmatrix
$$
is the matrix representation of a
linear transformation $T : P_3(x) to P_3(x)$ with respect to
bases $1-x,x(1-x),x(1+x)$ and $1,1+x,1+x^2$. Find T.

While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.

$T[1-x] = 1(1) -2(1+x) +1(1+x^2)$

$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$

$T[x(1+x)] = 2(1) -1(1+x) +3(1+x^2)$

Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$

Is this fine ?

linear-algebra matrices linear-transformations

share|cite|improve this question

edited Jul 21 at 16:06

mvw

30.4k22250

asked Jul 21 at 15:41

Sarkar

52

If
$$
A = beginpmatrix 1 & -1 & 2 \
-2 & 1 &-1 \ 1 & 2 & 3 endpmatrix
$$
is the matrix representation of a
linear transformation $T : P_3(x) to P_3(x)$ with respect to
bases $1-x,x(1-x),x(1+x)$ and $1,1+x,1+x^2$. Find T.

While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.

$T[1-x] = 1(1) -2(1+x) +1(1+x^2)$

$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$

$T[x(1+x)] = 2(1) -1(1+x) +3(1+x^2)$

Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$

Is this fine ?

linear-algebra matrices linear-transformations

share|cite|improve this question

edited Jul 21 at 16:06

mvw

30.4k22250

asked Jul 21 at 15:41

Sarkar

52

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 16:06

mvw

30.4k22250

edited Jul 21 at 16:06

mvw

30.4k22250

edited Jul 21 at 16:06

mvw

30.4k22250

30.4k22250

asked Jul 21 at 15:41

Sarkar

52

asked Jul 21 at 15:41

Sarkar

52

asked Jul 21 at 15:41

Sarkar

52

52

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
0
down vote

Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.

share|cite|improve this answer

answered Jul 21 at 15:44

Marcus M

8,1731847

add a comment | 

up vote
0
down vote

What you did is fine, but now you have to compute $T[alpha+beta x+gamma x^2]$ for arbitrary $alpha,beta,gammainmathbb R$. In order to do that, solve the equation$$alpha+beta x+gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$left{beginarrayla-b+2c=alpha\-2a+b-c=beta\a+2b+3c=gamma.endarrayright.$$

share|cite|improve this answer

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
  – Sarkar
  Jul 21 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
  – José Carlos Santos
  Jul 21 at 16:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
  – Sarkar
  Jul 21 at 16:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.

I would do it in a formal way first: denote $X, Y$, &c. the column vectors of polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.

We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),mathcal B_1)$ to $(P_2(x),mathcal B_2)$, i.e. we have a matrix relation
$$Y_2=AX_1$$
and asked for the matrix of this same linear transformation from $(P_2(x),mathcal B_textcanon)$ to itself, i.e. we're asked for the matrix $T$ such that
$$Y=TX.$$
Now that's easy, given the change of basis matrices:
$$P_1=beginbmatrix!!beginarrayrrc
1&0&0\-1&1&1\0&:llap-1&1
endarrayendbmatrix,qquad P_2=beginbmatrix
1&1&1\0&1&0\0&0&1
endbmatrix.$$
We have $Y=P_2Y_2$, $X=P_1X_1$, so
$$Y=P_2Y_2=P_2AX_1=(underbraceP_2AP_1^-1_T)X.$$
There remains to find the inverse of $P_1$, which is standard by row reduction.

share|cite|improve this answer

answered Jul 21 at 16:35

Bernard

110k635103

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.

share|cite|improve this answer

answered Jul 21 at 15:44

Marcus M

8,1731847

add a comment | 

up vote
0
down vote

Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.

share|cite|improve this answer

answered Jul 21 at 15:44

Marcus M

8,1731847

add a comment | 

up vote
0
down vote

up vote
0
down vote

Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.

share|cite|improve this answer

answered Jul 21 at 15:44

Marcus M

8,1731847

Everything you've written is correct, although I suspect the problem is asking you to find $T[a + bx + cx^2]$.

