Mixing
Finding $B^*$, the dual basis
$begingroup$
Find a basis $B$ for
$$V = left{ left[
begin{array}{cc}
x\
y\
z
end{array}
right] in mathbb{R}^3 vert x+y+z = 0right}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.
For example, if $V=mathbb{R}^2$ and $ B = $$left{ left[
begin{array}{cc}
3\
4
end{array}
right], left[
begin{array}{c}
5\
7
end{array}
right]right} $, then $ A = $$ left[
begin{array}{cc}
3&5\
4&7
end{array}
right]$, then $A^{-1} = $$ left[
begin{array}{cc}
7&-5\
-4&3
end{array}
right]$, and the dual basis $B^* = (l_1, l_2)$ when :
$ l_1= left( left[
begin{array}{cc}
x\
y\
end{array}
right]right) = 7x_1 - 5x_2 $
$ l_2= left(left[
begin{array}{cc}
x\
y\
end{array}
right]right) = -4x_1 + 3x_2 $
However, when finding a basis for $V$, I get to the following solution vector
$$left[begin{array}{cc}
-y-z\
y\
z
end{array}
right] = y left[begin{array}{cc}
-1\
1\
0
end{array}
right] + z left[begin{array}{cc}
-1\
0\
1
end{array}
right]$$
If I build a matrix out of the basis$ left{ left[begin{array}{cc}
-1\
1\
0
end{array}
right] , left[begin{array}{cc}
-1\
0\
1
end{array}
right] right}$ I will have a matrix that is $3times2$ and I cannot inverse this.
How to proceed from here?
linear-algebra dual-spaces
edited Jan 2 at 19:02
mechanodroid
27.1k62446
asked Jan 2 at 18:42
TegernakoTegernako
897
$endgroup$
add a comment |
$begingroup$
Find a basis $B$ for
$$V = left{ left[
begin{array}{cc}
x\
y\
z
end{array}
right] in mathbb{R}^3 vert x+y+z = 0right}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.
For example, if $V=mathbb{R}^2$ and $ B = $$left{ left[
begin{array}{cc}
3\
4
end{array}
right], left[
begin{array}{c}
5\
7
end{array}
right]right} $, then $ A = $$ left[
begin{array}{cc}
3&5\
4&7
end{array}
right]$, then $A^{-1} = $$ left[
begin{array}{cc}
7&-5\
-4&3
end{array}
right]$, and the dual basis $B^* = (l_1, l_2)$ when :
$ l_1= left( left[
begin{array}{cc}
x\
y\
end{array}
right]right) = 7x_1 - 5x_2 $
$ l_2= left(left[
begin{array}{cc}
x\
y\
end{array}
right]right) = -4x_1 + 3x_2 $
However, when finding a basis for $V$, I get to the following solution vector
$$left[begin{array}{cc}
-y-z\
y\
z
end{array}
right] = y left[begin{array}{cc}
-1\
1\
0
end{array}
right] + z left[begin{array}{cc}
-1\
0\
1
end{array}
right]$$
If I build a matrix out of the basis$ left{ left[begin{array}{cc}
-1\
1\
0
end{array}
right] , left[begin{array}{cc}
-1\
0\
1
end{array}
right] right}$ I will have a matrix that is $3times2$ and I cannot inverse this.
How to proceed from here?
linear-algebra dual-spaces
edited Jan 2 at 19:02
mechanodroid
27.1k62446
asked Jan 2 at 18:42
TegernakoTegernako
897
$endgroup$
1
$begingroup$
You are correct, this method breaks down here. Have you tried following Misguided's reasoning on math.stackexchange.com/questions/1286100/…
$endgroup$
– bounceback
Jan 2 at 18:52
$begingroup$
@bounceback explained very well, thanks!
$endgroup$
– Tegernako
Jan 2 at 19:44
add a comment |
$begingroup$
Find a basis $B$ for
$$V = left{ left[
begin{array}{cc}
x\
y\
z
end{array}
right] in mathbb{R}^3 vert x+y+z = 0right}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.
