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Finding matrix relative to a polynomial basis and basis in R2?
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Let U be the space of polynomials with basis E=[1,t,t^2], let V=R2 have the basis F=[(1,0),(1,1)] and define a linear map T: U->V by T(f)= (f(3),f'(3)).
Determine the matrix representing T relative to the basis E and F.
What is the Rank of T?
Exhibit a basis for the kernel of T.
I understand how to determine a matrix relative to a single basis in R2, but I've never done it relative to two basis, and never with a polynomial as a basis. Can anyone walk me though how to go about solving this type of problem?
linear-algebra matrices polynomials linear-transformations matrix-rank
asked Mar 18 at 23:45
Brandon Grothe
81
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Let U be the space of polynomials with basis E=[1,t,t^2], let V=R2 have the basis F=[(1,0),(1,1)] and define a linear map T: U->V by T(f)= (f(3),f'(3)).
Determine the matrix representing T relative to the basis E and F.
What is the Rank of T?
Exhibit a basis for the kernel of T.
I understand how to determine a matrix relative to a single basis in R2, but I've never done it relative to two basis, and never with a polynomial as a basis. Can anyone walk me though how to go about solving this type of problem?
linear-algebra matrices polynomials linear-transformations matrix-rank
asked Mar 18 at 23:45
Brandon Grothe
81
Please use mathjax to format your question
– K Split X
Mar 18 at 23:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let U be the space of polynomials with basis E=[1,t,t^2], let V=R2 have the basis F=[(1,0),(1,1)] and define a linear map T: U->V by T(f)= (f(3),f'(3)).
Determine the matrix representing T relative to the basis E and F.
What is the Rank of T?
Exhibit a basis for the kernel of T.
I understand how to determine a matrix relative to a single basis in R2, but I've never done it relative to two basis, and never with a polynomial as a basis. Can anyone walk me though how to go about solving this type of problem?
linear-algebra matrices polynomials linear-transformations matrix-rank
asked Mar 18 at 23:45
Brandon Grothe
81
Let U be the space of polynomials with basis E=[1,t,t^2], let V=R2 have the basis F=[(1,0),(1,1)] and define a linear map T: U->V by T(f)= (f(3),f'(3)).
Determine the matrix representing T relative to the basis E and F.
What is the Rank of T?
Exhibit a basis for the kernel of T.
I understand how to determine a matrix relative to a single basis in R2, but I've never done it relative to two basis, and never with a polynomial as a basis. Can anyone walk me though how to go about solving this type of problem?
linear-algebra matrices polynomials linear-transformations matrix-rank
linear-algebra matrices polynomials linear-transformations matrix-rank
asked Mar 18 at 23:45
Brandon Grothe
81
asked Mar 18 at 23:45
Brandon Grothe
81
asked Mar 18 at 23:45
Brandon Grothe
81
asked Mar 18 at 23:45
Brandon Grothe
81
asked Mar 18 at 23:45
Brandon Grothe
81
81
Please use mathjax to format your question
– K Split X
Mar 18 at 23:53
add a comment |
Please use mathjax to format your question
– K Split X
Mar 18 at 23:53
Please use mathjax to format your question
– K Split X
Mar 18 at 23:53
Please use mathjax to format your question
– K Split X
Mar 18 at 23:53
add a comment |
1 Answer
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0
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HINT
Polynomial of degree 2 in this case are a vector space in the sense that at each polynomial $f(t)=a+bt+ct^2$ we can associate the vector $v=(a,b,c)$.
The linear map $T$ maps a polynomial $f(t)=a+bt+ct^2$ in $(a+3b+9c,b+6c)$. Since we are requeste to use the basis F we need to write $(a+3b+9c,b+6c)$ in this basis that is
- $x+y=a+3b+9cimplies x=a+3b+9c-b-6c=a+2b+3c$
- $y= b+6c$
Then
$$T(a,b,c)=(a+2b+3c,b+6c)$$
Can you proceed from here?
answered Mar 18 at 23:59
gimusi
86.2k74392
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
HINT
Polynomial of degree 2 in this case are a vector space in the sense that at each polynomial $f(t)=a+bt+ct^2$ we can associate the vector $v=(a,b,c)$.
