Why isn't this a basis for $P^2$?
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Explain why S is not a basis for P2. S = {1 + x2, 1 - x2}
I am having a difficult time putting into words why I don't think S is a basis and I think it is due to my lack of understanding of polynomials. I understand from a matrix point of view, but not from a polynomial point of view.
Matrix View:
Transform polynomial into aug. matrix.
1 + 0x + x2 = 0
1 + 0x - x2 = 0
begin{bmatrix}1&0&1&0\1&0&-1&0end{bmatrix}
RREF
begin{bmatrix}1&0&1&0\0&0&-2&0end{bmatrix}
Clearly this transformed into RREF because there are too many elements in the matrix.
How is this connected to polynomials? Does it simply mean that there is a free variable (and which variable is free, x or x2?) Can either of the variables be free? This is where my understanding falls apart.
linear-algebra
edited Nov 12 at 18:10
idea
1,9132920
asked Nov 12 at 16:30
Evan Kim
266
|
show 3 more comments
up vote
0
down vote
favorite
Explain why S is not a basis for P2. S = {1 + x2, 1 - x2}
I am having a difficult time putting into words why I don't think S is a basis and I think it is due to my lack of understanding of polynomials. I understand from a matrix point of view, but not from a polynomial point of view.
Matrix View:
Transform polynomial into aug. matrix.
1 + 0x + x2 = 0
1 + 0x - x2 = 0
begin{bmatrix}1&0&1&0\1&0&-1&0end{bmatrix}
RREF
begin{bmatrix}1&0&1&0\0&0&-2&0end{bmatrix}
Clearly this transformed into RREF because there are too many elements in the matrix.
How is this connected to polynomials? Does it simply mean that there is a free variable (and which variable is free, x or x2?) Can either of the variables be free? This is where my understanding falls apart.
linear-algebra
edited Nov 12 at 18:10
idea
1,9132920
asked Nov 12 at 16:30
Evan Kim
266
What do the matrices stand for? ////I would prefer to explain that by definition of "basis", not others.
– xbh
Nov 12 at 16:34
That can't have been what $S$ is supposed to be.
– Randall
Nov 12 at 16:43
oops, there is a typo, second polynomial should have been -. I fixed it
– Evan Kim
Nov 12 at 16:53
The "matrix view" does not make sense to me.
– xbh
Nov 12 at 17:05
Nevermind, I figured it out. I was doing things wrong by making 1 + x<sup>2</sup> == 1 + 0x + x<sup>2</sup>. That is false. It is really 1 + 0 + x<sup>2</sup> == 1 + x<sup>2</sup>. I apologize for the confusion
– Evan Kim
Nov 12 at 17:12
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Explain why S is not a basis for P2. S = {1 + x2, 1 - x2}
I am having a difficult time putting into words why I don't think S is a basis and I think it is due to my lack of understanding of polynomials. I understand from a matrix point of view, but not from a polynomial point of view.
Matrix View:
Transform polynomial into aug. matrix.
1 + 0x + x2 = 0
1 + 0x - x2 = 0
begin{bmatrix}1&0&1&0\1&0&-1&0end{bmatrix}
RREF
begin{bmatrix}1&0&1&0\0&0&-2&0end{bmatrix}
Clearly this transformed into RREF because there are too many elements in the matrix.
How is this connected to polynomials? Does it simply mean that there is a free variable (and which variable is free, x or x2?) Can either of the variables be free? This is where my understanding falls apart.
linear-algebra
edited Nov 12 at 18:10
idea
1,9132920
asked Nov 12 at 16:30
Evan Kim
266
Explain why S is not a basis for P2. S = {1 + x2, 1 - x2}
I am having a difficult time putting into words why I don't think S is a basis and I think it is due to my lack of understanding of polynomials. I understand from a matrix point of view, but not from a polynomial point of view.
Matrix View:
Transform polynomial into aug. matrix.
1 + 0x + x2 = 0
1 + 0x - x2 = 0
begin{bmatrix}1&0&1&0\1&0&-1&0end{bmatrix}
RREF
begin{bmatrix}1&0&1&0\0&0&-2&0end{bmatrix}
Clearly this transformed into RREF because there are too many elements in the matrix.
How is this connected to polynomials? Does it simply mean that there is a free variable (and which variable is free, x or x2?) Can either of the variables be free? This is where my understanding falls apart.
linear-algebra
linear-algebra
edited Nov 12 at 18:10
idea
1,9132920
asked Nov 12 at 16:30
Evan Kim
266
edited Nov 12 at 18:10
idea
1,9132920
asked Nov 12 at 16:30
Evan Kim
266
edited Nov 12 at 18:10
idea
1,9132920
edited Nov 12 at 18:10
idea
1,9132920
edited Nov 12 at 18:10
idea
1,9132920
1,9132920
asked Nov 12 at 16:30
Evan Kim
266
asked Nov 12 at 16:30
Evan Kim
266
asked Nov 12 at 16:30
Evan Kim
266
266
What do the matrices stand for? ////I would prefer to explain that by definition of "basis", not others.
– xbh
Nov 12 at 16:34
That can't have been what $S$ is supposed to be.
