Let $ Y = 1 + x, 1 − x + x^2 , 1 + 3 x − x^2 subseteq P^2 ( mathbbR ).$
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Show that $Y$ does not span $P^2$ and find a basis for the span of $Y$.
I can row reduce the matrix so that the bottom row is all zeros is this enough to prove that it does not span? I find it easy to show that something does span but not to show that it does not span...
The basis I got was $ 1 + (1/2)x^2, x + (-1/2)x^2 $ is this correct?
linear-algebra polynomials vector-spaces
edited Aug 30 at 4:29
Robert Z
85.5k1055123
asked Aug 30 at 3:15
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up vote
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Show that $Y$ does not span $P^2$ and find a basis for the span of $Y$.
I can row reduce the matrix so that the bottom row is all zeros is this enough to prove that it does not span? I find it easy to show that something does span but not to show that it does not span...
The basis I got was $ 1 + (1/2)x^2, x + (-1/2)x^2 $ is this correct?
linear-algebra polynomials vector-spaces
edited Aug 30 at 4:29
Robert Z
85.5k1055123
asked Aug 30 at 3:15
Virtual mark
31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that $Y$ does not span $P^2$ and find a basis for the span of $Y$.
I can row reduce the matrix so that the bottom row is all zeros is this enough to prove that it does not span? I find it easy to show that something does span but not to show that it does not span...
The basis I got was $ 1 + (1/2)x^2, x + (-1/2)x^2 $ is this correct?
linear-algebra polynomials vector-spaces
edited Aug 30 at 4:29
Robert Z
85.5k1055123
asked Aug 30 at 3:15
Virtual mark
31
Show that $Y$ does not span $P^2$ and find a basis for the span of $Y$.
I can row reduce the matrix so that the bottom row is all zeros is this enough to prove that it does not span? I find it easy to show that something does span but not to show that it does not span...
The basis I got was $ 1 + (1/2)x^2, x + (-1/2)x^2 $ is this correct?
linear-algebra polynomials vector-spaces
linear-algebra polynomials vector-spaces
edited Aug 30 at 4:29
Robert Z
85.5k1055123
asked Aug 30 at 3:15
Virtual mark
31
edited Aug 30 at 4:29
Robert Z
85.5k1055123
asked Aug 30 at 3:15
Virtual mark
31
edited Aug 30 at 4:29
Robert Z
85.5k1055123
edited Aug 30 at 4:29
Robert Z
85.5k1055123
edited Aug 30 at 4:29
Robert Z
85.5k1055123
85.5k1055123
asked Aug 30 at 3:15
Virtual mark
31
asked Aug 30 at 3:15
Virtual mark
31
asked Aug 30 at 3:15
Virtual mark
31
31
add a comment |Â
add a comment |Â
2 Answers
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Yes, your calculations are correct and the basis for the span of $Y$ is good.
It would have been very helpful if you had your matrix, and its row reduced form written down, so the reader has an easier time to see your complete proof.
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
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Alternative approach:
For this particular question, we are told that it doesn't span $P^2(mathbbR)$ but we have to prove it.
It is easy to see that the first two polynomials are not multiple of each other, I shall try to prove that the third is a linear combination of them. By observing the coefficient of $x^2$ and the constant term. I can quickly write down that
$$2(1+x)-(1-x+x^2) = (1+3x-x^2)$$
Hence a basis is $ 1+x, 1-x+x^2 $.
To check that $ 1+ frac12 x^2, x-frac12 x^2 $ is a valid basis:
$$(1+ frac12 x^2)+ (x-frac12 x^2)=1+x$$
$$(1+ frac12 x^2)-(x-frac12 x^2)=1-x+x^2$$
Yup, it is correct.
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Yes, your calculations are correct and the basis for the span of $Y$ is good.
It would have been very helpful if you had your matrix, and its row reduced form written down, so the reader has an easier time to see your complete proof.