For example, if $V=mathbb{R}^2$ and $ B = $$left{ left[
begin{array}{cc}
3\
4
end{array}
right], left[
begin{array}{c}
5\
7
end{array}
right]right} $, then $ A = $$ left[
begin{array}{cc}
3&5\
4&7
end{array}
right]$, then $A^{-1} = $$ left[
begin{array}{cc}
7&-5\
-4&3
end{array}
right]$, and the dual basis $B^* = (l_1, l_2)$ when :
$ l_1= left( left[
begin{array}{cc}
x\
y\
end{array}
right]right) = 7x_1 - 5x_2 $
$ l_2= left(left[
begin{array}{cc}
x\
y\
end{array}
right]right) = -4x_1 + 3x_2 $
However, when finding a basis for $V$, I get to the following solution vector
$$left[begin{array}{cc}
-y-z\
y\
z
end{array}
right] = y left[begin{array}{cc}
-1\
1\
0
end{array}
right] + z left[begin{array}{cc}
-1\
0\
1
end{array}
right]$$
If I build a matrix out of the basis$ left{ left[begin{array}{cc}
-1\
1\
0
end{array}
right] , left[begin{array}{cc}
-1\
0\
1
end{array}
right] right}$ I will have a matrix that is $3times2$ and I cannot inverse this.
How to proceed from here?
linear-algebra dual-spaces
edited Jan 2 at 19:02
mechanodroid
27.1k62446
asked Jan 2 at 18:42
TegernakoTegernako
897
$endgroup$
Find a basis $B$ for
$$V = left{ left[
begin{array}{cc}
x\
y\
z
end{array}
right] in mathbb{R}^3 vert x+y+z = 0right}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{-1}$, and those build linear functionals that are the rows of the inverse matrix.
For example, if $V=mathbb{R}^2$ and $ B = $$left{ left[
begin{array}{cc}
3\
4
end{array}
right], left[
begin{array}{c}
5\
7
end{array}
right]right} $, then $ A = $$ left[
begin{array}{cc}
3&5\
4&7
end{array}
right]$, then $A^{-1} = $$ left[
begin{array}{cc}
7&-5\
-4&3
end{array}
right]$, and the dual basis $B^* = (l_1, l_2)$ when :
$ l_1= left( left[
begin{array}{cc}
x\
y\
end{array}
right]right) = 7x_1 - 5x_2 $
$ l_2= left(left[
begin{array}{cc}
x\
y\
end{array}
right]right) = -4x_1 + 3x_2 $
However, when finding a basis for $V$, I get to the following solution vector
$$left[begin{array}{cc}
-y-z\
y\
z
end{array}
right] = y left[begin{array}{cc}
-1\
1\
0
end{array}
right] + z left[begin{array}{cc}
-1\
0\
1
end{array}
right]$$
If I build a matrix out of the basis$ left{ left[begin{array}{cc}
-1\
1\
0
end{array}
right] , left[begin{array}{cc}
-1\
0\
1
end{array}
right] right}$ I will have a matrix that is $3times2$ and I cannot inverse this.
How to proceed from here?
linear-algebra dual-spaces
linear-algebra dual-spaces
edited Jan 2 at 19:02
mechanodroid
27.1k62446
asked Jan 2 at 18:42
TegernakoTegernako
897
edited Jan 2 at 19:02
mechanodroid
27.1k62446
asked Jan 2 at 18:42
TegernakoTegernako
897
edited Jan 2 at 19:02
mechanodroid
27.1k62446
edited Jan 2 at 19:02
mechanodroid
27.1k62446
edited Jan 2 at 19:02
mechanodroid
27.1k62446
27.1k62446
asked Jan 2 at 18:42
TegernakoTegernako
897
asked Jan 2 at 18:42
TegernakoTegernako
897
asked Jan 2 at 18:42
TegernakoTegernako
897
897
1
$begingroup$
You are correct, this method breaks down here. Have you tried following Misguided's reasoning on math.stackexchange.com/questions/1286100/…
$endgroup$
– bounceback
Jan 2 at 18:52
$begingroup$
@bounceback explained very well, thanks!
$endgroup$
– Tegernako
Jan 2 at 19:44
add a comment |
1
$begingroup$
You are correct, this method breaks down here. Have you tried following Misguided's reasoning on math.stackexchange.com/questions/1286100/…
$endgroup$
– bounceback
Jan 2 at 18:52
$begingroup$
@bounceback explained very well, thanks!
$endgroup$
– Tegernako
Jan 2 at 19:44
1
1
$begingroup$
You are correct, this method breaks down here. Have you tried following Misguided's reasoning on math.stackexchange.com/questions/1286100/…
$endgroup$
– bounceback
Jan 2 at 18:52
$begingroup$
You are correct, this method breaks down here. Have you tried following Misguided's reasoning on math.stackexchange.com/questions/1286100/…
$endgroup$
– bounceback
Jan 2 at 18:52
$begingroup$
@bounceback explained very well, thanks!
$endgroup$
– Tegernako
Jan 2 at 19:44
$begingroup$
@bounceback explained very well, thanks!
