The Matrix of a reflection (around abitrary plane)
$begingroup$
Let $Upsilon :mathbb{R}^3rightarrow mathbb{R}^3$ be a reflection across the plane: $pi : -x + y + 2z = 0 $. Find the matrix of this linear transformation using the standard basis vectors and the matrix which is diagonal.
Now first of, If I have this plane then for $Upsilon(x,y,z) = (-x,y,2z)$ I get this when passing any vector, so the matrix using standard basis vectors is:
$$Y=begin{pmatrix}-1&0&0\ 0&1&0\ 0&0&-2end{pmatrix};.$$
I feel like this is totally wrong and what If I had like $ax+by+cz=0$ an arbitrary plane?
linear-algebra linear-transformations
edited Jul 28 '16 at 14:41
joriki
171k10188349
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
$endgroup$
add a comment |
$begingroup$
Let $Upsilon :mathbb{R}^3rightarrow mathbb{R}^3$ be a reflection across the plane: $pi : -x + y + 2z = 0 $. Find the matrix of this linear transformation using the standard basis vectors and the matrix which is diagonal.
Now first of, If I have this plane then for $Upsilon(x,y,z) = (-x,y,2z)$ I get this when passing any vector, so the matrix using standard basis vectors is:
$$Y=begin{pmatrix}-1&0&0\ 0&1&0\ 0&0&-2end{pmatrix};.$$
I feel like this is totally wrong and what If I had like $ax+by+cz=0$ an arbitrary plane?
linear-algebra linear-transformations
edited Jul 28 '16 at 14:41
joriki
171k10188349
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
$endgroup$
$begingroup$
The matrix for that reflection in the standard basis isn't diagonal. Where did you encounter this problem? The phrase "and the matrix which is diagonal" doesn't really fit with the rest of the sentence. Perhaps it's meant to refer to another matrix?
$endgroup$
– joriki
Jul 28 '16 at 14:43
1
$begingroup$
@joriki Maybe he is supposed to find both the matrix w.r.t. the standard basis and also to diagonalize it (a reflection should be diagonalizable).
$endgroup$
– Gregory Grant
Jul 28 '16 at 14:46
$begingroup$
@joriki I meant two different matrices, the one that I provided in the example and another one that is diagonal.. however I am not sure is the one I calculated correct? If it is from then on I can find a diagonal matrice that is similar to it.. or is there a better way?
$endgroup$
– gvidoje
Jul 28 '16 at 14:50
$begingroup$
Before we can say what's correct and what might be a better way, you'll need to sort out the problem statement. You haven't edited it yet, and it doesn't make sense the way it stands. Your comment also doesn't make it much clearer. Gregory's guess seems to be the best attempt so far to make sense of it all, but since you didn't respond to that we don't know whether he's right. Please edit the question to clarify it.
$endgroup$
– joriki
Jul 28 '16 at 15:13
add a comment |
$begingroup$
Let $Upsilon :mathbb{R}^3rightarrow mathbb{R}^3$ be a reflection across the plane: $pi : -x + y + 2z = 0 $. Find the matrix of this linear transformation using the standard basis vectors and the matrix which is diagonal.
Now first of, If I have this plane then for $Upsilon(x,y,z) = (-x,y,2z)$ I get this when passing any vector, so the matrix using standard basis vectors is:
$$Y=begin{pmatrix}-1&0&0\ 0&1&0\ 0&0&-2end{pmatrix};.$$
I feel like this is totally wrong and what If I had like $ax+by+cz=0$ an arbitrary plane?
linear-algebra linear-transformations
edited Jul 28 '16 at 14:41
joriki
171k10188349
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
$endgroup$
Let $Upsilon :mathbb{R}^3rightarrow mathbb{R}^3$ be a reflection across the plane: $pi : -x + y + 2z = 0 $. Find the matrix of this linear transformation using the standard basis vectors and the matrix which is diagonal.
Now first of, If I have this plane then for $Upsilon(x,y,z) = (-x,y,2z)$ I get this when passing any vector, so the matrix using standard basis vectors is:
$$Y=begin{pmatrix}-1&0&0\ 0&1&0\ 0&0&-2end{pmatrix};.$$
I feel like this is totally wrong and what If I had like $ax+by+cz=0$ an arbitrary plane?
