Calculating matrix for linear transformation of orthogonal projection onto plane.
up vote
0
down vote
favorite
The question goes like this:
"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane
$$ 2x_1+2x_2+x_3^{}= 0" $$
So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:
$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$
That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$
Basically, what I will do is set up an equation
$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$
So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?
Thanks in advance.
linear-algebra linear-transformations
asked Nov 21 at 15:27
Synchrowave
375
add a comment |
up vote
0
down vote
favorite
The question goes like this:
"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane
$$ 2x_1+2x_2+x_3^{}= 0" $$
So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:
$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$
That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$
Basically, what I will do is set up an equation
$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$
So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?
Thanks in advance.
linear-algebra linear-transformations
asked Nov 21 at 15:27
Synchrowave
375
Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30
Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32
There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35
I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39
So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question goes like this:
"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane
$$ 2x_1+2x_2+x_3^{}= 0" $$
So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:
$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$
That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$
Basically, what I will do is set up an equation
$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$
So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?
Thanks in advance.
linear-algebra linear-transformations
asked Nov 21 at 15:27
Synchrowave
375
The question goes like this:
"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane
$$ 2x_1+2x_2+x_3^{}= 0" $$
So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:
$$ left[
begin{array}{cc|c}
2\
2\
1
end{array}
right] $$
That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
|e_1| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right], |e_2| =left[
begin{array}{cc|c}
0\
1\
0
end{array}
right], |e_3| =left[
begin{array}{cc|c}
0\
0\
1
end{array}
right]$$
Basically, what I will do is set up an equation
$$ Proj V_n + |n| = left[
begin{array}{cc|c}
1\
0\
0
end{array}
right]$$
So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?
Thanks in advance.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Nov 21 at 15:27
Synchrowave
375
asked Nov 21 at 15:27
Synchrowave
375
edited Nov 21 at 16:09
asked Nov 21 at 15:27
Synchrowave
375
asked Nov 21 at 15:27
Synchrowave
375
asked Nov 21 at 15:27
Synchrowave
375
375
Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30
Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32
There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35
I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39
So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59
add a comment |
Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30
Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32
There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35
I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39
So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59
Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30
Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30
Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32
Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32
There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35
There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35
I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39
I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39
So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59
So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.
The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.
answered Nov 21 at 21:26
amd
29k21050
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
add a comment |
up vote
0
down vote
Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.
So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that
$P(2,2,1)=(0,0,0)$;
$P(1,0,-2)=(1,0,-2)$;
$P(0,1,-2)=(0,1,-2)$.
A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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up vote
1
down vote
Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.
The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.
answered Nov 21 at 21:26
amd
29k21050
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
add a comment |
up vote
1
down vote
Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.
The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.
answered Nov 21 at 21:26
amd
29k21050
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
add a comment |
up vote
1
down vote
up vote
1
down vote
Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.
The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.
answered Nov 21 at 21:26
amd
29k21050
Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $mathbf e_1$ is equal to $mathbf e_1-mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.
The problem is that you’ve set the rejection of $mathbf e_1$ from the plane to be equal to $mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $Pmathbf e_1$ of $mathbf e_1$ onto the plane is $mathbf e_1-kmathbf n$ for some as-yet-undetermined scalar $k$. However, $kmathbf n$ here is simply the orthogonal projection of $mathbf e_1$ onto $mathbf n$, which I suspect that you know how to compute.
answered Nov 21 at 21:26
amd
29k21050
answered Nov 21 at 21:26
amd
29k21050
answered Nov 21 at 21:26
amd
29k21050
answered Nov 21 at 21:26
amd
29k21050
29k21050
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
add a comment |
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$ e_1 $$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane.
– Synchrowave
Nov 22 at 15:28
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
@Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22 at 19:25
add a comment |
up vote
0
down vote
Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.
So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that
$P(2,2,1)=(0,0,0)$;
$P(1,0,-2)=(1,0,-2)$;
$P(0,1,-2)=(0,1,-2)$.
A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
add a comment |
up vote
0
down vote
Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.
So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that
$P(2,2,1)=(0,0,0)$;
$P(1,0,-2)=(1,0,-2)$;
$P(0,1,-2)=(0,1,-2)$.
A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.
So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that
$P(2,2,1)=(0,0,0)$;
$P(1,0,-2)=(1,0,-2)$;
$P(0,1,-2)=(0,1,-2)$.
A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.
So, take the only linear map $Pcolonmathbb{R}^3longrightarrowmathbb{R}^3$ such that
$P(2,2,1)=(0,0,0)$;
$P(1,0,-2)=(1,0,-2)$;
$P(0,1,-2)=(0,1,-2)$.
A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$frac19begin{bmatrix}5 & -4 & -2 \ -4 & 5 & -2 \ -2 & -2 & 8end{bmatrix}.$$
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
edited Nov 21 at 16:24
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
answered Nov 21 at 16:11
José Carlos Santos
147k22117218
147k22117218
add a comment |
add a comment |
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Do you mean orthogonal projection?
– José Carlos Santos
Nov 21 at 15:30
Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane"
– Synchrowave
Nov 21 at 15:32
There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question.
– José Carlos Santos
Nov 21 at 15:35
I would say that they look for the projection that is parallell to the plane.
– Synchrowave
Nov 21 at 15:39
So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up!
– Synchrowave
Nov 21 at 15:59