Let $B={{x^2, x, 1}}$ and $S= {{x^2+x, 2x-1, x+1}}$ be two basis of $P_2$.
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Let $B={{x^2, x, 1}}$ and $S= {{x^2+x, 2x-1, x+1}}$ be two basis of $P_2$. Let $i_b$ and $i_s$ be the coordinates maps induced on $P_2$ by these two basis. Let$T: P_2to P_2$ be a liner transformation such that
$[T]_{B,S}$=$begin{bmatrix}1 & 2 &0 \-1&3&5\2 & 2&-2 end{bmatrix}$
a) Prove that $ker(T)$=$i_b^{-1}(Null([T]_{B,S}))$ and $i_s(Im(T))=Col([T])_{B,S})$.
I know that $[T(x)]_B=[T]_{B,S}[x]_B$
but what is $i_B$ and what is its $Im$??
I am stuck with finding the $ker(T)$, I would be appreciate if someone can walk me through this problem.
linear-algebra linear-transformations change-of-basis invariant-subspace
asked Nov 19 at 18:35
Anqi Luo
495
add a comment |
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0
down vote
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Let $B={{x^2, x, 1}}$ and $S= {{x^2+x, 2x-1, x+1}}$ be two basis of $P_2$. Let $i_b$ and $i_s$ be the coordinates maps induced on $P_2$ by these two basis. Let$T: P_2to P_2$ be a liner transformation such that
$[T]_{B,S}$=$begin{bmatrix}1 & 2 &0 \-1&3&5\2 & 2&-2 end{bmatrix}$
a) Prove that $ker(T)$=$i_b^{-1}(Null([T]_{B,S}))$ and $i_s(Im(T))=Col([T])_{B,S})$.
I know that $[T(x)]_B=[T]_{B,S}[x]_B$
but what is $i_B$ and what is its $Im$??
I am stuck with finding the $ker(T)$, I would be appreciate if someone can walk me through this problem.
linear-algebra linear-transformations change-of-basis invariant-subspace
asked Nov 19 at 18:35
Anqi Luo
495
Im(T) is the image or Range of the transformation.And i think for Ker(T) you just need to solve $[T]_{B,S}.bar x=bar 0$.
– emonhossain
Nov 19 at 18:56
@emonhossain is it the same as $Null([T]_{B,S})$ ?
– Anqi Luo
Nov 19 at 19:03
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $B={{x^2, x, 1}}$ and $S= {{x^2+x, 2x-1, x+1}}$ be two basis of $P_2$. Let $i_b$ and $i_s$ be the coordinates maps induced on $P_2$ by these two basis. Let$T: P_2to P_2$ be a liner transformation such that
$[T]_{B,S}$=$begin{bmatrix}1 & 2 &0 \-1&3&5\2 & 2&-2 end{bmatrix}$
a) Prove that $ker(T)$=$i_b^{-1}(Null([T]_{B,S}))$ and $i_s(Im(T))=Col([T])_{B,S})$.
I know that $[T(x)]_B=[T]_{B,S}[x]_B$
but what is $i_B$ and what is its $Im$??
I am stuck with finding the $ker(T)$, I would be appreciate if someone can walk me through this problem.
linear-algebra linear-transformations change-of-basis invariant-subspace
asked Nov 19 at 18:35
Anqi Luo
495
Let $B={{x^2, x, 1}}$ and $S= {{x^2+x, 2x-1, x+1}}$ be two basis of $P_2$. Let $i_b$ and $i_s$ be the coordinates maps induced on $P_2$ by these two basis. Let$T: P_2to P_2$ be a liner transformation such that
$[T]_{B,S}$=$begin{bmatrix}1 & 2 &0 \-1&3&5\2 & 2&-2 end{bmatrix}$
a) Prove that $ker(T)$=$i_b^{-1}(Null([T]_{B,S}))$ and $i_s(Im(T))=Col([T])_{B,S})$.
I know that $[T(x)]_B=[T]_{B,S}[x]_B$
but what is $i_B$ and what is its $Im$??
I am stuck with finding the $ker(T)$, I would be appreciate if someone can walk me through this problem.
linear-algebra linear-transformations change-of-basis invariant-subspace
linear-algebra linear-transformations change-of-basis invariant-subspace
asked Nov 19 at 18:35
Anqi Luo
495
asked Nov 19 at 18:35
Anqi Luo
495
edited Nov 19 at 18:48
asked Nov 19 at 18:35
Anqi Luo
495
asked Nov 19 at 18:35
Anqi Luo
495
asked Nov 19 at 18:35
Anqi Luo
495
495
Im(T) is the image or Range of the transformation.And i think for Ker(T) you just need to solve $[T]_{B,S}.bar x=bar 0$.
– emonhossain
Nov 19 at 18:56
@emonhossain is it the same as $Null([T]_{B,S})$ ?
– Anqi Luo
Nov 19 at 19:03
add a comment |
Im(T) is the image or Range of the transformation.And i think for Ker(T) you just need to solve $[T]_{B,S}.bar x=bar 0$.
– emonhossain
Nov 19 at 18:56
@emonhossain is it the same as $Null([T]_{B,S})$ ?
– Anqi Luo
Nov 19 at 19:03
Im(T) is the image or Range of the transformation.And i think for Ker(T) you just need to solve $[T]_{B,S}.bar x=bar 0$.
– emonhossain
Nov 19 at 18:56
Im(T) is the image or Range of the transformation.And i think for Ker(T) you just need to solve $[T]_{B,S}.bar x=bar 0$.
– emonhossain
Nov 19 at 18:56
@emonhossain is it the same as $Null([T]_{B,S})$ ?
– Anqi Luo
Nov 19 at 19:03
@emonhossain is it the same as $Null([T]_{B,S})$ ?
– Anqi Luo
Nov 19 at 19:03
add a comment |
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Im(T) is the image or Range of the transformation.And i think for Ker(T) you just need to solve $[T]_{B,S}.bar x=bar 0$.
– emonhossain
Nov 19 at 18:56
@emonhossain is it the same as $Null([T]_{B,S})$ ?
– Anqi Luo
Nov 19 at 19:03