Ways to find the orthogonal projection matrix
$begingroup$
I'm a bit lost trying to find the projection matrix for an orthogonal projection onto a plane defined by the normal vector $n = (1, 1, 1)^T$. Then I can find the basis C of the plain $C = ( (-1,0,1)^T (0,-1,1)^T)$.
Now i should be able to find the projection Matrix with $A(A^TA)^{-1}A^T$
Where $A:=begin{bmatrix}
-1 & 0\
0 & -1\
1 & 1end{bmatrix}$. Then my the projection matrix will look like this?
$A:=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$ Is this correct?
To which basis is this projection matrix? How can I change the matrix to a different basis?
There should be another way to find the matrix. Something like add to my basis $C$ a vector from my basis $B$ (which should not be the standard basis) in $mathbb{R^3}$, find the projection of the basis ( I only need to do this for the added basis vector from $B$ since the rest is already on the plane).
But how can I find the projection of the added basis vector?
After that what would be the matrix from basis B to B? The coefficients of the linear combinations $c_1,c_2,P(b_i)$ in B?
linear-algebra projection-matrices
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
$endgroup$
add a comment |
$begingroup$
I'm a bit lost trying to find the projection matrix for an orthogonal projection onto a plane defined by the normal vector $n = (1, 1, 1)^T$. Then I can find the basis C of the plain $C = ( (-1,0,1)^T (0,-1,1)^T)$.
Now i should be able to find the projection Matrix with $A(A^TA)^{-1}A^T$
Where $A:=begin{bmatrix}
-1 & 0\
0 & -1\
1 & 1end{bmatrix}$. Then my the projection matrix will look like this?
$A:=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$ Is this correct?
To which basis is this projection matrix? How can I change the matrix to a different basis?
There should be another way to find the matrix. Something like add to my basis $C$ a vector from my basis $B$ (which should not be the standard basis) in $mathbb{R^3}$, find the projection of the basis ( I only need to do this for the added basis vector from $B$ since the rest is already on the plane).
But how can I find the projection of the added basis vector?
After that what would be the matrix from basis B to B? The coefficients of the linear combinations $c_1,c_2,P(b_i)$ in B?
linear-algebra projection-matrices
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
$endgroup$
$begingroup$
Is clear to you the way to obtain the projection matrix?
$endgroup$
– gimusi
Dec 17 '17 at 13:36
$begingroup$
Not quite, I did it for a couple linear projections from R3 to R3. My approach was to find the basis of my vector space, project it, then the coefficients of the linear combination to my basis are the matrix.
$endgroup$
– JDizzle
Dec 17 '17 at 13:44
$begingroup$
I've added some explanation in the answer to explain the method.
$endgroup$
– gimusi
Dec 17 '17 at 13:52
$begingroup$
If something is not clear free feel to ask me every explanation.
$endgroup$
– gimusi
Dec 17 '17 at 13:57
$begingroup$
yes, thanks. I'll try to understand this, give me a couple of minutes more. I will comment your answer is i don't get it.
$endgroup$
– JDizzle
Dec 17 '17 at 13:58
add a comment |
$begingroup$
I'm a bit lost trying to find the projection matrix for an orthogonal projection onto a plane defined by the normal vector $n = (1, 1, 1)^T$. Then I can find the basis C of the plain $C = ( (-1,0,1)^T (0,-1,1)^T)$.
Now i should be able to find the projection Matrix with $A(A^TA)^{-1}A^T$
Where $A:=begin{bmatrix}
-1 & 0\
0 & -1\
1 & 1end{bmatrix}$. Then my the projection matrix will look like this?
$A:=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$ Is this correct?
To which basis is this projection matrix? How can I change the matrix to a different basis?
There should be another way to find the matrix. Something like add to my basis $C$ a vector from my basis $B$ (which should not be the standard basis) in $mathbb{R^3}$, find the projection of the basis ( I only need to do this for the added basis vector from $B$ since the rest is already on the plane).
But how can I find the projection of the added basis vector?
