Mixing
Show that $f$ is an orthogonal projection
$begingroup$
$A=M(f,B)=frac{1}{3}begin{pmatrix}
1 & -1 & 1\
-1& 1 & -1\
1& -1 & 1
end{pmatrix}$
Show that $f$ is an orthogonal projection on a line to be determined. How to solve ?
linear-algebra linear-transformations projection
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
$endgroup$
|
show 3 more comments
$begingroup$
$A=M(f,B)=frac{1}{3}begin{pmatrix}
1 & -1 & 1\
-1& 1 & -1\
1& -1 & 1
end{pmatrix}$
Show that $f$ is an orthogonal projection on a line to be determined. How to solve ?
linear-algebra linear-transformations projection
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
$endgroup$
$begingroup$
I assume that $M(f,B)$ means, “the matrix for $f$ with respect to the basis $B$”. Is this correct?
$endgroup$
– Jordan Green
Jan 29 at 2:14
$begingroup$
Yes correct and B is the canonical basis of R^3 equipped with the canonical euclidean inner product
$endgroup$
– Pedro Alvarès
Jan 29 at 2:21
$begingroup$
Ok, good. Thanks for clarifying.
$endgroup$
– Jordan Green
Jan 29 at 2:26
$begingroup$
Do you know what eigenvalues and eigenvectors are? Have you studied them before?
$endgroup$
– stressed out
Jan 29 at 2:54
$begingroup$
Yes I know, but why can't we prove that transpose of A is A and that A^2=A, so that f is an orthogonal projection
$endgroup$
– Pedro Alvarès
Jan 29 at 3:02
|
show 3 more comments
$begingroup$
$A=M(f,B)=frac{1}{3}begin{pmatrix}
1 & -1 & 1\
-1& 1 & -1\
1& -1 & 1
end{pmatrix}$
Show that $f$ is an orthogonal projection on a line to be determined. How to solve ?
linear-algebra linear-transformations projection
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
$endgroup$
$A=M(f,B)=frac{1}{3}begin{pmatrix}
1 & -1 & 1\
-1& 1 & -1\
1& -1 & 1
end{pmatrix}$
Show that $f$ is an orthogonal projection on a line to be determined. How to solve ?
linear-algebra linear-transformations projection
linear-algebra linear-transformations projection
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
asked Jan 29 at 2:04
Pedro AlvarèsPedro Alvarès
756
756
$begingroup$
I assume that $M(f,B)$ means, “the matrix for $f$ with respect to the basis $B$”. Is this correct?
$endgroup$
– Jordan Green
Jan 29 at 2:14
$begingroup$
Yes correct and B is the canonical basis of R^3 equipped with the canonical euclidean inner product
$endgroup$
– Pedro Alvarès
Jan 29 at 2:21
$begingroup$
Ok, good. Thanks for clarifying.
$endgroup$
– Jordan Green
Jan 29 at 2:26
$begingroup$
Do you know what eigenvalues and eigenvectors are? Have you studied them before?
$endgroup$
– stressed out
Jan 29 at 2:54
$begingroup$
Yes I know, but why can't we prove that transpose of A is A and that A^2=A, so that f is an orthogonal projection
$endgroup$
– Pedro Alvarès
Jan 29 at 3:02
|
show 3 more comments
$begingroup$
I assume that $M(f,B)$ means, “the matrix for $f$ with respect to the basis $B$”. Is this correct?
$endgroup$
– Jordan Green
Jan 29 at 2:14
$begingroup$
Yes correct and B is the canonical basis of R^3 equipped with the canonical euclidean inner product
$endgroup$
– Pedro Alvarès
Jan 29 at 2:21
$begingroup$
Ok, good. Thanks for clarifying.
$endgroup$
– Jordan Green
Jan 29 at 2:26
$begingroup$
Do you know what eigenvalues and eigenvectors are? Have you studied them before?
$endgroup$
– stressed out
Jan 29 at 2:54
$begingroup$
Yes I know, but why can't we prove that transpose of A is A and that A^2=A, so that f is an orthogonal projection
$endgroup$
– Pedro Alvarès
Jan 29 at 3:02
$begingroup$
I assume that $M(f,B)$ means, “the matrix for $f$ with respect to the basis $B$”. Is this correct?
