Mixing
Spectral theorem and Quadrics
Clash Royale CLAN TAG#URR8PPP
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The task is to convert a quadric equation to a normal form.
$Q equiv 3x_1^2 + 3x_2^2+3x_3^2+ 2x_1x_2+ 2x_1x_3 + 2x_2x_3+4x_1+4x_2+2x_3 +1=0$
What I've done is the following:
- Create a matrix A from the variables with degree 1 and 2
$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$
Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $
Determine the eigenvalues of $A- lambda I_3$
$E_2= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
$E_5= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
but $dim(E_2(A))+dim(E_5(A)) ne 3$. So a symmetrical matrix can't be constructed. Does this imply that there is no normal form? Can you maybe see if I made a mistake?
linear-algebra quadratic-forms
asked Jul 14 at 21:06
Anonymous I
804725
add a comment |Â
up vote
1
down vote
favorite
The task is to convert a quadric equation to a normal form.
$Q equiv 3x_1^2 + 3x_2^2+3x_3^2+ 2x_1x_2+ 2x_1x_3 + 2x_2x_3+4x_1+4x_2+2x_3 +1=0$
What I've done is the following:
- Create a matrix A from the variables with degree 1 and 2
$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$
Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $
Determine the eigenvalues of $A- lambda I_3$
$E_2= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
$E_5= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
but $dim(E_2(A))+dim(E_5(A)) ne 3$. So a symmetrical matrix can't be constructed. Does this imply that there is no normal form? Can you maybe see if I made a mistake?
linear-algebra quadratic-forms
asked Jul 14 at 21:06
Anonymous I
804725
Oops I forgot a term. I'll edit it.
– Anonymous I
Jul 14 at 22:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The task is to convert a quadric equation to a normal form.
$Q equiv 3x_1^2 + 3x_2^2+3x_3^2+ 2x_1x_2+ 2x_1x_3 + 2x_2x_3+4x_1+4x_2+2x_3 +1=0$
What I've done is the following:
- Create a matrix A from the variables with degree 1 and 2
$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$
Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $
Determine the eigenvalues of $A- lambda I_3$
$E_2= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
$E_5= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
but $dim(E_2(A))+dim(E_5(A)) ne 3$. So a symmetrical matrix can't be constructed. Does this imply that there is no normal form? Can you maybe see if I made a mistake?
linear-algebra quadratic-forms
asked Jul 14 at 21:06
Anonymous I
804725
The task is to convert a quadric equation to a normal form.
$Q equiv 3x_1^2 + 3x_2^2+3x_3^2+ 2x_1x_2+ 2x_1x_3 + 2x_2x_3+4x_1+4x_2+2x_3 +1=0$
What I've done is the following:
- Create a matrix A from the variables with degree 1 and 2
$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$
Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $
Determine the eigenvalues of $A- lambda I_3$
$E_2= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
$E_5= operatornamevectbeginpmatrix
1\
1\
1\
endpmatrix$
but $dim(E_2(A))+dim(E_5(A)) ne 3$. So a symmetrical matrix can't be constructed. Does this imply that there is no normal form? Can you maybe see if I made a mistake?
linear-algebra quadratic-forms
asked Jul 14 at 21:06
Anonymous I
804725
edited Jul 14 at 22:39
asked Jul 14 at 21:06
Anonymous I
804725
asked Jul 14 at 21:06
Anonymous I
804725
asked Jul 14 at 21:06
Anonymous I
804725
804725
Oops I forgot a term. I'll edit it.
– Anonymous I
Jul 14 at 22:38
add a comment |Â
Oops I forgot a term. I'll edit it.
– Anonymous I
Jul 14 at 22:38
Oops I forgot a term. I'll edit it.
– Anonymous I
Jul 14 at 22:38
Oops I forgot a term. I'll edit it.
– Anonymous I
Jul 14 at 22:38
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
You are wrong about $E_2$ and $E_5$:$$E_5=leftlanglebeginpmatrix1\1\1endpmatrixrightrangle$$and$$E_2=leftlanglebeginpmatrix-1\0\1endpmatrix,beginpmatrix-1\1\0endpmatrixrightrangle.$$So, $dim E_5=1$ and $dim E_2=2$.