After that what would be the matrix from basis B to B? The coefficients of the linear combinations $c_1,c_2,P(b_i)$ in B?
linear-algebra projection-matrices
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
$endgroup$
I'm a bit lost trying to find the projection matrix for an orthogonal projection onto a plane defined by the normal vector $n = (1, 1, 1)^T$. Then I can find the basis C of the plain $C = ( (-1,0,1)^T (0,-1,1)^T)$.
Now i should be able to find the projection Matrix with $A(A^TA)^{-1}A^T$
Where $A:=begin{bmatrix}
-1 & 0\
0 & -1\
1 & 1end{bmatrix}$. Then my the projection matrix will look like this?
$A:=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$ Is this correct?
To which basis is this projection matrix? How can I change the matrix to a different basis?
There should be another way to find the matrix. Something like add to my basis $C$ a vector from my basis $B$ (which should not be the standard basis) in $mathbb{R^3}$, find the projection of the basis ( I only need to do this for the added basis vector from $B$ since the rest is already on the plane).
But how can I find the projection of the added basis vector?
After that what would be the matrix from basis B to B? The coefficients of the linear combinations $c_1,c_2,P(b_i)$ in B?
linear-algebra projection-matrices
linear-algebra projection-matrices
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
edited Dec 4 '18 at 13:55
Martin Sleziak
44.7k10119272
44.7k10119272
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
asked Dec 17 '17 at 13:24
JDizzleJDizzle
1447
1447
$begingroup$
Is clear to you the way to obtain the projection matrix?
$endgroup$
– gimusi
Dec 17 '17 at 13:36
$begingroup$
Not quite, I did it for a couple linear projections from R3 to R3. My approach was to find the basis of my vector space, project it, then the coefficients of the linear combination to my basis are the matrix.
$endgroup$
– JDizzle
Dec 17 '17 at 13:44
$begingroup$
I've added some explanation in the answer to explain the method.
$endgroup$
– gimusi
Dec 17 '17 at 13:52
$begingroup$
If something is not clear free feel to ask me every explanation.
$endgroup$
– gimusi
Dec 17 '17 at 13:57
$begingroup$
yes, thanks. I'll try to understand this, give me a couple of minutes more. I will comment your answer is i don't get it.
$endgroup$
– JDizzle
Dec 17 '17 at 13:58
add a comment |
$begingroup$
Is clear to you the way to obtain the projection matrix?
$endgroup$
– gimusi
Dec 17 '17 at 13:36
$begingroup$
Not quite, I did it for a couple linear projections from R3 to R3. My approach was to find the basis of my vector space, project it, then the coefficients of the linear combination to my basis are the matrix.
$endgroup$
– JDizzle
Dec 17 '17 at 13:44
$begingroup$
I've added some explanation in the answer to explain the method.
$endgroup$
– gimusi
Dec 17 '17 at 13:52
$begingroup$
If something is not clear free feel to ask me every explanation.
$endgroup$
– gimusi
Dec 17 '17 at 13:57
$begingroup$
yes, thanks. I'll try to understand this, give me a couple of minutes more. I will comment your answer is i don't get it.
$endgroup$
– JDizzle
Dec 17 '17 at 13:58
$begingroup$
Is clear to you the way to obtain the projection matrix?
$endgroup$
– gimusi
Dec 17 '17 at 13:36
$begingroup$
Is clear to you the way to obtain the projection matrix?
$endgroup$
– gimusi
Dec 17 '17 at 13:36
$begingroup$
Not quite, I did it for a couple linear projections from R3 to R3. My approach was to find the basis of my vector space, project it, then the coefficients of the linear combination to my basis are the matrix.
$endgroup$
– JDizzle
Dec 17 '17 at 13:44
$begingroup$
Not quite, I did it for a couple linear projections from R3 to R3. My approach was to find the basis of my vector space, project it, then the coefficients of the linear combination to my basis are the matrix.
