Spectral theorem and Quadrics

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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:



  1. Create a matrix A from the variables with degree 1 and 2

$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$



  1. Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $


  2. 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




share|cite|improve this question





















  • Oops I forgot a term. I'll edit it.
    – Anonymous I
    Jul 14 at 22:38















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:



  1. Create a matrix A from the variables with degree 1 and 2

$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$



  1. Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $


  2. 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




share|cite|improve this question





















  • Oops I forgot a term. I'll edit it.
    – Anonymous I
    Jul 14 at 22:38













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:



  1. Create a matrix A from the variables with degree 1 and 2

$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$



  1. Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $


  2. 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




share|cite|improve this question













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:



  1. Create a matrix A from the variables with degree 1 and 2

$$beginbmatrix
x_1 & x_2 & x_3\
3 & 1 & 1 & x_1\
1 & 3 & 1& x_2\
1 & 1 & 3& x_3\
endbmatrix$$



  1. Determine the determinant of $A- lambda I_3 = -lambda^3 +9lambda^2 -24lambda+20= (lambda-2)^2(-lambda + 5) $


  2. 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





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 14 at 22:39
























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

















  • 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











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 $dim E_2=2$.






share|cite|improve this answer





















  • Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
    – Anonymous I
    Jul 14 at 21:30


















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.






share|cite|improve this answer























  • 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

















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 $$b=beginbmatrix4\4\2endbmatrix$$






share|cite|improve this answer





















  • 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











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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 $dim E_2=2$.






share|cite|improve this answer





















  • Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
    – Anonymous I
    Jul 14 at 21:30















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 $dim E_2=2$.






share|cite|improve this answer





















  • Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix.
    – Anonymous I
    Jul 14 at 21:30













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 $dim E_2=2$.






share|cite|improve this answer













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 $dim E_2=2$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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

















  • 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











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.






share|cite|improve this answer























  • 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














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.






share|cite|improve this answer























  • 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












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.






share|cite|improve this answer















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.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 14 at 22:43


























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
















  • 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










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 $$b=beginbmatrix4\4\2endbmatrix$$






share|cite|improve this answer





















  • 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















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 $$b=beginbmatrix4\4\2endbmatrix$$






share|cite|improve this answer





















  • 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













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 $$b=beginbmatrix4\4\2endbmatrix$$






share|cite|improve this answer













The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=beginbmatrix3&2&2\2&3&0\2&0&3endbmatrix$$and $$b=beginbmatrix4\4\2endbmatrix$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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

















  • 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













 

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