Mixing
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Show that $B$ is a basis of $mathbb{P}_{2}$
$begingroup$
Im asked to prove that $B$ is a basis of $mathbb{P}_{2}$. I know how to show that $B$ is a basis of $mathbb{R}^{3}$ but i don't know how to do it with polynomials. This is the exercise:
Show that $B = {1+x,2+x-x^2,1}$ is a basis of the vector space $mathbb{P}_{2} = $Span${1,x,x^2}$
linear-algebra
edited Jan 22 at 16:53
pwerth
3,265417
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
$endgroup$
add a comment |
$begingroup$
Im asked to prove that $B$ is a basis of $mathbb{P}_{2}$. I know how to show that $B$ is a basis of $mathbb{R}^{3}$ but i don't know how to do it with polynomials. This is the exercise:
Show that $B = {1+x,2+x-x^2,1}$ is a basis of the vector space $mathbb{P}_{2} = $Span${1,x,x^2}$
linear-algebra
edited Jan 22 at 16:53
pwerth
3,265417
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
$endgroup$
3
$begingroup$
It is equivalent to showing that $[1,1,0], [2,1,-1], [1,0,0]$ is a basis of $mathbb R^3$.
$endgroup$
– Robert Israel
Jan 22 at 16:45
$begingroup$
What do you mean $B$ is a basis of $mathbb{R}^{3}$? $B$ is not even a subset of $mathbb{R}^{3}$
$endgroup$
– pwerth
Jan 22 at 16:54
add a comment |
$begingroup$
Im asked to prove that $B$ is a basis of $mathbb{P}_{2}$. I know how to show that $B$ is a basis of $mathbb{R}^{3}$ but i don't know how to do it with polynomials. This is the exercise:
Show that $B = {1+x,2+x-x^2,1}$ is a basis of the vector space $mathbb{P}_{2} = $Span${1,x,x^2}$
linear-algebra
edited Jan 22 at 16:53
pwerth
3,265417
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
$endgroup$
Im asked to prove that $B$ is a basis of $mathbb{P}_{2}$. I know how to show that $B$ is a basis of $mathbb{R}^{3}$ but i don't know how to do it with polynomials. This is the exercise:
Show that $B = {1+x,2+x-x^2,1}$ is a basis of the vector space $mathbb{P}_{2} = $Span${1,x,x^2}$
linear-algebra
linear-algebra
edited Jan 22 at 16:53
pwerth
3,265417
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
edited Jan 22 at 16:53
pwerth
3,265417
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
edited Jan 22 at 16:53
pwerth
3,265417
edited Jan 22 at 16:53
pwerth
3,265417
edited Jan 22 at 16:53
pwerth
3,265417
3,265417
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
asked Jan 22 at 16:43
Juan HernándezJuan Hernández
32
32
3
$begingroup$
It is equivalent to showing that $[1,1,0], [2,1,-1], [1,0,0]$ is a basis of $mathbb R^3$.
$endgroup$
– Robert Israel
Jan 22 at 16:45
$begingroup$
What do you mean $B$ is a basis of $mathbb{R}^{3}$? $B$ is not even a subset of $mathbb{R}^{3}$
$endgroup$
– pwerth
Jan 22 at 16:54
add a comment |
3
$begingroup$
It is equivalent to showing that $[1,1,0], [2,1,-1], [1,0,0]$ is a basis of $mathbb R^3$.
$endgroup$
– Robert Israel
Jan 22 at 16:45
$begingroup$
What do you mean $B$ is a basis of $mathbb{R}^{3}$? $B$ is not even a subset of $mathbb{R}^{3}$
$endgroup$
– pwerth
Jan 22 at 16:54
3
3
$begingroup$
It is equivalent to showing that $[1,1,0], [2,1,-1], [1,0,0]$ is a basis of $mathbb R^3$.
$endgroup$
– Robert Israel
Jan 22 at 16:45
$begingroup$
It is equivalent to showing that $[1,1,0], [2,1,-1], [1,0,0]$ is a basis of $mathbb R^3$.
$endgroup$
– Robert Israel
Jan 22 at 16:45
$begingroup$
What do you mean $B$ is a basis of $mathbb{R}^{3}$? $B$ is not even a subset of $mathbb{R}^{3}$
$endgroup$
– pwerth
Jan 22 at 16:54
$begingroup$
What do you mean $B$ is a basis of $mathbb{R}^{3}$? $B$ is not even a subset of $mathbb{R}^{3}$
$endgroup$
– pwerth
Jan 22 at 16:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a,b,c in mathbb R$ and $a(1+x)+b(2+x-x^2)+c=0$ for all x$ in mathbb R$.
If you can show that $a=b=c=0$, then you are done.
answered Jan 22 at 16:54
FredFred
48k1849
$endgroup$
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a,b,c in mathbb R$ and $a(1+x)+b(2+x-x^2)+c=0$ for all x$ in mathbb R$.
If you can show that $a=b=c=0$, then you are done.
answered Jan 22 at 16:54
FredFred
48k1849
$endgroup$
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
add a comment |
$begingroup$
Let $a,b,c in mathbb R$ and $a(1+x)+b(2+x-x^2)+c=0$ for all x$ in mathbb R$.
If you can show that $a=b=c=0$, then you are done.
answered Jan 22 at 16:54
FredFred
48k1849
$endgroup$
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
add a comment |
$begingroup$
Let $a,b,c in mathbb R$ and $a(1+x)+b(2+x-x^2)+c=0$ for all x$ in mathbb R$.
If you can show that $a=b=c=0$, then you are done.
answered Jan 22 at 16:54
FredFred
48k1849
$endgroup$
Let $a,b,c in mathbb R$ and $a(1+x)+b(2+x-x^2)+c=0$ for all x$ in mathbb R$.
If you can show that $a=b=c=0$, then you are done.
answered Jan 22 at 16:54
FredFred
48k1849
answered Jan 22 at 16:54
FredFred
48k1849
answered Jan 22 at 16:54
FredFred
48k1849
answered Jan 22 at 16:54
FredFred
48k1849
48k1849
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
add a comment |
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
$begingroup$
Technically, one also has to show that the the polynomials span $Bbb P^2$.
$endgroup$
– Arthur
Jan 22 at 17:04
add a comment |
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$begingroup$
It is equivalent to showing that $[1,1,0], [2,1,-1], [1,0,0]$ is a basis of $mathbb R^3$.
$endgroup$
– Robert Israel
Jan 22 at 16:45
$begingroup$
What do you mean $B$ is a basis of $mathbb{R}^{3}$? $B$ is not even a subset of $mathbb{R}^{3}$
$endgroup$
– pwerth
Jan 22 at 16:54