$endgroup$
– JDizzle
Dec 17 '17 at 13:44
$begingroup$
I've added some explanation in the answer to explain the method.
$endgroup$
– gimusi
Dec 17 '17 at 13:52
$begingroup$
I've added some explanation in the answer to explain the method.
$endgroup$
– gimusi
Dec 17 '17 at 13:52
$begingroup$
If something is not clear free feel to ask me every explanation.
$endgroup$
– gimusi
Dec 17 '17 at 13:57
$begingroup$
If something is not clear free feel to ask me every explanation.
$endgroup$
– gimusi
Dec 17 '17 at 13:57
$begingroup$
yes, thanks. I'll try to understand this, give me a couple of minutes more. I will comment your answer is i don't get it.
$endgroup$
– JDizzle
Dec 17 '17 at 13:58
$begingroup$
yes, thanks. I'll try to understand this, give me a couple of minutes more. I will comment your answer is i don't get it.
$endgroup$
– JDizzle
Dec 17 '17 at 13:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can easily check for A considering the product by the basis vector of the plane, since $forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{mathcal{B}}=begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}$$
If w is a vector in the basis $mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mwimplies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{mathcal{B}}M^{-1}vimplies v_p=MP_{mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{mathcal{B}}M^{-1}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}begin{bmatrix}
-1 & frac13 & frac13\
frac13 & -1 & frac13\
frac13 & frac13 & frac13end{bmatrix}=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{mathcal{B}}wimplies v_p=MP_{mathcal{B}}w$$
Thus the matrix:
$$C=MP_{mathcal{B}}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}=begin{bmatrix}
-1 & 0 & 0\
0 & -1 & 0\
1 & 1 & 0end{bmatrix}$$
represent the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
$endgroup$
1
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
1
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
1
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
1
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
1
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
|
show 18 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can easily check for A considering the product by the basis vector of the plane, since $forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{mathcal{B}}=begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}$$
If w is a vector in the basis $mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mwimplies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{mathcal{B}}M^{-1}vimplies v_p=MP_{mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{mathcal{B}}M^{-1}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}begin{bmatrix}
-1 & frac13 & frac13\
frac13 & -1 & frac13\
frac13 & frac13 & frac13end{bmatrix}=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{mathcal{B}}wimplies v_p=MP_{mathcal{B}}w$$
Thus the matrix:
$$C=MP_{mathcal{B}}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}=begin{bmatrix}
-1 & 0 & 0\
0 & -1 & 0\
1 & 1 & 0end{bmatrix}$$
represent the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
$endgroup$
1
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
1
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
1
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
1
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
1
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
|
show 18 more comments
$begingroup$
You can easily check for A considering the product by the basis vector of the plane, since $forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{mathcal{B}}=begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}$$
If w is a vector in the basis $mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mwimplies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{mathcal{B}}M^{-1}vimplies v_p=MP_{mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{mathcal{B}}M^{-1}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}begin{bmatrix}
-1 & frac13 & frac13\
frac13 & -1 & frac13\
frac13 & frac13 & frac13end{bmatrix}=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{mathcal{B}}wimplies v_p=MP_{mathcal{B}}w$$
Thus the matrix:
$$C=MP_{mathcal{B}}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}=begin{bmatrix}
-1 & 0 & 0\
0 & -1 & 0\
1 & 1 & 0end{bmatrix}$$
represent the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
$endgroup$
1
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
1
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
1
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
1
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
1
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
|
show 18 more comments
$begingroup$
You can easily check for A considering the product by the basis vector of the plane, since $forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{mathcal{B}}=begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}$$
