Expressing a linear map $phi: >mathbb R^4tomathbb R^4$ in terms of a new base
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In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $phi: >mathbb R^4tomathbb R^4$ appears as
$$phi(x,y,z,t) = (t,x,y,z) .$$ Now we are given a new base $$R = (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) $$ of $mathbbR^4$ and should find the matrix $M_R(phi)$.
Is my thinking correct? As I understand I need to find the matrix created by the linear transormation:
$$
left( beginmatrix
0 & 0 & 0 & 1\
1 & 0 & 0 & 0\
0& 1& 0 & 0\
0& 0& 1 & 0\
endmatrix right)
$$
then, solve
$$ left[
beginarraycccc
1&0&0&0&a\
0&1&1&1&b\
1&0&1&1&c\
0&1&0&1&d\
endarray
right] $$
by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?
matrices linear-transformations
edited Sep 3 at 12:38
Christian Blatter
166k7110312
asked Sep 3 at 9:01
Puf Iniema
165
add a comment |Â
up vote
2
down vote
favorite
In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $phi: >mathbb R^4tomathbb R^4$ appears as
$$phi(x,y,z,t) = (t,x,y,z) .$$ Now we are given a new base $$R = (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) $$ of $mathbbR^4$ and should find the matrix $M_R(phi)$.
Is my thinking correct? As I understand I need to find the matrix created by the linear transormation:
$$
left( beginmatrix
0 & 0 & 0 & 1\
1 & 0 & 0 & 0\
0& 1& 0 & 0\
0& 0& 1 & 0\
endmatrix right)
$$
then, solve
$$ left[
beginarraycccc
1&0&0&0&a\
0&1&1&1&b\
1&0&1&1&c\
0&1&0&1&d\
endarray
right] $$
by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?
matrices linear-transformations
edited Sep 3 at 12:38
Christian Blatter
166k7110312
asked Sep 3 at 9:01
Puf Iniema
165
Is the third vector in $R$ supposed to be $(1,1,1,0)$?
– Arthur
Sep 3 at 9:11
Just to clarify, are you proposing to replace $(a,b,c,d)$ with the each of the results of applying $phi$ to each of the elements of $R$ In turn?
– amd
Sep 3 at 21:29
@amd yes, that was my initial idea
– Puf Iniema
Sep 3 at 22:55
Good. That’s essentially what Arthur’s answer describes.
– amd
Sep 3 at 23:17
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $phi: >mathbb R^4tomathbb R^4$ appears as
$$phi(x,y,z,t) = (t,x,y,z) .$$ Now we are given a new base $$R = (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) $$ of $mathbbR^4$ and should find the matrix $M_R(phi)$.
Is my thinking correct? As I understand I need to find the matrix created by the linear transormation:
$$
left( beginmatrix
0 & 0 & 0 & 1\
1 & 0 & 0 & 0\
0& 1& 0 & 0\
0& 0& 1 & 0\
endmatrix right)
$$
then, solve
$$ left[
beginarraycccc
1&0&0&0&a\
0&1&1&1&b\
1&0&1&1&c\
0&1&0&1&d\
endarray
right] $$
by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?
matrices linear-transformations
edited Sep 3 at 12:38
Christian Blatter
166k7110312
asked Sep 3 at 9:01
Puf Iniema
165
In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $phi: >mathbb R^4tomathbb R^4$ appears as
$$phi(x,y,z,t) = (t,x,y,z) .$$ Now we are given a new base $$R = (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) $$ of $mathbbR^4$ and should find the matrix $M_R(phi)$.
Is my thinking correct? As I understand I need to find the matrix created by the linear transormation:
$$
left( beginmatrix
0 & 0 & 0 & 1\
1 & 0 & 0 & 0\
0& 1& 0 & 0\
0& 0& 1 & 0\
endmatrix right)
$$
then, solve
$$ left[
beginarraycccc
1&0&0&0&a\
0&1&1&1&b\
1&0&1&1&c\
0&1&0&1&d\
endarray
right] $$
by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?
matrices linear-transformations
matrices linear-transformations
edited Sep 3 at 12:38
Christian Blatter
166k7110312
asked Sep 3 at 9:01
Puf Iniema
165
edited Sep 3 at 12:38
Christian Blatter
166k7110312
asked Sep 3 at 9:01
Puf Iniema
165
edited Sep 3 at 12:38
Christian Blatter
166k7110312
edited Sep 3 at 12:38
Christian Blatter
166k7110312
edited Sep 3 at 12:38
Christian Blatter
166k7110312
166k7110312
asked Sep 3 at 9:01
Puf Iniema
165
asked Sep 3 at 9:01
Puf Iniema
165
asked Sep 3 at 9:01
Puf Iniema
165
165
Is the third vector in $R$ supposed to be $(1,1,1,0)$?