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jul 28 '16 at 14:41
joriki
171k10188349
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
edited Jul 28 '16 at 14:41
joriki
171k10188349
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
edited Jul 28 '16 at 14:41
joriki
171k10188349
edited Jul 28 '16 at 14:41
joriki
171k10188349
edited Jul 28 '16 at 14:41
joriki
171k10188349
171k10188349
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
asked Jul 28 '16 at 14:37
gvidojegvidoje
398214
398214
$begingroup$
The matrix for that reflection in the standard basis isn't diagonal. Where did you encounter this problem? The phrase "and the matrix which is diagonal" doesn't really fit with the rest of the sentence. Perhaps it's meant to refer to another matrix?
$endgroup$
– joriki
Jul 28 '16 at 14:43
1
$begingroup$
@joriki Maybe he is supposed to find both the matrix w.r.t. the standard basis and also to diagonalize it (a reflection should be diagonalizable).
$endgroup$
– Gregory Grant
Jul 28 '16 at 14:46
$begingroup$
@joriki I meant two different matrices, the one that I provided in the example and another one that is diagonal.. however I am not sure is the one I calculated correct? If it is from then on I can find a diagonal matrice that is similar to it.. or is there a better way?
$endgroup$
– gvidoje
Jul 28 '16 at 14:50
$begingroup$
Before we can say what's correct and what might be a better way, you'll need to sort out the problem statement. You haven't edited it yet, and it doesn't make sense the way it stands. Your comment also doesn't make it much clearer. Gregory's guess seems to be the best attempt so far to make sense of it all, but since you didn't respond to that we don't know whether he's right. Please edit the question to clarify it.
$endgroup$
– joriki
Jul 28 '16 at 15:13
add a comment |
$begingroup$
The matrix for that reflection in the standard basis isn't diagonal. Where did you encounter this problem? The phrase "and the matrix which is diagonal" doesn't really fit with the rest of the sentence. Perhaps it's meant to refer to another matrix?
$endgroup$
– joriki
Jul 28 '16 at 14:43
1
$begingroup$
@joriki Maybe he is supposed to find both the matrix w.r.t. the standard basis and also to diagonalize it (a reflection should be diagonalizable).
$endgroup$
– Gregory Grant
Jul 28 '16 at 14:46
$begingroup$
@joriki I meant two different matrices, the one that I provided in the example and another one that is diagonal.. however I am not sure is the one I calculated correct? If it is from then on I can find a diagonal matrice that is similar to it.. or is there a better way?
$endgroup$
– gvidoje
Jul 28 '16 at 14:50
$begingroup$
Before we can say what's correct and what might be a better way, you'll need to sort out the problem statement. You haven't edited it yet, and it doesn't make sense the way it stands. Your comment also doesn't make it much clearer. Gregory's guess seems to be the best attempt so far to make sense of it all, but since you didn't respond to that we don't know whether he's right. Please edit the question to clarify it.
$endgroup$
– joriki
Jul 28 '16 at 15:13
$begingroup$
The matrix for that reflection in the standard basis isn't diagonal. Where did you encounter this problem? The phrase "and the matrix which is diagonal" doesn't really fit with the rest of the sentence. Perhaps it's meant to refer to another matrix?
$endgroup$
– joriki
Jul 28 '16 at 14:43
$begingroup$
The matrix for that reflection in the standard basis isn't diagonal. Where did you encounter this problem? The phrase "and the matrix which is diagonal" doesn't really fit with the rest of the sentence. Perhaps it's meant to refer to another matrix?
$endgroup$
– joriki
Jul 28 '16 at 14:43
1
1
$begingroup$
@joriki Maybe he is supposed to find both the matrix w.r.t. the standard basis and also to diagonalize it (a reflection should be diagonalizable).
$endgroup$
– Gregory Grant
Jul 28 '16 at 14:46
$begingroup$
@joriki Maybe he is supposed to find both the matrix w.r.t. the standard basis and also to diagonalize it (a reflection should be diagonalizable).