$endgroup$
– Jordan Green
Jan 29 at 2:14
$begingroup$
I assume that $M(f,B)$ means, “the matrix for $f$ with respect to the basis $B$”. Is this correct?
$endgroup$
– Jordan Green
Jan 29 at 2:14
$begingroup$
Yes correct and B is the canonical basis of R^3 equipped with the canonical euclidean inner product
$endgroup$
– Pedro Alvarès
Jan 29 at 2:21
$begingroup$
Yes correct and B is the canonical basis of R^3 equipped with the canonical euclidean inner product
$endgroup$
– Pedro Alvarès
Jan 29 at 2:21
$begingroup$
Ok, good. Thanks for clarifying.
$endgroup$
– Jordan Green
Jan 29 at 2:26
$begingroup$
Ok, good. Thanks for clarifying.
$endgroup$
– Jordan Green
Jan 29 at 2:26
$begingroup$
Do you know what eigenvalues and eigenvectors are? Have you studied them before?
$endgroup$
– stressed out
Jan 29 at 2:54
$begingroup$
Do you know what eigenvalues and eigenvectors are? Have you studied them before?
$endgroup$
– stressed out
Jan 29 at 2:54
$begingroup$
Yes I know, but why can't we prove that transpose of A is A and that A^2=A, so that f is an orthogonal projection
$endgroup$
– Pedro Alvarès
Jan 29 at 3:02
$begingroup$
Yes I know, but why can't we prove that transpose of A is A and that A^2=A, so that f is an orthogonal projection
$endgroup$
– Pedro Alvarès
Jan 29 at 3:02
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
It is geometrically obvious that the direction of the line must be an eigenvector of this transformation because any vector on this line must be projected to itself. Find this eigenvector first (there should be only one non-zero eigenvalue based on our geometric intuition). So, find the eigenvector associated with the non-zero eigenvalue of $A$ first. You can normalize the eigenvector if you want. In this particular case, the non-zero eigenvalue is equal to $1$. (I just checked it).
Once you've found it, let's call it $v$, we can decompose our space as follows $$mathbb{R}^3 = langle v rangle opluslangle v rangle ^ {perp}$$
To show that it's an orthogonal transformation, we should show that it sends $langle v rangle ^ {perp}$ to $0$. Find an orthogonal basis for $langle v rangle ^ {perp}$ and show that $$ker{A}=langle v rangle ^ {perp}$$
Note that since our matrix $A$ is symmetric, its eigenvectors are orthogonal. So, to find a basis for $langle v rangle ^ {perp}$, you can just find the eigenvectors associated to the eigenvalue $lambda=0$.
Edit: After some calculations as I explained, you can see that $v=frac{1}{sqrt{3}}(1,-1,1)$ and $langle v rangle ^ {perp}=langle n,mrangle$ where $$n=frac{1}{sqrt{6}}(1,2,1), m=frac{1}{sqrt{2}}(1,0,-1)$$
It is easily seen that $Av=v$ and $An=Am=0$.
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Let $textbf{a} = (1,-1,1)$. Then for any $textbf{v} in mathbb{R^3}$, we have
$$
A mathbf{v} = frac{textbf{a} cdot textbf{v}}{| textbf{a}|^2} textbf{a},
$$
which one can recognize as the formula for orthogonal projection onto the line spanned by $textbf{a}$. (To see that the projection is orthogonal, one can compute the dot product of $mathbf{v} - A mathbf{v}$ with $mathbf{a}$.)
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
$endgroup$
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
$begingroup$
Hint:
It is geometrically obvious that the direction of the line must be an eigenvector of this transformation because any vector on this line must be projected to itself. Find this eigenvector first (there should be only one non-zero eigenvalue based on our geometric intuition). So, find the eigenvector associated with the non-zero eigenvalue of $A$ first. You can normalize the eigenvector if you want. In this particular case, the non-zero eigenvalue is equal to $1$. (I just checked it).