– Randall
Nov 12 at 16:43
oops, there is a typo, second polynomial should have been -. I fixed it
– Evan Kim
Nov 12 at 16:53
The "matrix view" does not make sense to me.
– xbh
Nov 12 at 17:05
Nevermind, I figured it out. I was doing things wrong by making 1 + x<sup>2</sup> == 1 + 0x + x<sup>2</sup>. That is false. It is really 1 + 0 + x<sup>2</sup> == 1 + x<sup>2</sup>. I apologize for the confusion
– Evan Kim
Nov 12 at 17:12
|
show 3 more comments
What do the matrices stand for? ////I would prefer to explain that by definition of "basis", not others.
– xbh
Nov 12 at 16:34
That can't have been what $S$ is supposed to be.
– Randall
Nov 12 at 16:43
oops, there is a typo, second polynomial should have been -. I fixed it
– Evan Kim
Nov 12 at 16:53
The "matrix view" does not make sense to me.
– xbh
Nov 12 at 17:05
Nevermind, I figured it out. I was doing things wrong by making 1 + x<sup>2</sup> == 1 + 0x + x<sup>2</sup>. That is false. It is really 1 + 0 + x<sup>2</sup> == 1 + x<sup>2</sup>. I apologize for the confusion
– Evan Kim
Nov 12 at 17:12
What do the matrices stand for? ////I would prefer to explain that by definition of "basis", not others.
– xbh
Nov 12 at 16:34
What do the matrices stand for? ////I would prefer to explain that by definition of "basis", not others.
– xbh
Nov 12 at 16:34
That can't have been what $S$ is supposed to be.
– Randall
Nov 12 at 16:43
That can't have been what $S$ is supposed to be.
– Randall
Nov 12 at 16:43
oops, there is a typo, second polynomial should have been -. I fixed it
– Evan Kim
Nov 12 at 16:53
oops, there is a typo, second polynomial should have been -. I fixed it
– Evan Kim
Nov 12 at 16:53
The "matrix view" does not make sense to me.
– xbh
Nov 12 at 17:05
The "matrix view" does not make sense to me.
– xbh
Nov 12 at 17:05
Nevermind, I figured it out. I was doing things wrong by making 1 + x<sup>2</sup> == 1 + 0x + x<sup>2</sup>. That is false. It is really 1 + 0 + x<sup>2</sup> == 1 + x<sup>2</sup>. I apologize for the confusion
– Evan Kim
Nov 12 at 17:12
Nevermind, I figured it out. I was doing things wrong by making 1 + x<sup>2</sup> == 1 + 0x + x<sup>2</sup>. That is false. It is really 1 + 0 + x<sup>2</sup> == 1 + x<sup>2</sup>. I apologize for the confusion
– Evan Kim
Nov 12 at 17:12
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
$P^2$ is three dimensional. But $S$ only has two elements.
Alternatively, $xnotinoperatorname{span} S$. Try to prove this...
answered Nov 12 at 17:13
Chris Custer
8,4292623
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$P^2$ is three dimensional. But $S$ only has two elements.
Alternatively, $xnotinoperatorname{span} S$. Try to prove this...
answered Nov 12 at 17:13
Chris Custer
8,4292623
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
add a comment |
up vote
2
down vote
$P^2$ is three dimensional. But $S$ only has two elements.
Alternatively, $xnotinoperatorname{span} S$. Try to prove this...
answered Nov 12 at 17:13
Chris Custer
8,4292623
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
add a comment |
up vote
2
down vote
up vote
2
down vote
$P^2$ is three dimensional. But $S$ only has two elements.
Alternatively, $xnotinoperatorname{span} S$. Try to prove this...
answered Nov 12 at 17:13
Chris Custer
8,4292623
$P^2$ is three dimensional. But $S$ only has two elements.
Alternatively, $xnotinoperatorname{span} S$. Try to prove this...
answered Nov 12 at 17:13
Chris Custer
8,4292623
answered Nov 12 at 17:13
Chris Custer
8,4292623
answered Nov 12 at 17:13
Chris Custer
8,4292623
answered Nov 12 at 17:13
Chris Custer
8,4292623
8,4292623
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
add a comment |
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
I see my mistake. I assumed S has 3 elements by thinking that 1 + x2 == 1 + 0x + x2
– Evan Kim
Nov 12 at 17:16
add a comment |
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What do the matrices stand for? ////I would prefer to explain that by definition of "basis", not others.
– xbh
Nov 12 at 16:34
That can't have been what $S$ is supposed to be.
– Randall
Nov 12 at 16:43
oops, there is a typo, second polynomial should have been -. I fixed it
– Evan Kim
Nov 12 at 16:53
The "matrix view" does not make sense to me.
– xbh
Nov 12 at 17:05
Nevermind, I figured it out. I was doing things wrong by making 1 + x<sup>2</sup> == 1 + 0x + x<sup>2</sup>. That is false. It is really 1 + 0 + x<sup>2</sup> == 1 + x<sup>2</sup>. I apologize for the confusion
– Evan Kim
Nov 12 at 17:12