$endgroup$
– Gregory Grant
Jul 28 '16 at 14:46
$begingroup$
@joriki I meant two different matrices, the one that I provided in the example and another one that is diagonal.. however I am not sure is the one I calculated correct? If it is from then on I can find a diagonal matrice that is similar to it.. or is there a better way?
$endgroup$
– gvidoje
Jul 28 '16 at 14:50
$begingroup$
@joriki I meant two different matrices, the one that I provided in the example and another one that is diagonal.. however I am not sure is the one I calculated correct? If it is from then on I can find a diagonal matrice that is similar to it.. or is there a better way?
$endgroup$
– gvidoje
Jul 28 '16 at 14:50
$begingroup$
Before we can say what's correct and what might be a better way, you'll need to sort out the problem statement. You haven't edited it yet, and it doesn't make sense the way it stands. Your comment also doesn't make it much clearer. Gregory's guess seems to be the best attempt so far to make sense of it all, but since you didn't respond to that we don't know whether he's right. Please edit the question to clarify it.
$endgroup$
– joriki
Jul 28 '16 at 15:13
$begingroup$
Before we can say what's correct and what might be a better way, you'll need to sort out the problem statement. You haven't edited it yet, and it doesn't make sense the way it stands. Your comment also doesn't make it much clearer. Gregory's guess seems to be the best attempt so far to make sense of it all, but since you didn't respond to that we don't know whether he's right. Please edit the question to clarify it.
$endgroup$
– joriki
Jul 28 '16 at 15:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $n$ is the plane's unit normal vector (here $(-1,1,2)/sqrt6$), then the reflection of any vector $p$ across the plane is given by
$$p-2langle n,prangle n.$$
Now plug in the vectors of the standard basis to get the column vectors of the desired matrix.
Facit: avoid coordinates.
Edit: if $n$ is not a unit vector we have
$$p-frac{2langle n,prangle}{|n|^2}n.$$
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
$endgroup$
add a comment |
$begingroup$
I’m not quite sure what you’re asking, but here’s a way to construct the matrix of the reflection via a diagonal matrix:
A reflection of a vector across a plane (more generally, across an $(n-1)$-dimensional subspace of an $n$-dimensional space) reverses the component of the vector that’s orthogonal to the plane and leaves fixed its component in the plane. Another way of putting this is that the reflection is the identity on vectors in the plane and multiplication by $-1$ on vectors orthogonal to it. So, take a basis for the plane and extend it by adding a vector normal to it. Relative to this basis, the matrix of the reflection is simply $$Y=pmatrix{1&0&0\0&1&0\0&0&-1}$$ from which you can get $Upsilon$ via a change of basis: $BYB^{-1}$. Note that right-multiplication by $Y$ just changes the sign of the third column of a matrix.
In your case, the equation of the plane immediately gives us the normal $(-1,1,2)^T$. A basis for the plane is easily generated from this vector: ${(1,1,0)^T,(2,0,1)^T}$. Putting this all together gives $$Upsilon=pmatrix{1&2&1\1&0&-1\0&1&-2}pmatrix{1&2&-1\1&0&1\0&1&2}^{-1}=pmatrix{frac23&frac13&frac23\frac13&frac23&-frac23\frac23&-frac23&-frac13}.$$
The construction in Michael Hoppe’s answer is easier to calculate, though, especially in higher-dimensional spaces.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $n$ is the plane's unit normal vector (here $(-1,1,2)/sqrt6$), then the reflection of any vector $p$ across the plane is given by
$$p-2langle n,prangle n.$$
Now plug in the vectors of the standard basis to get the column vectors of the desired matrix.
Facit: avoid coordinates.
Edit: if $n$ is not a unit vector we have
$$p-frac{2langle n,prangle}{|n|^2}n.$$
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
$endgroup$
add a comment |
$begingroup$
If $n$ is the plane's unit normal vector (here $(-1,1,2)/sqrt6$), then the reflection of any vector $p$ across the plane is given by
$$p-2langle n,prangle n.$$
Now plug in the vectors of the standard basis to get the column vectors of the desired matrix.