$endgroup$
– Tegernako
Jan 2 at 19:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By definition, the dual basis functionals $f_1, f_2$ are given on your basis ${v_1, v_2}$ as $$f_1(alpha_1v_1 + alpha_2v_2) = alpha_1, quad f_2(alpha_1v_1 + alpha_2v_2) = alpha_2$$
Now $$f_1left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_1left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_1left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_1(yv_1 + zv_2) = y$$
$$f_2left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_2left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_2left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_2(yv_1 + zv_2) = z$$
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
Be
begin{equation}
V = leftlbrace
left[begin{array}{c}
x \
y \
z \
end{array}right] in mathbb{R}^3 : x+y+z = 0 rightrbrace
end{equation}
Then $z=-x-y$, thus basis $B$ is:
begin{equation}
B = leftlbrace
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
,
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right] rightrbrace
end{equation}
Where $B={v_1,v_2}$. The dual basis is given by $f_i(v_j)=delta_{ij}$ where $f_iin B^{*}$, $v_jin B$ and $delta_{ij}$ is the Kronecker delta since:
begin{equation}
delta_{ij}=begin{cases}
1 quadtextrm{ if } i=j \
0 quadtextrm{ if } ineq j \
end{cases}
end{equation}
Then for $f_1$:
begin{eqnarray}
f_1left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_1left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_1left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
x
end{eqnarray}
And $f_2$:
begin{eqnarray}
f_2left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_2left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_2left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
y
end{eqnarray}
Then $B^{*}={f_1,f_2}$
answered Jan 2 at 19:25
El PastaEl Pasta
35115
$endgroup$
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
$begingroup$
By definition, the dual basis functionals $f_1, f_2$ are given on your basis ${v_1, v_2}$ as $$f_1(alpha_1v_1 + alpha_2v_2) = alpha_1, quad f_2(alpha_1v_1 + alpha_2v_2) = alpha_2$$
Now $$f_1left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_1left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_1left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_1(yv_1 + zv_2) = y$$
$$f_2left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_2left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_2left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_2(yv_1 + zv_2) = z$$
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
By definition, the dual basis functionals $f_1, f_2$ are given on your basis ${v_1, v_2}$ as $$f_1(alpha_1v_1 + alpha_2v_2) = alpha_1, quad f_2(alpha_1v_1 + alpha_2v_2) = alpha_2$$
Now $$f_1left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_1left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_1left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_1(yv_1 + zv_2) = y$$
$$f_2left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_2left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_2left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_2(yv_1 + zv_2) = z$$
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
By definition, the dual basis functionals $f_1, f_2$ are given on your basis ${v_1, v_2}$ as $$f_1(alpha_1v_1 + alpha_2v_2) = alpha_1, quad f_2(alpha_1v_1 + alpha_2v_2) = alpha_2$$
Now $$f_1left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_1left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_1left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_1(yv_1 + zv_2) = y$$
$$f_2left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_2left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_2left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_2(yv_1 + zv_2) = z$$
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
$endgroup$
By definition, the dual basis functionals $f_1, f_2$ are given on your basis ${v_1, v_2}$ as $$f_1(alpha_1v_1 + alpha_2v_2) = alpha_1, quad f_2(alpha_1v_1 + alpha_2v_2) = alpha_2$$
Now $$f_1left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_1left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_1left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_1(yv_1 + zv_2) = y$$
$$f_2left(begin{bmatrix} x \ y \ zend{bmatrix}right) = f_2left(begin{bmatrix} -y-z \ y \ zend{bmatrix}right) = f_2left(ybegin{bmatrix} -1 \ 1 \ 0end{bmatrix} + zbegin{bmatrix} -1 \ 0 \ -1end{bmatrix}right) = f_2(yv_1 + zv_2) = z$$
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
answered Jan 2 at 18:59
mechanodroidmechanodroid
27.1k62446
27.1k62446
add a comment |
add a comment |
$begingroup$
Be
begin{equation}
V = leftlbrace
left[begin{array}{c}
x \
y \
z \
end{array}right] in mathbb{R}^3 : x+y+z = 0 rightrbrace
end{equation}
Then $z=-x-y$, thus basis $B$ is:
begin{equation}
B = leftlbrace
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
,
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right] rightrbrace
