Linear Algebra : Basis of quadratic equations?
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I am struggling with Linear Algebra lately , specifically , finding the basis of a set of quadratic equations.
question goes like this : B = {P1 = (1+X^3) ; P2 = (2X+X^2+X^3) P3 = (1-X^2) P4 = (2+X+X^3) , find if B is a basis in the vectorial space of R^<=3
Basically finding if it's components are linearly independent, but how do I treat/transform P1,P2,P3,P4 into vectors ?
I've looked online and couldn't find anything similar.
Thanks in advance.
linear-algebra
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
$endgroup$
add a comment |
$begingroup$
I am struggling with Linear Algebra lately , specifically , finding the basis of a set of quadratic equations.
question goes like this : B = {P1 = (1+X^3) ; P2 = (2X+X^2+X^3) P3 = (1-X^2) P4 = (2+X+X^3) , find if B is a basis in the vectorial space of R^<=3
Basically finding if it's components are linearly independent, but how do I treat/transform P1,P2,P3,P4 into vectors ?
I've looked online and couldn't find anything similar.
Thanks in advance.
linear-algebra
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
$endgroup$
1
$begingroup$
Welcome to Math.SE. It would improve your Question to clarify what the vector space is that $B$ might be a basis for. Mathematical notation can be properly "typeset" in posts here, which make them more readable.
$endgroup$
– hardmath
Jan 24 at 16:49
1
$begingroup$
Note that $P_1$ and $P_4$ are not quadratic.
$endgroup$
– Servaes
Jan 24 at 16:58
$begingroup$
Sorry , I forgot to mention , set of quadratic/cubic equations.
$endgroup$
– Ionut Aurelian Craciun
Jan 24 at 17:01
$begingroup$
They are vectors in the coefficients, not the variable. I think of polynomial vectors as an alternative labeling for the coeeffecients. So if I have some vector $(a,b,c)$ I could instead write this as $a + bx + cx^2$ where the power of $x$ indicated the position of the coefficients. Changing basis can result in different polynomials but the basic are essentially the same.
$endgroup$
– CyclotomicField
Jan 24 at 17:09
add a comment |
$begingroup$
I am struggling with Linear Algebra lately , specifically , finding the basis of a set of quadratic equations.
question goes like this : B = {P1 = (1+X^3) ; P2 = (2X+X^2+X^3) P3 = (1-X^2) P4 = (2+X+X^3) , find if B is a basis in the vectorial space of R^<=3
Basically finding if it's components are linearly independent, but how do I treat/transform P1,P2,P3,P4 into vectors ?
I've looked online and couldn't find anything similar.
Thanks in advance.
linear-algebra
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
$endgroup$
I am struggling with Linear Algebra lately , specifically , finding the basis of a set of quadratic equations.
question goes like this : B = {P1 = (1+X^3) ; P2 = (2X+X^2+X^3) P3 = (1-X^2) P4 = (2+X+X^3) , find if B is a basis in the vectorial space of R^<=3
Basically finding if it's components are linearly independent, but how do I treat/transform P1,P2,P3,P4 into vectors ?
I've looked online and couldn't find anything similar.
Thanks in advance.
linear-algebra
linear-algebra
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
edited Jan 24 at 17:02
Ionut Aurelian Craciun
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
asked Jan 24 at 16:44
Ionut Aurelian CraciunIonut Aurelian Craciun
11
11
1
$begingroup$
Welcome to Math.SE. It would improve your Question to clarify what the vector space is that $B$ might be a basis for. Mathematical notation can be properly "typeset" in posts here, which make them more readable.
$endgroup$
– hardmath
Jan 24 at 16:49
1
$begingroup$
Note that $P_1$ and $P_4$ are not quadratic.
$endgroup$
– Servaes
Jan 24 at 16:58
$begingroup$
Sorry , I forgot to mention , set of quadratic/cubic equations.