Facit: avoid coordinates.
Edit: if $n$ is not a unit vector we have
$$p-frac{2langle n,prangle}{|n|^2}n.$$
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
$endgroup$
add a comment |
$begingroup$
If $n$ is the plane's unit normal vector (here $(-1,1,2)/sqrt6$), then the reflection of any vector $p$ across the plane is given by
$$p-2langle n,prangle n.$$
Now plug in the vectors of the standard basis to get the column vectors of the desired matrix.
Facit: avoid coordinates.
Edit: if $n$ is not a unit vector we have
$$p-frac{2langle n,prangle}{|n|^2}n.$$
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
$endgroup$
If $n$ is the plane's unit normal vector (here $(-1,1,2)/sqrt6$), then the reflection of any vector $p$ across the plane is given by
$$p-2langle n,prangle n.$$
Now plug in the vectors of the standard basis to get the column vectors of the desired matrix.
Facit: avoid coordinates.
Edit: if $n$ is not a unit vector we have
$$p-frac{2langle n,prangle}{|n|^2}n.$$
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
answered Jul 28 '16 at 16:37
Michael HoppeMichael Hoppe
11.1k31836
11.1k31836
add a comment |
add a comment |
$begingroup$
I’m not quite sure what you’re asking, but here’s a way to construct the matrix of the reflection via a diagonal matrix:
A reflection of a vector across a plane (more generally, across an $(n-1)$-dimensional subspace of an $n$-dimensional space) reverses the component of the vector that’s orthogonal to the plane and leaves fixed its component in the plane. Another way of putting this is that the reflection is the identity on vectors in the plane and multiplication by $-1$ on vectors orthogonal to it. So, take a basis for the plane and extend it by adding a vector normal to it. Relative to this basis, the matrix of the reflection is simply $$Y=pmatrix{1&0&0\0&1&0\0&0&-1}$$ from which you can get $Upsilon$ via a change of basis: $BYB^{-1}$. Note that right-multiplication by $Y$ just changes the sign of the third column of a matrix.
In your case, the equation of the plane immediately gives us the normal $(-1,1,2)^T$. A basis for the plane is easily generated from this vector: ${(1,1,0)^T,(2,0,1)^T}$. Putting this all together gives $$Upsilon=pmatrix{1&2&1\1&0&-1\0&1&-2}pmatrix{1&2&-1\1&0&1\0&1&2}^{-1}=pmatrix{frac23&frac13&frac23\frac13&frac23&-frac23\frac23&-frac23&-frac13}.$$
The construction in Michael Hoppe’s answer is easier to calculate, though, especially in higher-dimensional spaces.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
I’m not quite sure what you’re asking, but here’s a way to construct the matrix of the reflection via a diagonal matrix:
A reflection of a vector across a plane (more generally, across an $(n-1)$-dimensional subspace of an $n$-dimensional space) reverses the component of the vector that’s orthogonal to the plane and leaves fixed its component in the plane. Another way of putting this is that the reflection is the identity on vectors in the plane and multiplication by $-1$ on vectors orthogonal to it. So, take a basis for the plane and extend it by adding a vector normal to it. Relative to this basis, the matrix of the reflection is simply $$Y=pmatrix{1&0&0\0&1&0\0&0&-1}$$ from which you can get $Upsilon$ via a change of basis: $BYB^{-1}$. Note that right-multiplication by $Y$ just changes the sign of the third column of a matrix.