end{equation}
Where $B={v_1,v_2}$. The dual basis is given by $f_i(v_j)=delta_{ij}$ where $f_iin B^{*}$, $v_jin B$ and $delta_{ij}$ is the Kronecker delta since:
begin{equation}
delta_{ij}=begin{cases}
1 quadtextrm{ if } i=j \
0 quadtextrm{ if } ineq j \
end{cases}
end{equation}
Then for $f_1$:
begin{eqnarray}
f_1left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_1left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_1left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
x
end{eqnarray}
And $f_2$:
begin{eqnarray}
f_2left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_2left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_2left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
y
end{eqnarray}
Then $B^{*}={f_1,f_2}$
answered Jan 2 at 19:25
El PastaEl Pasta
35115
$endgroup$
add a comment |
$begingroup$
Be
begin{equation}
V = leftlbrace
left[begin{array}{c}
x \
y \
z \
end{array}right] in mathbb{R}^3 : x+y+z = 0 rightrbrace
end{equation}
Then $z=-x-y$, thus basis $B$ is:
begin{equation}
B = leftlbrace
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
,
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right] rightrbrace
end{equation}
Where $B={v_1,v_2}$. The dual basis is given by $f_i(v_j)=delta_{ij}$ where $f_iin B^{*}$, $v_jin B$ and $delta_{ij}$ is the Kronecker delta since:
begin{equation}
delta_{ij}=begin{cases}
1 quadtextrm{ if } i=j \
0 quadtextrm{ if } ineq j \
end{cases}
end{equation}
Then for $f_1$:
begin{eqnarray}
f_1left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_1left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_1left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
x
end{eqnarray}
And $f_2$:
begin{eqnarray}
f_2left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_2left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_2left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
y
end{eqnarray}
Then $B^{*}={f_1,f_2}$
answered Jan 2 at 19:25
El PastaEl Pasta
35115
$endgroup$
add a comment |
$begingroup$
Be
begin{equation}
V = leftlbrace
left[begin{array}{c}
x \
y \
z \
end{array}right] in mathbb{R}^3 : x+y+z = 0 rightrbrace
end{equation}
Then $z=-x-y$, thus basis $B$ is:
begin{equation}
B = leftlbrace
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
,
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right] rightrbrace
end{equation}
Where $B={v_1,v_2}$. The dual basis is given by $f_i(v_j)=delta_{ij}$ where $f_iin B^{*}$, $v_jin B$ and $delta_{ij}$ is the Kronecker delta since:
begin{equation}
delta_{ij}=begin{cases}
1 quadtextrm{ if } i=j \
0 quadtextrm{ if } ineq j \
end{cases}
end{equation}
Then for $f_1$:
begin{eqnarray}
f_1left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_1left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_1left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
x
end{eqnarray}
And $f_2$:
begin{eqnarray}
f_2left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_2left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_2left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
y
end{eqnarray}
Then $B^{*}={f_1,f_2}$
answered Jan 2 at 19:25
El PastaEl Pasta
35115
$endgroup$
Be
begin{equation}
V = leftlbrace
left[begin{array}{c}
x \
y \
z \
end{array}right] in mathbb{R}^3 : x+y+z = 0 rightrbrace
end{equation}
Then $z=-x-y$, thus basis $B$ is:
begin{equation}
B = leftlbrace
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
,
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right] rightrbrace
end{equation}
Where $B={v_1,v_2}$. The dual basis is given by $f_i(v_j)=delta_{ij}$ where $f_iin B^{*}$, $v_jin B$ and $delta_{ij}$ is the Kronecker delta since:
begin{equation}
delta_{ij}=begin{cases}
1 quadtextrm{ if } i=j \
0 quadtextrm{ if } ineq j \
end{cases}
end{equation}
Then for $f_1$:
begin{eqnarray}
f_1left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_1left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_1left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_1left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
x
end{eqnarray}
And $f_2$:
begin{eqnarray}
f_2left(
left[begin{array}{c}
x \
y \
z \
end{array}right]
right)
=
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
xf_2left(
left[begin{array}{c}
1 \
0 \
-1 \
end{array}right]
right)
+
yf_2left(
left[begin{array}{c}
0 \
1 \
-1 \
end{array}right]
right) \
f_2left(
left[begin{array}{c}
x \
y \
-x-y \
end{array}right]
right)
&=&
y
end{eqnarray}
Then $B^{*}={f_1,f_2}$
answered Jan 2 at 19:25
El PastaEl Pasta
35115
answered Jan 2 at 19:25
El PastaEl Pasta
35115
answered Jan 2 at 19:25
El PastaEl Pasta
35115
answered Jan 2 at 19:25
El PastaEl Pasta
35115
35115
add a comment |
add a comment |
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1
$begingroup$
You are correct, this method breaks down here. Have you tried following Misguided's reasoning on math.stackexchange.com/questions/1286100/…
$endgroup$
– bounceback
Jan 2 at 18:52
$begingroup$
@bounceback explained very well, thanks!
$endgroup$
– Tegernako
Jan 2 at 19:44