share|cite|improve this answer

answered Jul 21 at 15:44

Marcus M

8,1731847

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 15:44

Marcus M

8,1731847

answered Jul 21 at 15:44

Marcus M

8,1731847

answered Jul 21 at 15:44

Marcus M

8,1731847

8,1731847

add a comment | 

add a comment | 

up vote
0
down vote

What you did is fine, but now you have to compute $T[alpha+beta x+gamma x^2]$ for arbitrary $alpha,beta,gammainmathbb R$. In order to do that, solve the equation$$alpha+beta x+gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$left{beginarrayla-b+2c=alpha\-2a+b-c=beta\a+2b+3c=gamma.endarrayright.$$

share|cite|improve this answer

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
  – Sarkar
  Jul 21 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
  – José Carlos Santos
  Jul 21 at 16:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
  – Sarkar
  Jul 21 at 16:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

What you did is fine, but now you have to compute $T[alpha+beta x+gamma x^2]$ for arbitrary $alpha,beta,gammainmathbb R$. In order to do that, solve the equation$$alpha+beta x+gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$left{beginarrayla-b+2c=alpha\-2a+b-c=beta\a+2b+3c=gamma.endarrayright.$$

share|cite|improve this answer

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
  – Sarkar
  Jul 21 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
  – José Carlos Santos
  Jul 21 at 16:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
  – Sarkar
  Jul 21 at 16:29
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

What you did is fine, but now you have to compute $T[alpha+beta x+gamma x^2]$ for arbitrary $alpha,beta,gammainmathbb R$. In order to do that, solve the equation$$alpha+beta x+gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$left{beginarrayla-b+2c=alpha\-2a+b-c=beta\a+2b+3c=gamma.endarrayright.$$

share|cite|improve this answer

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

What you did is fine, but now you have to compute $T[alpha+beta x+gamma x^2]$ for arbitrary $alpha,beta,gammainmathbb R$. In order to do that, solve the equation$$alpha+beta x+gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$left{beginarrayla-b+2c=alpha\-2a+b-c=beta\a+2b+3c=gamma.endarrayright.$$

share|cite|improve this answer

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

answered Jul 21 at 15:46

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
  – Sarkar
  Jul 21 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
  – José Carlos Santos
  Jul 21 at 16:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
  – Sarkar
  Jul 21 at 16:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
  – Sarkar
  Jul 21 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
  – José Carlos Santos
  Jul 21 at 16:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
  – Sarkar
  Jul 21 at 16:29
  

  
  
  
  

then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
– Sarkar
Jul 21 at 16:12

then it becomes a $ T:[(1,x,x^2) -> (1,1+x,1+x^2)]$ ?
– Sarkar
Jul 21 at 16:12

This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
– José Carlos Santos
Jul 21 at 16:19

This doesn't make sense. The elements of the domain and of the codomain of $T$ are polynomials with degree smaller than or equal to $2$.
– José Carlos Santos
Jul 21 at 16:19

Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
– Sarkar
Jul 21 at 16:29

Okay. i get it. So now the coefficients of basis $(1,1+x,1+x^2)$ is standardised using standard basis$(1,x,x^2)$ coefficients
– Sarkar
Jul 21 at 16:29

add a comment | 

up vote
0
down vote

I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.

I would do it in a formal way first: denote $X, Y$, &c. the column vectors of polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.

We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),mathcal B_1)$ to $(P_2(x),mathcal B_2)$, i.e. we have a matrix relation
$$Y_2=AX_1$$
and asked for the matrix of this same linear transformation from $(P_2(x),mathcal B_textcanon)$ to itself, i.e. we're asked for the matrix $T$ such that
$$Y=TX.$$
Now that's easy, given the change of basis matrices:
$$P_1=beginbmatrix!!beginarrayrrc
1&0&0\-1&1&1\0&:llap-1&1
endarrayendbmatrix,qquad P_2=beginbmatrix
1&1&1\0&1&0\0&0&1
endbmatrix.$$
We have $Y=P_2Y_2$, $X=P_1X_1$, so
$$Y=P_2Y_2=P_2AX_1=(underbraceP_2AP_1^-1_T)X.$$
There remains to find the inverse of $P_1$, which is standard by row reduction.

share|cite|improve this answer

answered Jul 21 at 16:35

Bernard

110k635103

add a comment | 

up vote
0
down vote

I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.