In your case, the equation of the plane immediately gives us the normal $(-1,1,2)^T$. A basis for the plane is easily generated from this vector: ${(1,1,0)^T,(2,0,1)^T}$. Putting this all together gives $$Upsilon=pmatrix{1&2&1\1&0&-1\0&1&-2}pmatrix{1&2&-1\1&0&1\0&1&2}^{-1}=pmatrix{frac23&frac13&frac23\frac13&frac23&-frac23\frac23&-frac23&-frac13}.$$
The construction in Michael Hoppe’s answer is easier to calculate, though, especially in higher-dimensional spaces.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
I’m not quite sure what you’re asking, but here’s a way to construct the matrix of the reflection via a diagonal matrix:
A reflection of a vector across a plane (more generally, across an $(n-1)$-dimensional subspace of an $n$-dimensional space) reverses the component of the vector that’s orthogonal to the plane and leaves fixed its component in the plane. Another way of putting this is that the reflection is the identity on vectors in the plane and multiplication by $-1$ on vectors orthogonal to it. So, take a basis for the plane and extend it by adding a vector normal to it. Relative to this basis, the matrix of the reflection is simply $$Y=pmatrix{1&0&0\0&1&0\0&0&-1}$$ from which you can get $Upsilon$ via a change of basis: $BYB^{-1}$. Note that right-multiplication by $Y$ just changes the sign of the third column of a matrix.
In your case, the equation of the plane immediately gives us the normal $(-1,1,2)^T$. A basis for the plane is easily generated from this vector: ${(1,1,0)^T,(2,0,1)^T}$. Putting this all together gives $$Upsilon=pmatrix{1&2&1\1&0&-1\0&1&-2}pmatrix{1&2&-1\1&0&1\0&1&2}^{-1}=pmatrix{frac23&frac13&frac23\frac13&frac23&-frac23\frac23&-frac23&-frac13}.$$
The construction in Michael Hoppe’s answer is easier to calculate, though, especially in higher-dimensional spaces.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
$endgroup$
I’m not quite sure what you’re asking, but here’s a way to construct the matrix of the reflection via a diagonal matrix:
A reflection of a vector across a plane (more generally, across an $(n-1)$-dimensional subspace of an $n$-dimensional space) reverses the component of the vector that’s orthogonal to the plane and leaves fixed its component in the plane. Another way of putting this is that the reflection is the identity on vectors in the plane and multiplication by $-1$ on vectors orthogonal to it. So, take a basis for the plane and extend it by adding a vector normal to it. Relative to this basis, the matrix of the reflection is simply $$Y=pmatrix{1&0&0\0&1&0\0&0&-1}$$ from which you can get $Upsilon$ via a change of basis: $BYB^{-1}$. Note that right-multiplication by $Y$ just changes the sign of the third column of a matrix.
In your case, the equation of the plane immediately gives us the normal $(-1,1,2)^T$. A basis for the plane is easily generated from this vector: ${(1,1,0)^T,(2,0,1)^T}$. Putting this all together gives $$Upsilon=pmatrix{1&2&1\1&0&-1\0&1&-2}pmatrix{1&2&-1\1&0&1\0&1&2}^{-1}=pmatrix{frac23&frac13&frac23\frac13&frac23&-frac23\frac23&-frac23&-frac13}.$$
The construction in Michael Hoppe’s answer is easier to calculate, though, especially in higher-dimensional spaces.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
answered Jul 28 '16 at 18:08
amdamd
30.4k21050
30.4k21050
add a comment |
add a comment |
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$begingroup$
The matrix for that reflection in the standard basis isn't diagonal. Where did you encounter this problem? The phrase "and the matrix which is diagonal" doesn't really fit with the rest of the sentence. Perhaps it's meant to refer to another matrix?
$endgroup$
– joriki
Jul 28 '16 at 14:43
1
$begingroup$
@joriki Maybe he is supposed to find both the matrix w.r.t. the standard basis and also to diagonalize it (a reflection should be diagonalizable).
$endgroup$
– Gregory Grant
Jul 28 '16 at 14:46
$begingroup$
@joriki I meant two different matrices, the one that I provided in the example and another one that is diagonal.. however I am not sure is the one I calculated correct? If it is from then on I can find a diagonal matrice that is similar to it.. or is there a better way?
$endgroup$
– gvidoje
Jul 28 '16 at 14:50
$begingroup$
Before we can say what's correct and what might be a better way, you'll need to sort out the problem statement. You haven't edited it yet, and it doesn't make sense the way it stands. Your comment also doesn't make it much clearer. Gregory's guess seems to be the best attempt so far to make sense of it all, but since you didn't respond to that we don't know whether he's right. Please edit the question to clarify it.
$endgroup$
– joriki
Jul 28 '16 at 15:13