Mixing
Determine matrix A with respect to standard basis $f:U->R^2$
$begingroup$
I am having trouble understanding my problem and what to calculate
I have been given the subspace
$U={x=$$
begin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
$
and the linear transformation $f: U rightarrow F^2 $
$fbegin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
=begin{vmatrix}
x_1\
x_2+x_3\
end{vmatrix}$
The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$
My attempt:
I have calculated the basis $B={begin{vmatrix}
1\
0\
-1\
end{vmatrix},begin{vmatrix}
0\
1\
-1\
end{vmatrix}}$
And my matrix A calculated from the linear transformation
$begin{vmatrix}
1&0&0\
0&1&1\
end{vmatrix}$
I'm aware the standard basis are $e_1=begin{vmatrix}
1\
0\
end{vmatrix} ,
e_2=begin{vmatrix}
0\
1\
end{vmatrix}$
I'm not sure about the next step. Do I calculate:
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
-1\
end{vmatrix}$
and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
0\
end{vmatrix}$
and
$fbegin{vmatrix}
0\
1\
-1\
end{vmatrix}
=begin{vmatrix}
0\
1\
end{vmatrix}$
I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?
linear-transformations
asked Jan 2 at 22:39
mahmamahma
11
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding my problem and what to calculate
I have been given the subspace
$U={x=$$
begin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
$
and the linear transformation $f: U rightarrow F^2 $
$fbegin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
=begin{vmatrix}
x_1\
x_2+x_3\
end{vmatrix}$
The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$
My attempt:
I have calculated the basis $B={begin{vmatrix}
1\
0\
-1\
end{vmatrix},begin{vmatrix}
0\
1\
-1\
end{vmatrix}}$
And my matrix A calculated from the linear transformation
$begin{vmatrix}
1&0&0\
0&1&1\
end{vmatrix}$
I'm aware the standard basis are $e_1=begin{vmatrix}
1\
0\
end{vmatrix} ,
e_2=begin{vmatrix}
0\
1\
end{vmatrix}$
I'm not sure about the next step. Do I calculate:
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
-1\
end{vmatrix}$
and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
0\
end{vmatrix}$
and
$fbegin{vmatrix}
0\
1\
-1\
end{vmatrix}
=begin{vmatrix}
0\
1\
end{vmatrix}$
I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?
linear-transformations
asked Jan 2 at 22:39
mahmamahma
11
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding my problem and what to calculate
I have been given the subspace
$U={x=$$
begin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
$
and the linear transformation $f: U rightarrow F^2 $
$fbegin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
=begin{vmatrix}
x_1\
x_2+x_3\
end{vmatrix}$
The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$
My attempt:
I have calculated the basis $B={begin{vmatrix}
1\
0\
-1\
end{vmatrix},begin{vmatrix}
0\
1\
-1\
end{vmatrix}}$
And my matrix A calculated from the linear transformation
$begin{vmatrix}
1&0&0\
0&1&1\
end{vmatrix}$
I'm aware the standard basis are $e_1=begin{vmatrix}
1\
0\
end{vmatrix} ,
e_2=begin{vmatrix}
0\
1\
end{vmatrix}$
I'm not sure about the next step. Do I calculate:
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
-1\
end{vmatrix}$
and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
0\
end{vmatrix}$
and
$fbegin{vmatrix}
0\
1\
-1\
end{vmatrix}
=begin{vmatrix}
0\
1\
end{vmatrix}$
I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?