– Arthur
Sep 3 at 9:11
Just to clarify, are you proposing to replace $(a,b,c,d)$ with the each of the results of applying $phi$ to each of the elements of $R$ In turn?
– amd
Sep 3 at 21:29
@amd yes, that was my initial idea
– Puf Iniema
Sep 3 at 22:55
Good. That’s essentially what Arthur’s answer describes.
– amd
Sep 3 at 23:17
add a comment |Â
Is the third vector in $R$ supposed to be $(1,1,1,0)$?
– Arthur
Sep 3 at 9:11
Just to clarify, are you proposing to replace $(a,b,c,d)$ with the each of the results of applying $phi$ to each of the elements of $R$ In turn?
– amd
Sep 3 at 21:29
@amd yes, that was my initial idea
– Puf Iniema
Sep 3 at 22:55
Good. That’s essentially what Arthur’s answer describes.
– amd
Sep 3 at 23:17
Is the third vector in $R$ supposed to be $(1,1,1,0)$?
– Arthur
Sep 3 at 9:11
Is the third vector in $R$ supposed to be $(1,1,1,0)$?
– Arthur
Sep 3 at 9:11
Just to clarify, are you proposing to replace $(a,b,c,d)$ with the each of the results of applying $phi$ to each of the elements of $R$ In turn?
– amd
Sep 3 at 21:29
Just to clarify, are you proposing to replace $(a,b,c,d)$ with the each of the results of applying $phi$ to each of the elements of $R$ In turn?
– amd
Sep 3 at 21:29
@amd yes, that was my initial idea
– Puf Iniema
Sep 3 at 22:55
@amd yes, that was my initial idea
– Puf Iniema
Sep 3 at 22:55
Good. That’s essentially what Arthur’s answer describes.
– amd
Sep 3 at 23:17
Good. That’s essentially what Arthur’s answer describes.
– amd
Sep 3 at 23:17
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
The most straight-forward way (as I see it) would be to follow this:
The columns of the matrix representation of a linear transformation are the images of the basis vectors.
So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(phi)$ is $phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.
answered Sep 3 at 9:15
Arthur
102k795176
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
1
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
 |Â
show 1 more comment
up vote
1
down vote
The matrix $P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(phi) = P^-1M_E(phi) P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix^-1
beginbmatrix
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
endbmatrixbeginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix = beginbmatrix
0 & 1 & 0 & 1 \
1 & 0 & -1 & 0 \
0 & 0 & -1 & -1 \
0 & 0 & 2 & 1 \
endbmatrix$$
answered Sep 3 at 12:54
mechanodroid
24.3k62245
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The most straight-forward way (as I see it) would be to follow this:
The columns of the matrix representation of a linear transformation are the images of the basis vectors.
So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(phi)$ is $phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.
answered Sep 3 at 9:15
Arthur
102k795176
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
1
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
 |Â
show 1 more comment
up vote
2
down vote
The most straight-forward way (as I see it) would be to follow this:
The columns of the matrix representation of a linear transformation are the images of the basis vectors.
So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(phi)$ is $phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.
answered Sep 3 at 9:15
Arthur
102k795176
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
1
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
The most straight-forward way (as I see it) would be to follow this:
The columns of the matrix representation of a linear transformation are the images of the basis vectors.
So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(phi)$ is $phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.
answered Sep 3 at 9:15
Arthur
102k795176
The most straight-forward way (as I see it) would be to follow this:
The columns of the matrix representation of a linear transformation are the images of the basis vectors.