If w is a vector in the basis $mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mwimplies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{mathcal{B}}M^{-1}vimplies v_p=MP_{mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{mathcal{B}}M^{-1}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}begin{bmatrix}
-1 & frac13 & frac13\
frac13 & -1 & frac13\
frac13 & frac13 & frac13end{bmatrix}=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{mathcal{B}}wimplies v_p=MP_{mathcal{B}}w$$
Thus the matrix:
$$C=MP_{mathcal{B}}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}=begin{bmatrix}
-1 & 0 & 0\
0 & -1 & 0\
1 & 1 & 0end{bmatrix}$$
represent the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
$endgroup$
You can easily check for A considering the product by the basis vector of the plane, since $forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{mathcal{B}}=begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}$$
If w is a vector in the basis $mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mwimplies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{mathcal{B}}M^{-1}vimplies v_p=MP_{mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{mathcal{B}}M^{-1}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}begin{bmatrix}
-1 & frac13 & frac13\
frac13 & -1 & frac13\
frac13 & frac13 & frac13end{bmatrix}=begin{bmatrix}
2/3 & -1/3 & -1/3\
-1/3 & 2/3 & -1/3\
-1/3 & -1/3 & 2/3end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $mathcal{B}$ is given by:
$$w_p=P_{mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{mathcal{B}}wimplies v_p=MP_{mathcal{B}}w$$
Thus the matrix:
$$C=MP_{mathcal{B}}=$$
$$=begin{bmatrix}
-1 & 0 & 1\
0 & -1 & 1\
1 & 1 & 1end{bmatrix}begin{bmatrix}
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 0end{bmatrix}=begin{bmatrix}
-1 & 0 & 0\
0 & -1 & 0\
1 & 1 & 0end{bmatrix}$$
represent the projection matrix from the base $mathcal{B}$ to the canonical $mathcal{C}$.
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
edited Dec 17 '17 at 21:27
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
answered Dec 17 '17 at 13:33
gimusigimusi
92.9k84494
92.9k84494
1
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
1
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
1
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
1
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
1
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
|
show 18 more comments
1
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
1
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
1
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
1
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
1
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
1
1
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
$begingroup$
$P_B$ works on the basis B , you plug in a vector in the basis B and obtain the projection in the same basis. EG in the basis B $c_1$ is (1,0,0), $c_2$ is (0,1,0) and $n$ is (0,0,1).
$endgroup$
– gimusi
Dec 17 '17 at 14:20
1
1
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
$begingroup$
With respect to the basis B the projection is trivial. Note that $P_Bc_1=c_1$, $P_Bc_2=c_2$, $P_Bn=0$. In this way every vector is projected in the plane. Once you have the trivial $P_B$ with the change of basis you can find the projection matrix A with respect to the canonical.
$endgroup$
– gimusi
Dec 17 '17 at 14:33
1
1
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
$begingroup$
By definition of projection matrix with respect to the plane, $n$ is the normal vector to the plane thus we want that $P_Bn=0$. And the same properties is true also for An=0.
$endgroup$
– gimusi
Dec 17 '17 at 14:38
1
1
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
$begingroup$
when you calculate with $P_B$ you are using basis $B$ thus $n$ is (0,0,1) and $P_B(0,0,1)=(0,0,0)$
$endgroup$
– gimusi
Dec 17 '17 at 14:47
1
1
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
$begingroup$
Hi, have you found $A=MP_{mathcal{B}}M^{-1}$?
$endgroup$
– gimusi
Dec 17 '17 at 20:43
|
show 18 more comments
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$begingroup$
Is clear to you the way to obtain the projection matrix?
$endgroup$
– gimusi
Dec 17 '17 at 13:36
$begingroup$
Not quite, I did it for a couple linear projections from R3 to R3. My approach was to find the basis of my vector space, project it, then the coefficients of the linear combination to my basis are the matrix.
$endgroup$
– JDizzle
Dec 17 '17 at 13:44
$begingroup$
I've added some explanation in the answer to explain the method.
$endgroup$
– gimusi
Dec 17 '17 at 13:52
$begingroup$
If something is not clear free feel to ask me every explanation.
$endgroup$
– gimusi
Dec 17 '17 at 13:57
$begingroup$
yes, thanks. I'll try to understand this, give me a couple of minutes more. I will comment your answer is i don't get it.
$endgroup$
– JDizzle
Dec 17 '17 at 13:58