Show that the linear map $T:P_3(Bbb R)→Bbb R^4, p↦(p(0),p(1),p′(0),p′(1))$ is an isomorphism.
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Show that the linear map
$$T:P_3(Bbb R)→Bbb R^4,quad p↦(p(0),p(1),p′(0),p′(1))$$
is an isomorphism.
what I did following shows it's not a isomorphism ,what's wrong with it?
linear-algebra linear-transformations
edited Nov 16 at 22:58
Tianlalu
2,594632
asked Nov 16 at 22:50
DORCT
405
add a comment |
up vote
0
down vote
favorite
Show that the linear map
$$T:P_3(Bbb R)→Bbb R^4,quad p↦(p(0),p(1),p′(0),p′(1))$$
is an isomorphism.
what I did following shows it's not a isomorphism ,what's wrong with it?
linear-algebra linear-transformations
edited Nov 16 at 22:58
Tianlalu
2,594632
asked Nov 16 at 22:50
DORCT
405
Hint for a shorter solution: it's sufficient to show $T$ has trivial kernel (why?). Now, if $p(0) = p'(0) = 0$ then $x^2 mid p$ and if $p(1) = p'(1) = 0$ then $(x-1)^2 mid p$.
– Daniel Schepler
Nov 16 at 23:05
You should explicitly state what $P_3(mathbb{R})$ is. It becomes clear if one looks at the attached photo, but a question should be asked in a form that its essence becomes transparent at first glance.
– Paul Frost
Nov 16 at 23:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that the linear map
$$T:P_3(Bbb R)→Bbb R^4,quad p↦(p(0),p(1),p′(0),p′(1))$$
is an isomorphism.
what I did following shows it's not a isomorphism ,what's wrong with it?
linear-algebra linear-transformations
edited Nov 16 at 22:58
Tianlalu
2,594632
asked Nov 16 at 22:50
DORCT
405
Show that the linear map
$$T:P_3(Bbb R)→Bbb R^4,quad p↦(p(0),p(1),p′(0),p′(1))$$
is an isomorphism.
what I did following shows it's not a isomorphism ,what's wrong with it?
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Nov 16 at 22:58
Tianlalu
2,594632
asked Nov 16 at 22:50
DORCT
405
edited Nov 16 at 22:58
Tianlalu
2,594632
asked Nov 16 at 22:50
DORCT
405
edited Nov 16 at 22:58
Tianlalu
2,594632
edited Nov 16 at 22:58
Tianlalu
2,594632
edited Nov 16 at 22:58
Tianlalu
2,594632
2,594632
asked Nov 16 at 22:50
DORCT
405
asked Nov 16 at 22:50
DORCT
405
asked Nov 16 at 22:50
DORCT
405
405
Hint for a shorter solution: it's sufficient to show $T$ has trivial kernel (why?). Now, if $p(0) = p'(0) = 0$ then $x^2 mid p$ and if $p(1) = p'(1) = 0$ then $(x-1)^2 mid p$.
– Daniel Schepler
Nov 16 at 23:05
You should explicitly state what $P_3(mathbb{R})$ is. It becomes clear if one looks at the attached photo, but a question should be asked in a form that its essence becomes transparent at first glance.
– Paul Frost
Nov 16 at 23:14
add a comment |
Hint for a shorter solution: it's sufficient to show $T$ has trivial kernel (why?). Now, if $p(0) = p'(0) = 0$ then $x^2 mid p$ and if $p(1) = p'(1) = 0$ then $(x-1)^2 mid p$.
– Daniel Schepler
Nov 16 at 23:05
You should explicitly state what $P_3(mathbb{R})$ is. It becomes clear if one looks at the attached photo, but a question should be asked in a form that its essence becomes transparent at first glance.
– Paul Frost
Nov 16 at 23:14
Hint for a shorter solution: it's sufficient to show $T$ has trivial kernel (why?). Now, if $p(0) = p'(0) = 0$ then $x^2 mid p$ and if $p(1) = p'(1) = 0$ then $(x-1)^2 mid p$.
– Daniel Schepler
Nov 16 at 23:05
Hint for a shorter solution: it's sufficient to show $T$ has trivial kernel (why?). Now, if $p(0) = p'(0) = 0$ then $x^2 mid p$ and if $p(1) = p'(1) = 0$ then $(x-1)^2 mid p$.
– Daniel Schepler
Nov 16 at 23:05
You should explicitly state what $P_3(mathbb{R})$ is. It becomes clear if one looks at the attached photo, but a question should be asked in a form that its essence becomes transparent at first glance.
– Paul Frost
Nov 16 at 23:14
You should explicitly state what $P_3(mathbb{R})$ is. It becomes clear if one looks at the attached photo, but a question should be asked in a form that its essence becomes transparent at first glance.
– Paul Frost
Nov 16 at 23:14
add a comment |
1 Answer
1
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up vote
1
down vote
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Note that, if $p(x)=ax^3+bx^2+cx+din P_3(mathbb R)$, then we have
$$begin{align}&(1)qquad p(0)=d
\&(2)qquad p(1)=a+b+c+d
\&(3)qquad p'(0)=c
\&(4)qquad p'(1)=3a+2b+cend{align}$$
Therefore, your map $T$ sends
$$begin{align}&(1)qquad color{red}{1mapsto (1,1,0,0)}
\&(2)qquad xmapsto (0,1,1,1)
\&(3)qquad x^2mapsto (0,1,0,2)
\&(4)qquad x^3mapsto (0,1,0,3)end{align}$$
so you mapped $1$ incorrectly under $T$. You need to remember that $p'(x)$ loses the constant coefficient from $p(x)$.
