coordinate vector relative to basis for $P_2$
This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.
Question:
Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$
$p_1 = 1$
$p_2 = 1+x$
$p_3 = 1+x+x^2$
My attempt:
Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.
p = $(2,6,7)$
$p_1= $ $(1,0,0)$
$p_2= $ $(1,1,0)$
$p_3= $ $(1,1,1)$
Transposed of $p_n$ have:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
*$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$
I reversed p because it was reversed relative to $p_n$
linear-algebra
asked Nov 28 '18 at 1:38
Forextrader
346
add a comment |
This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.
Question:
Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$
$p_1 = 1$
$p_2 = 1+x$
$p_3 = 1+x+x^2$
My attempt:
Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.
p = $(2,6,7)$
$p_1= $ $(1,0,0)$
$p_2= $ $(1,1,0)$
$p_3= $ $(1,1,1)$
Transposed of $p_n$ have:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
*$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$
I reversed p because it was reversed relative to $p_n$
linear-algebra
asked Nov 28 '18 at 1:38
Forextrader
346
add a comment |
This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.
Question:
Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$
$p_1 = 1$
$p_2 = 1+x$
$p_3 = 1+x+x^2$
My attempt:
Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.
p = $(2,6,7)$
$p_1= $ $(1,0,0)$
$p_2= $ $(1,1,0)$
$p_3= $ $(1,1,1)$
Transposed of $p_n$ have:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
*$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$
I reversed p because it was reversed relative to $p_n$
linear-algebra
asked Nov 28 '18 at 1:38
Forextrader
346
This was a question in one of my online assignments that I got wrong but have no idea why. Unfortunately I do not have the option to re-enter other attempts to see if they are correct, hence stack exchange.
Question:
Let p $=2x^2+6x+7$ .Find the coordinate vector of p relative to the following basis for $P_2$
$p_1 = 1$
$p_2 = 1+x$
$p_3 = 1+x+x^2$
My attempt:
Plugging in for simplicity $x=1$ for p , $p_1$, $p_2$, and $p_3$ results in the following vectors.
p = $(2,6,7)$
$p_1= $ $(1,0,0)$
$p_2= $ $(1,1,0)$
$p_3= $ $(1,1,1)$
Transposed of $p_n$ have:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
*$begin{bmatrix}7\6\2end{bmatrix} = $$begin{bmatrix}15\8\2end{bmatrix}$
I reversed p because it was reversed relative to $p_n$
linear-algebra
linear-algebra
asked Nov 28 '18 at 1:38
Forextrader
346
asked Nov 28 '18 at 1:38
Forextrader
346
asked Nov 28 '18 at 1:38
Forextrader
346
asked Nov 28 '18 at 1:38
Forextrader
346
asked Nov 28 '18 at 1:38
Forextrader
346
346
add a comment |
add a comment |
1 Answer
1
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oldest
votes
This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.
Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.
Now you can read off the solution. For starters, $z=2$.
You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.
Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.
Now you can read off the solution. For starters, $z=2$.
You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
add a comment |
This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.
Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.
Now you can read off the solution. For starters, $z=2$.
You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
add a comment |
This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.
Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.
Now you can read off the solution. For starters, $z=2$.
You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
This is incorrect because you have simply taken the image of your polynomial under the matrix whose columns are the basis vectors.
Instead, you want to augment that same matrix with the vector you were given: $left (begin{array}{rrr|r}1&1&1&7\0&1&1&6\0&0&1&2end{array}right)$.
Now you can read off the solution. For starters, $z=2$.
You are looking for the linear combination of the basis vectors which will give you $(7,6,2)^t$.
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
edited Nov 28 '18 at 2:00
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
answered Nov 28 '18 at 1:58
Chris Custer
10.8k3824
10.8k3824
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
add a comment |
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
Ah ah ah, I see now. Thank you!
– Forextrader
Nov 28 '18 at 1:59
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
But I was correct in changing (2,6,7) to (7,6,2) right?
– Forextrader
Nov 28 '18 at 2:00
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
Yes. As the standard basis is${1,x,x^2}$.
– Chris Custer
Nov 28 '18 at 2:08
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
In other words, you need to multiply the vector by the inverse of this matrix.
– amd
Nov 28 '18 at 3:01
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
Yes that would do it.
– Chris Custer
Nov 28 '18 at 3:16
add a comment |
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