The linear map $T$ maps a polynomial $f(t)=a+bt+ct^2$ in $(a+3b+9c,b+6c)$. Since we are requeste to use the basis F we need to write $(a+3b+9c,b+6c)$ in this basis that is
- $x+y=a+3b+9cimplies x=a+3b+9c-b-6c=a+2b+3c$
- $y= b+6c$
Then
$$T(a,b,c)=(a+2b+3c,b+6c)$$
Can you proceed from here?
answered Mar 18 at 23:59
gimusi
86.2k74392
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
add a comment |
up vote
0
down vote
accepted
HINT
Polynomial of degree 2 in this case are a vector space in the sense that at each polynomial $f(t)=a+bt+ct^2$ we can associate the vector $v=(a,b,c)$.
The linear map $T$ maps a polynomial $f(t)=a+bt+ct^2$ in $(a+3b+9c,b+6c)$. Since we are requeste to use the basis F we need to write $(a+3b+9c,b+6c)$ in this basis that is
- $x+y=a+3b+9cimplies x=a+3b+9c-b-6c=a+2b+3c$
- $y= b+6c$
Then
$$T(a,b,c)=(a+2b+3c,b+6c)$$
Can you proceed from here?
answered Mar 18 at 23:59
gimusi
86.2k74392
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
HINT
Polynomial of degree 2 in this case are a vector space in the sense that at each polynomial $f(t)=a+bt+ct^2$ we can associate the vector $v=(a,b,c)$.
The linear map $T$ maps a polynomial $f(t)=a+bt+ct^2$ in $(a+3b+9c,b+6c)$. Since we are requeste to use the basis F we need to write $(a+3b+9c,b+6c)$ in this basis that is
- $x+y=a+3b+9cimplies x=a+3b+9c-b-6c=a+2b+3c$
- $y= b+6c$
Then
$$T(a,b,c)=(a+2b+3c,b+6c)$$
Can you proceed from here?
answered Mar 18 at 23:59
gimusi
86.2k74392
HINT
Polynomial of degree 2 in this case are a vector space in the sense that at each polynomial $f(t)=a+bt+ct^2$ we can associate the vector $v=(a,b,c)$.
The linear map $T$ maps a polynomial $f(t)=a+bt+ct^2$ in $(a+3b+9c,b+6c)$. Since we are requeste to use the basis F we need to write $(a+3b+9c,b+6c)$ in this basis that is
- $x+y=a+3b+9cimplies x=a+3b+9c-b-6c=a+2b+3c$
- $y= b+6c$
Then
$$T(a,b,c)=(a+2b+3c,b+6c)$$
Can you proceed from here?
answered Mar 18 at 23:59
gimusi
86.2k74392
answered Mar 18 at 23:59
gimusi
86.2k74392
answered Mar 18 at 23:59
gimusi
86.2k74392
answered Mar 18 at 23:59
gimusi
86.2k74392
86.2k74392
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
add a comment |
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Thank you, but I'm having trouble wrapping my head around it... could you go a step further?
– Brandon Grothe
Mar 19 at 0:11
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
Can you derive the matrix for T from the last step?
– gimusi
Mar 19 at 9:12
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
I know I need to turn that into a system of linear equations but I'm confused by the x + y part and how to use it to get to the matrix
– Brandon Grothe
Mar 19 at 12:04
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
@BrandonGrothe Did you solve the problem? the matrix representation in the standard basis is $begin{pmatrix}1&3&9\0&1&6end{pmatrix}$ with the bais F is $begin{pmatrix}1&2&3\0&1&6end{pmatrix}$
– gimusi
Mar 23 at 13:32
add a comment |
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– K Split X
Mar 18 at 23:53