Mixing
Find a basis for Vector space of polynomials
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I would like to find a basis for the vector space of Polynomials of degree 3 or less over the reals satisfying the following 2 properties: $$p(1)=0$$ $$p(x)=p(-x)$$
I started with a generic polynomial in the vector space: $$a_0 + a_1x+a_2x^2+a_3x^3$$ and tried to make it fit both conditions:$$a_0 + a_1+_2+a_3 = 0$$ $$a_0 + a_1x+a_2x^2+a_3x^3=a_0 -a_1x+a_2x^2-a_3x^3$$
the second equations becomes $$a_1x+a_3x^3=0$$ thus $a_1$ and $a_3$ must be constantly equal to 0.
Plugging back into the first equation we get $$a_0 = -a_2$$ thus $p=a_0 -a_0x^2$.
Then $1-x^2$ would be a basis?
Is my method correct? If not, how would one solve this type of problem.
Thanks in advance
linear-algebra polynomials vector-spaces
asked Jan 23 at 14:59
mmmmommmmo
1227
$endgroup$
add a comment |
$begingroup$
I would like to find a basis for the vector space of Polynomials of degree 3 or less over the reals satisfying the following 2 properties: $$p(1)=0$$ $$p(x)=p(-x)$$
I started with a generic polynomial in the vector space: $$a_0 + a_1x+a_2x^2+a_3x^3$$ and tried to make it fit both conditions:$$a_0 + a_1+_2+a_3 = 0$$ $$a_0 + a_1x+a_2x^2+a_3x^3=a_0 -a_1x+a_2x^2-a_3x^3$$
the second equations becomes $$a_1x+a_3x^3=0$$ thus $a_1$ and $a_3$ must be constantly equal to 0.
Plugging back into the first equation we get $$a_0 = -a_2$$ thus $p=a_0 -a_0x^2$.
Then $1-x^2$ would be a basis?
Is my method correct? If not, how would one solve this type of problem.
Thanks in advance
linear-algebra polynomials vector-spaces
asked Jan 23 at 14:59
mmmmommmmo
1227
$endgroup$
1
$begingroup$
Yes, it is quite correct.
$endgroup$
– Bernard
Jan 23 at 15:07
add a comment |
$begingroup$
I would like to find a basis for the vector space of Polynomials of degree 3 or less over the reals satisfying the following 2 properties: $$p(1)=0$$ $$p(x)=p(-x)$$
I started with a generic polynomial in the vector space: $$a_0 + a_1x+a_2x^2+a_3x^3$$ and tried to make it fit both conditions:$$a_0 + a_1+_2+a_3 = 0$$ $$a_0 + a_1x+a_2x^2+a_3x^3=a_0 -a_1x+a_2x^2-a_3x^3$$
the second equations becomes $$a_1x+a_3x^3=0$$ thus $a_1$ and $a_3$ must be constantly equal to 0.
Plugging back into the first equation we get $$a_0 = -a_2$$ thus $p=a_0 -a_0x^2$.
Then $1-x^2$ would be a basis?
Is my method correct? If not, how would one solve this type of problem.
Thanks in advance
linear-algebra polynomials vector-spaces
asked Jan 23 at 14:59
mmmmommmmo
1227
$endgroup$
I would like to find a basis for the vector space of Polynomials of degree 3 or less over the reals satisfying the following 2 properties: $$p(1)=0$$ $$p(x)=p(-x)$$
I started with a generic polynomial in the vector space: $$a_0 + a_1x+a_2x^2+a_3x^3$$ and tried to make it fit both conditions:$$a_0 + a_1+_2+a_3 = 0$$ $$a_0 + a_1x+a_2x^2+a_3x^3=a_0 -a_1x+a_2x^2-a_3x^3$$
the second equations becomes $$a_1x+a_3x^3=0$$ thus $a_1$ and $a_3$ must be constantly equal to 0.
Plugging back into the first equation we get $$a_0 = -a_2$$ thus $p=a_0 -a_0x^2$.
Then $1-x^2$ would be a basis?
Is my method correct? If not, how would one solve this type of problem.
Thanks in advance
linear-algebra polynomials vector-spaces
linear-algebra polynomials vector-spaces
asked Jan 23 at 14:59
mmmmommmmo
1227
asked Jan 23 at 14:59
mmmmommmmo
1227
asked Jan 23 at 14:59
mmmmommmmo
1227
asked Jan 23 at 14:59
mmmmommmmo
1227
asked Jan 23 at 14:59
mmmmommmmo
1227
1227
1
$begingroup$
Yes, it is quite correct.
$endgroup$
– Bernard
Jan 23 at 15:07
add a comment |
1
$begingroup$
Yes, it is quite correct.
$endgroup$
– Bernard
Jan 23 at 15:07
1
1
$begingroup$
Yes, it is quite correct.
$endgroup$
– Bernard
Jan 23 at 15:07
$begingroup$
Yes, it is quite correct.
$endgroup$
– Bernard
Jan 23 at 15:07
add a comment |
1 Answer
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$begingroup$
Yes, that is correct. The vector space has dimension $1$ and ${1-x^2}$ is a basis.
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
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$begingroup$
Yes, that is correct. The vector space has dimension $1$ and ${1-x^2}$ is a basis.
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
$endgroup$
add a comment |
$begingroup$
Yes, that is correct. The vector space has dimension $1$ and ${1-x^2}$ is a basis.
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
$endgroup$
add a comment |
$begingroup$
Yes, that is correct. The vector space has dimension $1$ and ${1-x^2}$ is a basis.
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
$endgroup$
Yes, that is correct. The vector space has dimension $1$ and ${1-x^2}$ is a basis.
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
answered Jan 23 at 15:10
gandalf61gandalf61
9,011825
9,011825
add a comment |
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Yes, it is quite correct.
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– Bernard
Jan 23 at 15:07