$endgroup$
– Ionut Aurelian Craciun
Jan 24 at 17:01
$begingroup$
They are vectors in the coefficients, not the variable. I think of polynomial vectors as an alternative labeling for the coeeffecients. So if I have some vector $(a,b,c)$ I could instead write this as $a + bx + cx^2$ where the power of $x$ indicated the position of the coefficients. Changing basis can result in different polynomials but the basic are essentially the same.
$endgroup$
– CyclotomicField
Jan 24 at 17:09
add a comment |
1
$begingroup$
Welcome to Math.SE. It would improve your Question to clarify what the vector space is that $B$ might be a basis for. Mathematical notation can be properly "typeset" in posts here, which make them more readable.
$endgroup$
– hardmath
Jan 24 at 16:49
1
$begingroup$
Note that $P_1$ and $P_4$ are not quadratic.
$endgroup$
– Servaes
Jan 24 at 16:58
$begingroup$
Sorry , I forgot to mention , set of quadratic/cubic equations.
$endgroup$
– Ionut Aurelian Craciun
Jan 24 at 17:01
$begingroup$
They are vectors in the coefficients, not the variable. I think of polynomial vectors as an alternative labeling for the coeeffecients. So if I have some vector $(a,b,c)$ I could instead write this as $a + bx + cx^2$ where the power of $x$ indicated the position of the coefficients. Changing basis can result in different polynomials but the basic are essentially the same.
$endgroup$
– CyclotomicField
Jan 24 at 17:09
1
1
$begingroup$
Welcome to Math.SE. It would improve your Question to clarify what the vector space is that $B$ might be a basis for. Mathematical notation can be properly "typeset" in posts here, which make them more readable.
$endgroup$
– hardmath
Jan 24 at 16:49
$begingroup$
Welcome to Math.SE. It would improve your Question to clarify what the vector space is that $B$ might be a basis for. Mathematical notation can be properly "typeset" in posts here, which make them more readable.
$endgroup$
– hardmath
Jan 24 at 16:49
1
1
$begingroup$
Note that $P_1$ and $P_4$ are not quadratic.
$endgroup$
– Servaes
Jan 24 at 16:58
$begingroup$
Note that $P_1$ and $P_4$ are not quadratic.
$endgroup$
– Servaes
Jan 24 at 16:58
$begingroup$
Sorry , I forgot to mention , set of quadratic/cubic equations.
$endgroup$
– Ionut Aurelian Craciun
Jan 24 at 17:01
$begingroup$
Sorry , I forgot to mention , set of quadratic/cubic equations.
$endgroup$
– Ionut Aurelian Craciun
Jan 24 at 17:01
$begingroup$
They are vectors in the coefficients, not the variable. I think of polynomial vectors as an alternative labeling for the coeeffecients. So if I have some vector $(a,b,c)$ I could instead write this as $a + bx + cx^2$ where the power of $x$ indicated the position of the coefficients. Changing basis can result in different polynomials but the basic are essentially the same.
$endgroup$
– CyclotomicField
Jan 24 at 17:09
$begingroup$
They are vectors in the coefficients, not the variable. I think of polynomial vectors as an alternative labeling for the coeeffecients. So if I have some vector $(a,b,c)$ I could instead write this as $a + bx + cx^2$ where the power of $x$ indicated the position of the coefficients. Changing basis can result in different polynomials but the basic are essentially the same.
$endgroup$
– CyclotomicField
Jan 24 at 17:09
add a comment |
1 Answer
1
active
oldest
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$begingroup$
How do I transform polynomials into vectors?
Polynomials ARE vectors!
If a collection of objects satisfy the axioms of a vector space (i.e. have a well defined concepts of addition and scalar multiplication) they form a vector space.
However, you might look at the coeficients of the polynomial and it will look very much like the vectors you are familiar with in $mathbb R^4$
i.e.
$P_1 = (1,0,0,1), P_2 = (0,2,1,1), P_3 = (1,0,-1,0), P_4 = (2,1,0,1)$
Whether you convert them to this form or not.