Once you've found it, let's call it $v$, we can decompose our space as follows $$mathbb{R}^3 = langle v rangle opluslangle v rangle ^ {perp}$$
To show that it's an orthogonal transformation, we should show that it sends $langle v rangle ^ {perp}$ to $0$. Find an orthogonal basis for $langle v rangle ^ {perp}$ and show that $$ker{A}=langle v rangle ^ {perp}$$
Note that since our matrix $A$ is symmetric, its eigenvectors are orthogonal. So, to find a basis for $langle v rangle ^ {perp}$, you can just find the eigenvectors associated to the eigenvalue $lambda=0$.
Edit: After some calculations as I explained, you can see that $v=frac{1}{sqrt{3}}(1,-1,1)$ and $langle v rangle ^ {perp}=langle n,mrangle$ where $$n=frac{1}{sqrt{6}}(1,2,1), m=frac{1}{sqrt{2}}(1,0,-1)$$
It is easily seen that $Av=v$ and $An=Am=0$.
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Hint:
It is geometrically obvious that the direction of the line must be an eigenvector of this transformation because any vector on this line must be projected to itself. Find this eigenvector first (there should be only one non-zero eigenvalue based on our geometric intuition). So, find the eigenvector associated with the non-zero eigenvalue of $A$ first. You can normalize the eigenvector if you want. In this particular case, the non-zero eigenvalue is equal to $1$. (I just checked it).
Once you've found it, let's call it $v$, we can decompose our space as follows $$mathbb{R}^3 = langle v rangle opluslangle v rangle ^ {perp}$$
To show that it's an orthogonal transformation, we should show that it sends $langle v rangle ^ {perp}$ to $0$. Find an orthogonal basis for $langle v rangle ^ {perp}$ and show that $$ker{A}=langle v rangle ^ {perp}$$
Note that since our matrix $A$ is symmetric, its eigenvectors are orthogonal. So, to find a basis for $langle v rangle ^ {perp}$, you can just find the eigenvectors associated to the eigenvalue $lambda=0$.
Edit: After some calculations as I explained, you can see that $v=frac{1}{sqrt{3}}(1,-1,1)$ and $langle v rangle ^ {perp}=langle n,mrangle$ where $$n=frac{1}{sqrt{6}}(1,2,1), m=frac{1}{sqrt{2}}(1,0,-1)$$
It is easily seen that $Av=v$ and $An=Am=0$.
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Hint:
It is geometrically obvious that the direction of the line must be an eigenvector of this transformation because any vector on this line must be projected to itself. Find this eigenvector first (there should be only one non-zero eigenvalue based on our geometric intuition). So, find the eigenvector associated with the non-zero eigenvalue of $A$ first. You can normalize the eigenvector if you want. In this particular case, the non-zero eigenvalue is equal to $1$. (I just checked it).
Once you've found it, let's call it $v$, we can decompose our space as follows $$mathbb{R}^3 = langle v rangle opluslangle v rangle ^ {perp}$$
To show that it's an orthogonal transformation, we should show that it sends $langle v rangle ^ {perp}$ to $0$. Find an orthogonal basis for $langle v rangle ^ {perp}$ and show that $$ker{A}=langle v rangle ^ {perp}$$
Note that since our matrix $A$ is symmetric, its eigenvectors are orthogonal. So, to find a basis for $langle v rangle ^ {perp}$, you can just find the eigenvectors associated to the eigenvalue $lambda=0$.
Edit: After some calculations as I explained, you can see that $v=frac{1}{sqrt{3}}(1,-1,1)$ and $langle v rangle ^ {perp}=langle n,mrangle$ where $$n=frac{1}{sqrt{6}}(1,2,1), m=frac{1}{sqrt{2}}(1,0,-1)$$
It is easily seen that $Av=v$ and $An=Am=0$.
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
$endgroup$
Hint:
It is geometrically obvious that the direction of the line must be an eigenvector of this transformation because any vector on this line must be projected to itself. Find this eigenvector first (there should be only one non-zero eigenvalue based on our geometric intuition). So, find the eigenvector associated with the non-zero eigenvalue of $A$ first. You can normalize the eigenvector if you want. In this particular case, the non-zero eigenvalue is equal to $1$. (I just checked it).