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
add a comment |Â
up vote
1
down vote
You should’ve noticed for yourself that you’d done something wrong when you ended up with the same eigenvector for two different eigenvalues. $E_5$ is clearly correct by inspection—multiplying $A$ by this vector sums the rows—and let’s assume that you’re correct about that eigenspace being one-dimensional. By using the fact that the eigenspaces of a symmetric real matrix are mutually orthogonal, you can produce eigenvectors of $2$ with almost no work: $(-1,1,0)^T$, $(0,-1,1)^T$ and $(1,0,-1)^T$ are all clearly orthogonal to $(1,1,1)^T$. Any two of these, together with $E_5$, will do for diagonalizing the matrix. If you want an orthogonal basis, pick one of these and generate the other vector via a cross product, further normalizing everything if you’d like to end up with an orthonormal basis.
I’ll note that it looks like you’re trying to construct the canonical form rather than a normal form. The latter doesn’t really require finding eigenvalues/eigenvectors.
answered Jul 14 at 22:27
amd
26k2944
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
add a comment |Â
up vote
0
down vote
The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=beginbmatrix3&2&2\2&3&0\2&0&3endbmatrix$$and $$b=beginbmatrix4\4\2endbmatrix$$
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are wrong about $E_2$ and $E_5$:$$E_5=leftlanglebeginpmatrix1\1\1endpmatrixrightrangle$$and$$E_2=leftlanglebeginpmatrix-1\0\1endpmatrix,beginpmatrix-1\1\0endpmatrixrightrangle.$$So, $dim E_5=1$ and $dim E_2=2$.
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
add a comment |Â
up vote
1
down vote
accepted
You are wrong about $E_2$ and $E_5$:$$E_5=leftlanglebeginpmatrix1\1\1endpmatrixrightrangle$$and$$E_2=leftlanglebeginpmatrix-1\0\1endpmatrix,beginpmatrix-1\1\0endpmatrixrightrangle.$$So, $dim E_5=1$ and $dim E_2=2$.
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are wrong about $E_2$ and $E_5$:$$E_5=leftlanglebeginpmatrix1\1\1endpmatrixrightrangle$$and$$E_2=leftlanglebeginpmatrix-1\0\1endpmatrix,beginpmatrix-1\1\0endpmatrixrightrangle.$$So, $dim E_5=1$ and $dim E_2=2$.
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
You are wrong about $E_2$ and $E_5$:$$E_5=leftlanglebeginpmatrix1\1\1endpmatrixrightrangle$$and$$E_2=leftlanglebeginpmatrix-1\0\1endpmatrix,beginpmatrix-1\1\0endpmatrixrightrangle.$$So, $dim E_5=1$ and $dim E_2=2$.
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
answered Jul 14 at 21:16
José Carlos Santos
114k1698177
114k1698177
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
add a comment |Â
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
– Anonymous I
Jul 14 at 21:30
add a comment |Â
up vote
1
down vote
You should’ve noticed for yourself that you’d done something wrong when you ended up with the same eigenvector for two different eigenvalues. $E_5$ is clearly correct by inspection—multiplying $A$ by this vector sums the rows—and let’s assume that you’re correct about that eigenspace being one-dimensional. By using the fact that the eigenspaces of a symmetric real matrix are mutually orthogonal, you can produce eigenvectors of $2$ with almost no work: $(-1,1,0)^T$, $(0,-1,1)^T$ and $(1,0,-1)^T$ are all clearly orthogonal to $(1,1,1)^T$. Any two of these, together with $E_5$, will do for diagonalizing the matrix. If you want an orthogonal basis, pick one of these and generate the other vector via a cross product, further normalizing everything if you’d like to end up with an orthonormal basis.