So, since the first basis vector in $R$ is $(1,0,1,0)$, the first column of $M_R(phi)$ is $phi(1, 0, 1, 0) = (0,1,0,1)$, but expressed using the basis $R$ rather than the standard basis. And so on.
answered Sep 3 at 9:15
Arthur
102k795176
answered Sep 3 at 9:15
Arthur
102k795176
answered Sep 3 at 9:15
Arthur
102k795176
answered Sep 3 at 9:15
Arthur
102k795176
102k795176
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
1
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
 |Â
show 1 more comment
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
1
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
So, the final matrix is this?: beginmatrix 0 & 1& 0 & 1\ 1 & 0 & 0 & 0\ 0& 1& 1 & 1\ 1& 0& 1 & 1\ endmatrix
– Puf Iniema
Sep 3 at 9:29
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
@PufIniema Almost. But you have expressed the vectors in the standard basis. You have to express them in the basis $R$. So the first column should be $(0,1,0,0)^T$ because that's how you express $(0,1,0,1)_textstandard$ in the basis $R$.
– Arthur
Sep 3 at 10:07
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
I'm not able to post an image as a comment, so could you plase take a look at what I've written?
– Puf Iniema
Sep 3 at 12:11
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
@PufIniema The first two columns are correct. The third and fourth are not. It is not the case that $(0,0,1,1)$ (the standard representation of $phi(R_3)$) is equal to $1cdot (0,1,1,0) + 2cdot(0,1,1,1)$ (which is the standard representation of $(0,0,1,2)_R$). Instead, we have $(0,0,1,1) = (-1)cdot (0,1,0,1) + (-1)cdot(0,1,1,0) + 2cdot(0,1,1,1)$, meaning that the $R$-representation of $(0,0,1,1)$ is $(0,-1, -1, 2)_R$, so that is what the third column should be (transposed, of course). Smimilarily, the fourth column should be $(1, 0, -1, 1)^T$.
– Arthur
Sep 3 at 12:36
1
1
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
I found a basic arithmetic mistake in my calculations. After correcting it I got the same answer as you. Thank you for your time.
– Puf Iniema
Sep 3 at 13:29
 |Â
show 1 more comment
up vote
1
down vote
The matrix $P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(phi) = P^-1M_E(phi) P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix^-1
beginbmatrix
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
endbmatrixbeginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix = beginbmatrix
0 & 1 & 0 & 1 \
1 & 0 & -1 & 0 \
0 & 0 & -1 & -1 \
0 & 0 & 2 & 1 \
endbmatrix$$
answered Sep 3 at 12:54
mechanodroid
24.3k62245
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
add a comment |Â
up vote
1
down vote
The matrix $P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(phi) = P^-1M_E(phi) P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix^-1
beginbmatrix
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
endbmatrixbeginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix = beginbmatrix
0 & 1 & 0 & 1 \
1 & 0 & -1 & 0 \
0 & 0 & -1 & -1 \
0 & 0 & 2 & 1 \
endbmatrix$$
answered Sep 3 at 12:54
mechanodroid
24.3k62245
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The matrix $P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(phi) = P^-1M_E(phi) P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix^-1
beginbmatrix
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
endbmatrixbeginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix = beginbmatrix
0 & 1 & 0 & 1 \
1 & 0 & -1 & 0 \
0 & 0 & -1 & -1 \
0 & 0 & 2 & 1 \
endbmatrix$$
answered Sep 3 at 12:54
mechanodroid
24.3k62245
The matrix $P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(phi) = P^-1M_E(phi) P = beginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix^-1
beginbmatrix
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
endbmatrixbeginbmatrix
1 & 0 & 0 & 0 \
0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
0 & 1 & 0 & 1 \
endbmatrix = beginbmatrix
0 & 1 & 0 & 1 \
1 & 0 & -1 & 0 \
0 & 0 & -1 & -1 \
0 & 0 & 2 & 1 \
endbmatrix$$
answered Sep 3 at 12:54
mechanodroid
24.3k62245
answered Sep 3 at 12:54
mechanodroid
24.3k62245
answered Sep 3 at 12:54
mechanodroid
24.3k62245
answered Sep 3 at 12:54
mechanodroid
24.3k62245
24.3k62245
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
add a comment |Â
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
Thank you so much!
– Puf Iniema
Sep 3 at 13:12
add a comment |Â
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Is the third vector in $R$ supposed to be $(1,1,1,0)$?
– Arthur
Sep 3 at 9:11
Just to clarify, are you proposing to replace $(a,b,c,d)$ with the each of the results of applying $phi$ to each of the elements of $R$ In turn?
– amd
Sep 3 at 21:29
@amd yes, that was my initial idea
– Puf Iniema
Sep 3 at 22:55
Good. That’s essentially what Arthur’s answer describes.
– amd
Sep 3 at 23:17