Once you've found it, let's call it $v$, we can decompose our space as follows $$mathbb{R}^3 = langle v rangle opluslangle v rangle ^ {perp}$$
To show that it's an orthogonal transformation, we should show that it sends $langle v rangle ^ {perp}$ to $0$. Find an orthogonal basis for $langle v rangle ^ {perp}$ and show that $$ker{A}=langle v rangle ^ {perp}$$
Note that since our matrix $A$ is symmetric, its eigenvectors are orthogonal. So, to find a basis for $langle v rangle ^ {perp}$, you can just find the eigenvectors associated to the eigenvalue $lambda=0$.
Edit: After some calculations as I explained, you can see that $v=frac{1}{sqrt{3}}(1,-1,1)$ and $langle v rangle ^ {perp}=langle n,mrangle$ where $$n=frac{1}{sqrt{6}}(1,2,1), m=frac{1}{sqrt{2}}(1,0,-1)$$
It is easily seen that $Av=v$ and $An=Am=0$.
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
edited Jan 29 at 2:51
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
answered Jan 29 at 2:20
stressed outstressed out
6,5831939
6,5831939
add a comment |
add a comment |
$begingroup$
Let $textbf{a} = (1,-1,1)$. Then for any $textbf{v} in mathbb{R^3}$, we have
$$
A mathbf{v} = frac{textbf{a} cdot textbf{v}}{| textbf{a}|^2} textbf{a},
$$
which one can recognize as the formula for orthogonal projection onto the line spanned by $textbf{a}$. (To see that the projection is orthogonal, one can compute the dot product of $mathbf{v} - A mathbf{v}$ with $mathbf{a}$.)
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
$endgroup$
add a comment |
$begingroup$
Let $textbf{a} = (1,-1,1)$. Then for any $textbf{v} in mathbb{R^3}$, we have
$$
A mathbf{v} = frac{textbf{a} cdot textbf{v}}{| textbf{a}|^2} textbf{a},
$$
which one can recognize as the formula for orthogonal projection onto the line spanned by $textbf{a}$. (To see that the projection is orthogonal, one can compute the dot product of $mathbf{v} - A mathbf{v}$ with $mathbf{a}$.)
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
$endgroup$
add a comment |
$begingroup$
Let $textbf{a} = (1,-1,1)$. Then for any $textbf{v} in mathbb{R^3}$, we have
$$
A mathbf{v} = frac{textbf{a} cdot textbf{v}}{| textbf{a}|^2} textbf{a},
$$
which one can recognize as the formula for orthogonal projection onto the line spanned by $textbf{a}$. (To see that the projection is orthogonal, one can compute the dot product of $mathbf{v} - A mathbf{v}$ with $mathbf{a}$.)
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
$endgroup$
Let $textbf{a} = (1,-1,1)$. Then for any $textbf{v} in mathbb{R^3}$, we have
$$
A mathbf{v} = frac{textbf{a} cdot textbf{v}}{| textbf{a}|^2} textbf{a},
$$
which one can recognize as the formula for orthogonal projection onto the line spanned by $textbf{a}$. (To see that the projection is orthogonal, one can compute the dot product of $mathbf{v} - A mathbf{v}$ with $mathbf{a}$.)
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
answered Jan 29 at 2:29
Jordan GreenJordan Green
1,146410
1,146410
add a comment |
add a comment |
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$begingroup$
I assume that $M(f,B)$ means, “the matrix for $f$ with respect to the basis $B$”. Is this correct?
$endgroup$
– Jordan Green
Jan 29 at 2:14
$begingroup$
Yes correct and B is the canonical basis of R^3 equipped with the canonical euclidean inner product
$endgroup$
– Pedro Alvarès
Jan 29 at 2:21
$begingroup$
Ok, good. Thanks for clarifying.
$endgroup$
– Jordan Green
Jan 29 at 2:26
$begingroup$
Do you know what eigenvalues and eigenvectors are? Have you studied them before?
$endgroup$
– stressed out
Jan 29 at 2:54
$begingroup$
Yes I know, but why can't we prove that transpose of A is A and that A^2=A, so that f is an orthogonal projection
$endgroup$
– Pedro Alvarès
Jan 29 at 3:02