I’ll note that it looks like you’re trying to construct the canonical form rather than a normal form. The latter doesn’t really require finding eigenvalues/eigenvectors.
answered Jul 14 at 22:27
amd
26k2944
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
add a comment |Â
up vote
1
down vote
You should’ve noticed for yourself that you’d done something wrong when you ended up with the same eigenvector for two different eigenvalues. $E_5$ is clearly correct by inspection—multiplying $A$ by this vector sums the rows—and let’s assume that you’re correct about that eigenspace being one-dimensional. By using the fact that the eigenspaces of a symmetric real matrix are mutually orthogonal, you can produce eigenvectors of $2$ with almost no work: $(-1,1,0)^T$, $(0,-1,1)^T$ and $(1,0,-1)^T$ are all clearly orthogonal to $(1,1,1)^T$. Any two of these, together with $E_5$, will do for diagonalizing the matrix. If you want an orthogonal basis, pick one of these and generate the other vector via a cross product, further normalizing everything if you’d like to end up with an orthonormal basis.
I’ll note that it looks like you’re trying to construct the canonical form rather than a normal form. The latter doesn’t really require finding eigenvalues/eigenvectors.
answered Jul 14 at 22:27
amd
26k2944
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You should’ve noticed for yourself that you’d done something wrong when you ended up with the same eigenvector for two different eigenvalues. $E_5$ is clearly correct by inspection—multiplying $A$ by this vector sums the rows—and let’s assume that you’re correct about that eigenspace being one-dimensional. By using the fact that the eigenspaces of a symmetric real matrix are mutually orthogonal, you can produce eigenvectors of $2$ with almost no work: $(-1,1,0)^T$, $(0,-1,1)^T$ and $(1,0,-1)^T$ are all clearly orthogonal to $(1,1,1)^T$. Any two of these, together with $E_5$, will do for diagonalizing the matrix. If you want an orthogonal basis, pick one of these and generate the other vector via a cross product, further normalizing everything if you’d like to end up with an orthonormal basis.
I’ll note that it looks like you’re trying to construct the canonical form rather than a normal form. The latter doesn’t really require finding eigenvalues/eigenvectors.
answered Jul 14 at 22:27
amd
26k2944
You should’ve noticed for yourself that you’d done something wrong when you ended up with the same eigenvector for two different eigenvalues. $E_5$ is clearly correct by inspection—multiplying $A$ by this vector sums the rows—and let’s assume that you’re correct about that eigenspace being one-dimensional. By using the fact that the eigenspaces of a symmetric real matrix are mutually orthogonal, you can produce eigenvectors of $2$ with almost no work: $(-1,1,0)^T$, $(0,-1,1)^T$ and $(1,0,-1)^T$ are all clearly orthogonal to $(1,1,1)^T$. Any two of these, together with $E_5$, will do for diagonalizing the matrix. If you want an orthogonal basis, pick one of these and generate the other vector via a cross product, further normalizing everything if you’d like to end up with an orthonormal basis.
I’ll note that it looks like you’re trying to construct the canonical form rather than a normal form. The latter doesn’t really require finding eigenvalues/eigenvectors.
answered Jul 14 at 22:27
amd
26k2944
edited Jul 14 at 22:43
answered Jul 14 at 22:27
amd
26k2944
answered Jul 14 at 22:27
amd
26k2944
answered Jul 14 at 22:27
amd
26k2944
26k2944
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
add a comment |Â
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough.
– Anonymous I
Jul 14 at 22:37
add a comment |Â
up vote
0
down vote
The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=beginbmatrix3&2&2\2&3&0\2&0&3endbmatrix$$and $$b=beginbmatrix4\4\2endbmatrix$$
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
add a comment |Â
up vote
0
down vote
The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=beginbmatrix3&2&2\2&3&0\2&0&3endbmatrix$$and $$b=beginbmatrix4\4\2endbmatrix$$
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=beginbmatrix3&2&2\2&3&0\2&0&3endbmatrix$$and $$b=beginbmatrix4\4\2endbmatrix$$
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=beginbmatrix3&2&2\2&3&0\2&0&3endbmatrix$$and $$b=beginbmatrix4\4\2endbmatrix$$
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
answered Jul 14 at 21:35
Mostafa Ayaz
8,6573630
8,6573630
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
add a comment |Â
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$
– Anonymous I
Jul 14 at 21:37
add a comment |Â
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Oops I forgot a term. I'll edit it.
– Anonymous I
Jul 14 at 22:38