linear-transformations
asked Jan 2 at 22:39
mahmamahma
11
$endgroup$
I am having trouble understanding my problem and what to calculate
I have been given the subspace
$U={x=$$
begin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
in F^3 | x_1 + x_2 + x_3 = 0} subset F^3
$
and the linear transformation $f: U rightarrow F^2 $
$fbegin{vmatrix}
x_1\
x_2\
x_3\
end{vmatrix}
=begin{vmatrix}
x_1\
x_2+x_3\
end{vmatrix}$
The question is to determine a matrix A that represent $f: U rightarrow F^2 $ with respect to the basis for U and standard basis $(e_1,e_2)$ for $F^2$
My attempt:
I have calculated the basis $B={begin{vmatrix}
1\
0\
-1\
end{vmatrix},begin{vmatrix}
0\
1\
-1\
end{vmatrix}}$
And my matrix A calculated from the linear transformation
$begin{vmatrix}
1&0&0\
0&1&1\
end{vmatrix}$
I'm aware the standard basis are $e_1=begin{vmatrix}
1\
0\
end{vmatrix} ,
e_2=begin{vmatrix}
0\
1\
end{vmatrix}$
I'm not sure about the next step. Do I calculate:
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
-1\
end{vmatrix}$
and do the same thing for the other basis or am I suppose to find a matrix A that's going to give me
$fbegin{vmatrix}
1\
0\
-1\
end{vmatrix}
=begin{vmatrix}
1\
0\
end{vmatrix}$
and
$fbegin{vmatrix}
0\
1\
-1\
end{vmatrix}
=begin{vmatrix}
0\
1\
end{vmatrix}$
I just don't understand the question. How am I suppose to find a matrix A with respect to the basis of U and the standard basis for $F^2$?
linear-transformations
linear-transformations
asked Jan 2 at 22:39
mahmamahma
11
asked Jan 2 at 22:39
mahmamahma
11
edited Jan 2 at 23:07
mahma
asked Jan 2 at 22:39
mahmamahma
11
asked Jan 2 at 22:39
mahmamahma
11
asked Jan 2 at 22:39
mahmamahma
11
11
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The image of the first vector in the basis is
$$
begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
$$
and the image of the second vector is
$$
begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
$$
so the matrix is
$$
begin{bmatrix}
1 & 0 \
-1 & 0
end{bmatrix}
$$
Note that the map $f$ can be more easily described as
$$
fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
$$
answered Jan 2 at 23:57
egregegreg
180k1485202
$endgroup$
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The image of the first vector in the basis is
$$
begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
$$
and the image of the second vector is
$$
begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
$$
so the matrix is
$$
begin{bmatrix}
1 & 0 \
-1 & 0
end{bmatrix}
$$
Note that the map $f$ can be more easily described as
$$
fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
$$
answered Jan 2 at 23:57
egregegreg
180k1485202
$endgroup$
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
add a comment |
$begingroup$
The image of the first vector in the basis is
$$
begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
$$
and the image of the second vector is
$$
begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
$$
so the matrix is
$$
begin{bmatrix}
1 & 0 \
-1 & 0
end{bmatrix}
$$
Note that the map $f$ can be more easily described as
$$
fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
$$
answered Jan 2 at 23:57
egregegreg
180k1485202
$endgroup$
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
add a comment |
$begingroup$
The image of the first vector in the basis is
$$
begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
$$
and the image of the second vector is
$$
begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
$$
so the matrix is
$$
begin{bmatrix}
1 & 0 \
-1 & 0
end{bmatrix}
$$
Note that the map $f$ can be more easily described as
$$
fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
$$
answered Jan 2 at 23:57
egregegreg
180k1485202
$endgroup$
The image of the first vector in the basis is
$$
begin{bmatrix} 1 \ 0+(-1) end{bmatrix}
$$
and the image of the second vector is
$$
begin{bmatrix} 0 \ 1+(-1) end{bmatrix}
$$
so the matrix is
$$
begin{bmatrix}
1 & 0 \
-1 & 0
end{bmatrix}
$$
Note that the map $f$ can be more easily described as
$$
fbegin{bmatrix} x_1 \ x_2 \ x_3 end{bmatrix} = begin{bmatrix} x_1 \ -x_1 end{bmatrix}
$$
answered Jan 2 at 23:57
egregegreg
180k1485202
answered Jan 2 at 23:57
egregegreg
180k1485202
answered Jan 2 at 23:57
egregegreg
180k1485202
answered Jan 2 at 23:57
egregegreg
180k1485202
180k1485202
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
add a comment |
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
Thats it? I'm afraid I don't fully understand how this is a representation of the standard basis despite it being simple and straightforward. So that means the matrix you have written is the representing matrix and the answer to the problem and not my matrix A calculated from the linear transformation?
$endgroup$
– mahma
Jan 3 at 0:08
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
$begingroup$
@mahma The map is $fcolon Uto F^2$ and the domain has dimension $2$, as well as the codomain; so the representing matrix must be $2times2$.
$endgroup$
– egreg
Jan 3 at 0:47
add a comment |
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