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
1
down vote
Yes, your calculations are correct and the basis for the span of $Y$ is good.
It would have been very helpful if you had your matrix, and its row reduced form written down, so the reader has an easier time to see your complete proof.
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, your calculations are correct and the basis for the span of $Y$ is good.
It would have been very helpful if you had your matrix, and its row reduced form written down, so the reader has an easier time to see your complete proof.
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
Yes, your calculations are correct and the basis for the span of $Y$ is good.
It would have been very helpful if you had your matrix, and its row reduced form written down, so the reader has an easier time to see your complete proof.
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
answered Aug 30 at 3:31
Mohammad Riazi-Kermani
31k41853
31k41853
add a comment |Â
add a comment |Â
up vote
0
down vote
Alternative approach:
For this particular question, we are told that it doesn't span $P^2(mathbbR)$ but we have to prove it.
It is easy to see that the first two polynomials are not multiple of each other, I shall try to prove that the third is a linear combination of them. By observing the coefficient of $x^2$ and the constant term. I can quickly write down that
$$2(1+x)-(1-x+x^2) = (1+3x-x^2)$$
Hence a basis is $ 1+x, 1-x+x^2 $.
To check that $ 1+ frac12 x^2, x-frac12 x^2 $ is a valid basis:
$$(1+ frac12 x^2)+ (x-frac12 x^2)=1+x$$
$$(1+ frac12 x^2)-(x-frac12 x^2)=1-x+x^2$$
Yup, it is correct.
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
add a comment |Â
up vote
0
down vote
Alternative approach:
For this particular question, we are told that it doesn't span $P^2(mathbbR)$ but we have to prove it.
It is easy to see that the first two polynomials are not multiple of each other, I shall try to prove that the third is a linear combination of them. By observing the coefficient of $x^2$ and the constant term. I can quickly write down that
$$2(1+x)-(1-x+x^2) = (1+3x-x^2)$$
Hence a basis is $ 1+x, 1-x+x^2 $.
To check that $ 1+ frac12 x^2, x-frac12 x^2 $ is a valid basis:
$$(1+ frac12 x^2)+ (x-frac12 x^2)=1+x$$
$$(1+ frac12 x^2)-(x-frac12 x^2)=1-x+x^2$$
Yup, it is correct.
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alternative approach:
For this particular question, we are told that it doesn't span $P^2(mathbbR)$ but we have to prove it.
It is easy to see that the first two polynomials are not multiple of each other, I shall try to prove that the third is a linear combination of them. By observing the coefficient of $x^2$ and the constant term. I can quickly write down that
$$2(1+x)-(1-x+x^2) = (1+3x-x^2)$$
Hence a basis is $ 1+x, 1-x+x^2 $.
To check that $ 1+ frac12 x^2, x-frac12 x^2 $ is a valid basis:
$$(1+ frac12 x^2)+ (x-frac12 x^2)=1+x$$
$$(1+ frac12 x^2)-(x-frac12 x^2)=1-x+x^2$$
Yup, it is correct.
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
Alternative approach:
For this particular question, we are told that it doesn't span $P^2(mathbbR)$ but we have to prove it.
It is easy to see that the first two polynomials are not multiple of each other, I shall try to prove that the third is a linear combination of them. By observing the coefficient of $x^2$ and the constant term. I can quickly write down that
$$2(1+x)-(1-x+x^2) = (1+3x-x^2)$$
Hence a basis is $ 1+x, 1-x+x^2 $.
To check that $ 1+ frac12 x^2, x-frac12 x^2 $ is a valid basis:
$$(1+ frac12 x^2)+ (x-frac12 x^2)=1+x$$
$$(1+ frac12 x^2)-(x-frac12 x^2)=1-x+x^2$$
Yup, it is correct.
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
edited Aug 30 at 4:24
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
answered Aug 30 at 4:16
Siong Thye Goh
81.2k1453103
81.2k1453103
add a comment |Â
add a comment |Â
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