I would do it in a formal way first: denote $X, Y$, &c. the column vectors of polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.

We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),mathcal B_1)$ to $(P_2(x),mathcal B_2)$, i.e. we have a matrix relation
$$Y_2=AX_1$$
and asked for the matrix of this same linear transformation from $(P_2(x),mathcal B_textcanon)$ to itself, i.e. we're asked for the matrix $T$ such that
$$Y=TX.$$
Now that's easy, given the change of basis matrices:
$$P_1=beginbmatrix!!beginarrayrrc
1&0&0\-1&1&1\0&:llap-1&1
endarrayendbmatrix,qquad P_2=beginbmatrix
1&1&1\0&1&0\0&0&1
endbmatrix.$$
We have $Y=P_2Y_2$, $X=P_1X_1$, so
$$Y=P_2Y_2=P_2AX_1=(underbraceP_2AP_1^-1_T)X.$$
There remains to find the inverse of $P_1$, which is standard by row reduction.

share|cite|improve this answer

answered Jul 21 at 16:35

Bernard

110k635103

add a comment | 

up vote
0
down vote

up vote
0
down vote

I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.

I would do it in a formal way first: denote $X, Y$, &c. the column vectors of polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.

We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),mathcal B_1)$ to $(P_2(x),mathcal B_2)$, i.e. we have a matrix relation
$$Y_2=AX_1$$
and asked for the matrix of this same linear transformation from $(P_2(x),mathcal B_textcanon)$ to itself, i.e. we're asked for the matrix $T$ such that
$$Y=TX.$$
Now that's easy, given the change of basis matrices:
$$P_1=beginbmatrix!!beginarrayrrc
1&0&0\-1&1&1\0&:llap-1&1
endarrayendbmatrix,qquad P_2=beginbmatrix
1&1&1\0&1&0\0&0&1
endbmatrix.$$
We have $Y=P_2Y_2$, $X=P_1X_1$, so
$$Y=P_2Y_2=P_2AX_1=(underbraceP_2AP_1^-1_T)X.$$
There remains to find the inverse of $P_1$, which is standard by row reduction.

share|cite|improve this answer

answered Jul 21 at 16:35

Bernard

110k635103

I also suspect the real question is to find the image of $a+bx+cx^2$ in the canonical basis.

I would do it in a formal way first: denote $X, Y$, &c. the column vectors of polynomials in the canonical basis, $X_1, Y_1$, &c. their column vectors in the first basis and $X_2, Y_2$, &c. their column vectors in the second basis.

We're given the matrix $A$ of a linear transformation $T$ from $(P_2(x),mathcal B_1)$ to $(P_2(x),mathcal B_2)$, i.e. we have a matrix relation
$$Y_2=AX_1$$
and asked for the matrix of this same linear transformation from $(P_2(x),mathcal B_textcanon)$ to itself, i.e. we're asked for the matrix $T$ such that
$$Y=TX.$$
Now that's easy, given the change of basis matrices:
$$P_1=beginbmatrix!!beginarrayrrc
1&0&0\-1&1&1\0&:llap-1&1
endarrayendbmatrix,qquad P_2=beginbmatrix
1&1&1\0&1&0\0&0&1
endbmatrix.$$
We have $Y=P_2Y_2$, $X=P_1X_1$, so
$$Y=P_2Y_2=P_2AX_1=(underbraceP_2AP_1^-1_T)X.$$
There remains to find the inverse of $P_1$, which is standard by row reduction.

share|cite|improve this answer

answered Jul 21 at 16:35

Bernard

110k635103

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 16:35

Bernard

110k635103

answered Jul 21 at 16:35

Bernard

110k635103

answered Jul 21 at 16:35

Bernard

110k635103

110k635103

add a comment | 

add a comment | 

 

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