They form a basis if:
Polynomials of degree 3 or less indeed form a vector space. The proposed basis vectors span the space. And, they are linearly independent.
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
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add a comment |
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$begingroup$
How do I transform polynomials into vectors?
Polynomials ARE vectors!
If a collection of objects satisfy the axioms of a vector space (i.e. have a well defined concepts of addition and scalar multiplication) they form a vector space.
However, you might look at the coeficients of the polynomial and it will look very much like the vectors you are familiar with in $mathbb R^4$
i.e.
$P_1 = (1,0,0,1), P_2 = (0,2,1,1), P_3 = (1,0,-1,0), P_4 = (2,1,0,1)$
Whether you convert them to this form or not.
They form a basis if:
Polynomials of degree 3 or less indeed form a vector space. The proposed basis vectors span the space. And, they are linearly independent.
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
How do I transform polynomials into vectors?
Polynomials ARE vectors!
If a collection of objects satisfy the axioms of a vector space (i.e. have a well defined concepts of addition and scalar multiplication) they form a vector space.
However, you might look at the coeficients of the polynomial and it will look very much like the vectors you are familiar with in $mathbb R^4$
i.e.
$P_1 = (1,0,0,1), P_2 = (0,2,1,1), P_3 = (1,0,-1,0), P_4 = (2,1,0,1)$
Whether you convert them to this form or not.
They form a basis if:
Polynomials of degree 3 or less indeed form a vector space. The proposed basis vectors span the space. And, they are linearly independent.
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
How do I transform polynomials into vectors?
Polynomials ARE vectors!
If a collection of objects satisfy the axioms of a vector space (i.e. have a well defined concepts of addition and scalar multiplication) they form a vector space.
However, you might look at the coeficients of the polynomial and it will look very much like the vectors you are familiar with in $mathbb R^4$
i.e.
$P_1 = (1,0,0,1), P_2 = (0,2,1,1), P_3 = (1,0,-1,0), P_4 = (2,1,0,1)$
Whether you convert them to this form or not.
They form a basis if:
Polynomials of degree 3 or less indeed form a vector space. The proposed basis vectors span the space. And, they are linearly independent.
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
$endgroup$
How do I transform polynomials into vectors?
Polynomials ARE vectors!
If a collection of objects satisfy the axioms of a vector space (i.e. have a well defined concepts of addition and scalar multiplication) they form a vector space.
However, you might look at the coeficients of the polynomial and it will look very much like the vectors you are familiar with in $mathbb R^4$
i.e.
$P_1 = (1,0,0,1), P_2 = (0,2,1,1), P_3 = (1,0,-1,0), P_4 = (2,1,0,1)$
Whether you convert them to this form or not.
They form a basis if:
Polynomials of degree 3 or less indeed form a vector space. The proposed basis vectors span the space. And, they are linearly independent.
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
answered Jan 24 at 17:04
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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$begingroup$
Welcome to Math.SE. It would improve your Question to clarify what the vector space is that $B$ might be a basis for. Mathematical notation can be properly "typeset" in posts here, which make them more readable.
$endgroup$
– hardmath
Jan 24 at 16:49
1
$begingroup$
Note that $P_1$ and $P_4$ are not quadratic.
$endgroup$
– Servaes
Jan 24 at 16:58
$begingroup$
Sorry , I forgot to mention , set of quadratic/cubic equations.
$endgroup$
– Ionut Aurelian Craciun
Jan 24 at 17:01
$begingroup$
They are vectors in the coefficients, not the variable. I think of polynomial vectors as an alternative labeling for the coeeffecients. So if I have some vector $(a,b,c)$ I could instead write this as $a + bx + cx^2$ where the power of $x$ indicated the position of the coefficients. Changing basis can result in different polynomials but the basic are essentially the same.
$endgroup$
– CyclotomicField
Jan 24 at 17:09