Why is the change of basis formula counter-intuitive? [See details] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)Change of basis = similarity?Change of Basis vs. Linear TransformationMatrices for change of basis linear transformationsConfusion about change of basis matrixIntuitive understanding of the $BAB^-1$ formula for changing basis in linear transformations.Standard Basis and Change of Basis MatrixStandard matrix linear transformation - change of basisHard change of basis/ linear transformation problemChange of basis difference between linear and bilinear transformationChange of basis of the kernel of a rectangular matrix
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Why is the change of basis formula counter-intuitive? [See details]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)Change of basis = similarity?Change of Basis vs. Linear TransformationMatrices for change of basis linear transformationsConfusion about change of basis matrixIntuitive understanding of the $BAB^-1$ formula for changing basis in linear transformations.Standard Basis and Change of Basis MatrixStandard matrix linear transformation - change of basisHard change of basis/ linear transformation problemChange of basis difference between linear and bilinear transformationChange of basis of the kernel of a rectangular matrix
$begingroup$
The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.
I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?
linear-algebra
edited 2 days ago
Carmeister
2,8792924
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
$endgroup$
add a comment |
$begingroup$
The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.
I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?
linear-algebra
edited 2 days ago
Carmeister
2,8792924
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
$endgroup$
1
$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
Apr 20 at 21:01
add a comment |
$begingroup$
The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.
I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?
linear-algebra
edited 2 days ago
Carmeister
2,8792924
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
$endgroup$
The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.
I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?
linear-algebra
linear-algebra
edited 2 days ago
Carmeister
2,8792924
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
edited 2 days ago
Carmeister
2,8792924
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
edited 2 days ago
Carmeister
2,8792924
edited 2 days ago
Carmeister
2,8792924
edited 2 days ago
Carmeister
2,8792924
2,8792924
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
asked Apr 20 at 20:31
Dr.StoneDr.Stone
877
877
1
$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
Apr 20 at 21:01
add a comment |
1
$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
Apr 20 at 21:01
1
1
$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
Apr 20 at 21:01
$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
Apr 20 at 21:01
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
$endgroup$
add a comment |
$begingroup$
Write $B=e_1,...,e_n, B' =e_1',...,e_n'$
If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.
So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula
answered Apr 20 at 20:37
MaxMax
16.8k11144
$endgroup$
add a comment |
$begingroup$
Remember that $T_B'$ is a function from $B'$ to $B'$.
Consider the following diagram:
The arrow $T_B'$ can be found by the following steps:
- Follow the arrow $P_B' to B$.
- Follow the arrow $T_B$.
- Follow the arrow $P_B to B'$.
Then $T_B'$ is simply the composition of these 3 arrows.
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
$endgroup$
add a comment |
$begingroup$
Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
$endgroup$
add a comment |
$begingroup$
Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
$endgroup$
Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
answered Apr 20 at 21:11
littleOlittleO
30.7k649111
30.7k649111
add a comment |
add a comment |
$begingroup$
Write $B=e_1,...,e_n, B' =e_1',...,e_n'$
If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.
So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula
answered Apr 20 at 20:37
MaxMax
16.8k11144
$endgroup$
add a comment |
$begingroup$
Write $B=e_1,...,e_n, B' =e_1',...,e_n'$
If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.
So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula
answered Apr 20 at 20:37
MaxMax
16.8k11144
$endgroup$
add a comment |
$begingroup$
Write $B=e_1,...,e_n, B' =e_1',...,e_n'$
If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.
So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula
answered Apr 20 at 20:37
MaxMax
16.8k11144
$endgroup$
Write $B=e_1,...,e_n, B' =e_1',...,e_n'$
If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.
So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula
answered Apr 20 at 20:37
MaxMax
16.8k11144
answered Apr 20 at 20:37
MaxMax
16.8k11144
answered Apr 20 at 20:37
MaxMax
16.8k11144
answered Apr 20 at 20:37
MaxMax
16.8k11144
16.8k11144
add a comment |
add a comment |
$begingroup$
Remember that $T_B'$ is a function from $B'$ to $B'$.
Consider the following diagram:
The arrow $T_B'$ can be found by the following steps:
- Follow the arrow $P_B' to B$.
- Follow the arrow $T_B$.
- Follow the arrow $P_B to B'$.
Then $T_B'$ is simply the composition of these 3 arrows.
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
$endgroup$
add a comment |
$begingroup$
Remember that $T_B'$ is a function from $B'$ to $B'$.
Consider the following diagram:
The arrow $T_B'$ can be found by the following steps:
- Follow the arrow $P_B' to B$.
- Follow the arrow $T_B$.
- Follow the arrow $P_B to B'$.
Then $T_B'$ is simply the composition of these 3 arrows.
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
$endgroup$
add a comment |
$begingroup$
Remember that $T_B'$ is a function from $B'$ to $B'$.
Consider the following diagram:
The arrow $T_B'$ can be found by the following steps:
- Follow the arrow $P_B' to B$.
- Follow the arrow $T_B$.
- Follow the arrow $P_B to B'$.
Then $T_B'$ is simply the composition of these 3 arrows.
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
$endgroup$
Remember that $T_B'$ is a function from $B'$ to $B'$.
Consider the following diagram:
The arrow $T_B'$ can be found by the following steps:
- Follow the arrow $P_B' to B$.
- Follow the arrow $T_B$.
- Follow the arrow $P_B to B'$.
Then $T_B'$ is simply the composition of these 3 arrows.
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
answered 2 days ago
Mateen UlhaqMateen Ulhaq
76821228
76821228
add a comment |
add a comment |
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$begingroup$
@littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
$endgroup$
– Dr.Stone
Apr 20 at 21:01