But for your argument about mapping constant polynomials to $(0,0,0,0)$, this is false. $T$ maps constant polynomials to a scalar multiple of $(1,1,0,0)$. You should see that the matrix of $T$ relative to these standard bases is invertible, and therefore $T$ is an isomorphism (it is clearly linear).
answered Nov 16 at 23:03
Dave
8,39811033
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that, if $p(x)=ax^3+bx^2+cx+din P_3(mathbb R)$, then we have
$$begin{align}&(1)qquad p(0)=d
\&(2)qquad p(1)=a+b+c+d
\&(3)qquad p'(0)=c
\&(4)qquad p'(1)=3a+2b+cend{align}$$
Therefore, your map $T$ sends
$$begin{align}&(1)qquad color{red}{1mapsto (1,1,0,0)}
\&(2)qquad xmapsto (0,1,1,1)
\&(3)qquad x^2mapsto (0,1,0,2)
\&(4)qquad x^3mapsto (0,1,0,3)end{align}$$
so you mapped $1$ incorrectly under $T$. You need to remember that $p'(x)$ loses the constant coefficient from $p(x)$.
But for your argument about mapping constant polynomials to $(0,0,0,0)$, this is false. $T$ maps constant polynomials to a scalar multiple of $(1,1,0,0)$. You should see that the matrix of $T$ relative to these standard bases is invertible, and therefore $T$ is an isomorphism (it is clearly linear).
answered Nov 16 at 23:03
Dave
8,39811033
add a comment |
up vote
1
down vote
accepted
Note that, if $p(x)=ax^3+bx^2+cx+din P_3(mathbb R)$, then we have
$$begin{align}&(1)qquad p(0)=d
\&(2)qquad p(1)=a+b+c+d
\&(3)qquad p'(0)=c
\&(4)qquad p'(1)=3a+2b+cend{align}$$
Therefore, your map $T$ sends
$$begin{align}&(1)qquad color{red}{1mapsto (1,1,0,0)}
\&(2)qquad xmapsto (0,1,1,1)
\&(3)qquad x^2mapsto (0,1,0,2)
\&(4)qquad x^3mapsto (0,1,0,3)end{align}$$
so you mapped $1$ incorrectly under $T$. You need to remember that $p'(x)$ loses the constant coefficient from $p(x)$.
But for your argument about mapping constant polynomials to $(0,0,0,0)$, this is false. $T$ maps constant polynomials to a scalar multiple of $(1,1,0,0)$. You should see that the matrix of $T$ relative to these standard bases is invertible, and therefore $T$ is an isomorphism (it is clearly linear).
answered Nov 16 at 23:03
Dave
8,39811033
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that, if $p(x)=ax^3+bx^2+cx+din P_3(mathbb R)$, then we have
$$begin{align}&(1)qquad p(0)=d
\&(2)qquad p(1)=a+b+c+d
\&(3)qquad p'(0)=c
\&(4)qquad p'(1)=3a+2b+cend{align}$$
Therefore, your map $T$ sends
$$begin{align}&(1)qquad color{red}{1mapsto (1,1,0,0)}
\&(2)qquad xmapsto (0,1,1,1)
\&(3)qquad x^2mapsto (0,1,0,2)
\&(4)qquad x^3mapsto (0,1,0,3)end{align}$$
so you mapped $1$ incorrectly under $T$. You need to remember that $p'(x)$ loses the constant coefficient from $p(x)$.
But for your argument about mapping constant polynomials to $(0,0,0,0)$, this is false. $T$ maps constant polynomials to a scalar multiple of $(1,1,0,0)$. You should see that the matrix of $T$ relative to these standard bases is invertible, and therefore $T$ is an isomorphism (it is clearly linear).
answered Nov 16 at 23:03
Dave
8,39811033
Note that, if $p(x)=ax^3+bx^2+cx+din P_3(mathbb R)$, then we have
$$begin{align}&(1)qquad p(0)=d
\&(2)qquad p(1)=a+b+c+d
\&(3)qquad p'(0)=c
\&(4)qquad p'(1)=3a+2b+cend{align}$$
Therefore, your map $T$ sends
$$begin{align}&(1)qquad color{red}{1mapsto (1,1,0,0)}
\&(2)qquad xmapsto (0,1,1,1)
\&(3)qquad x^2mapsto (0,1,0,2)
\&(4)qquad x^3mapsto (0,1,0,3)end{align}$$
so you mapped $1$ incorrectly under $T$. You need to remember that $p'(x)$ loses the constant coefficient from $p(x)$.
But for your argument about mapping constant polynomials to $(0,0,0,0)$, this is false. $T$ maps constant polynomials to a scalar multiple of $(1,1,0,0)$. You should see that the matrix of $T$ relative to these standard bases is invertible, and therefore $T$ is an isomorphism (it is clearly linear).
answered Nov 16 at 23:03
Dave
8,39811033
answered Nov 16 at 23:03
Dave
8,39811033
answered Nov 16 at 23:03
Dave
8,39811033
answered Nov 16 at 23:03
Dave
8,39811033
8,39811033
add a comment |
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Hint for a shorter solution: it's sufficient to show $T$ has trivial kernel (why?). Now, if $p(0) = p'(0) = 0$ then $x^2 mid p$ and if $p(1) = p'(1) = 0$ then $(x-1)^2 mid p$.
– Daniel Schepler
Nov 16 at 23:05
You should explicitly state what $P_3(mathbb{R})$ is. It becomes clear if one looks at the attached photo, but a question should be asked in a form that its essence becomes transparent at first glance.
– Paul Frost
Nov 16 at 23:14