Computing the matrix $T$ using the basis $beta$
I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.
Thanks in advance for the help :)
linear-algebra
edited Nov 29 at 15:59
mike65535
110116
asked Nov 29 at 15:32
Vlad
34
add a comment |
I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.
Thanks in advance for the help :)
linear-algebra
edited Nov 29 at 15:59
mike65535
110116
asked Nov 29 at 15:32
Vlad
34
add a comment |
I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.
Thanks in advance for the help :)
linear-algebra
edited Nov 29 at 15:59
mike65535
110116
asked Nov 29 at 15:32
Vlad
34
I have $T(f(x)) = f(1) + f'(0)x + (f'(0)+f''(0))x^2$, and I need to compute the matrix of this linear transformation in base $beta$, which is the standard ordered basis of $P_2(R)$ (namely $(1,x,x^2)$).
I have tried doing it but I get the wrong coordinates and now I don't really know how to properly calculate them.
Thanks in advance for the help :)
linear-algebra
linear-algebra
edited Nov 29 at 15:59
mike65535
110116
asked Nov 29 at 15:32
Vlad
34
edited Nov 29 at 15:59
mike65535
110116
asked Nov 29 at 15:32
Vlad
34
edited Nov 29 at 15:59
mike65535
110116
edited Nov 29 at 15:59
mike65535
110116
edited Nov 29 at 15:59
mike65535
110116
110116
asked Nov 29 at 15:32
Vlad
34
asked Nov 29 at 15:32
Vlad
34
asked Nov 29 at 15:32
Vlad
34
34
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,
$A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.
answered Nov 29 at 16:05
Chris Custer
10.7k3724
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,
$A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.
answered Nov 29 at 16:05
Chris Custer
10.7k3724
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
add a comment |
Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,
$A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.
answered Nov 29 at 16:05
Chris Custer
10.7k3724
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
add a comment |
Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,
$A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.
answered Nov 29 at 16:05
Chris Custer
10.7k3724
Compute $T(1)=1$, $T(x)=1+x+x^2$ and $T(x^2)=2+2x^2$. These are the columns. That is,
$A=begin{pmatrix} 1&1&2\0&1&0\0&1&2end{pmatrix}$.
answered Nov 29 at 16:05
Chris Custer
10.7k3724
answered Nov 29 at 16:05
Chris Custer
10.7k3724
answered Nov 29 at 16:05
Chris Custer
10.7k3724
answered Nov 29 at 16:05
Chris Custer
10.7k3724
10.7k3724
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
add a comment |
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Could you show me how you get to those results? I only got it right for $T(1)$.
– Vlad
Nov 29 at 16:07
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Yeah. When $f(x)=x$, you need $f'(x)=1$ and $f''(x)=0$. Similarly $f(x)=x^2implies f'(x)=2x$ and $f''(x)=2$.
– Chris Custer
Nov 29 at 16:12
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
Oh, I get it now, thank you so much.
– Vlad
Nov 29 at 16:14
add a comment |
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