Mixing
How to express a quadric equation from canonical form to a different basis.
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I have the quadric $3X^2-Y^2-Z^2=0$ expressed in the canonical form, and the matrix of change of basis from a basis B to the canonical form is
$$P=begin{bmatrix}frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & 0\ frac{1}{sqrt{2}} & -frac{1}{sqrt{2}} &0\ 0 & 0 &-1end{bmatrix}$$
Now, I would like to express the quadric in respect to the basis B. How do I do that?
linear-algebra quadrics
asked Jan 20 at 22:15
KevinKevin
16211
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add a comment |
$begingroup$
I have the quadric $3X^2-Y^2-Z^2=0$ expressed in the canonical form, and the matrix of change of basis from a basis B to the canonical form is
$$P=begin{bmatrix}frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & 0\ frac{1}{sqrt{2}} & -frac{1}{sqrt{2}} &0\ 0 & 0 &-1end{bmatrix}$$
Now, I would like to express the quadric in respect to the basis B. How do I do that?
linear-algebra quadrics
asked Jan 20 at 22:15
KevinKevin
16211
$endgroup$
$begingroup$
If $Px_B=x_c$ (the subscripts B and c meaning that the object is considered with basis B or canonical) then, being $A$ the matrix associated to the quadratic form, $(Px_B)^tA_c(Px_B)=x_B^t(P^tA_cP)x_B=x_B^t(A_B)x_B$
$endgroup$
– gabriele cassese
Jan 20 at 22:21
$begingroup$
@gabrielecassese You should write that up as an answer instead of a comment.
$endgroup$
– amd
Jan 20 at 23:45
add a comment |
$begingroup$
I have the quadric $3X^2-Y^2-Z^2=0$ expressed in the canonical form, and the matrix of change of basis from a basis B to the canonical form is
$$P=begin{bmatrix}frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & 0\ frac{1}{sqrt{2}} & -frac{1}{sqrt{2}} &0\ 0 & 0 &-1end{bmatrix}$$
Now, I would like to express the quadric in respect to the basis B. How do I do that?
linear-algebra quadrics
asked Jan 20 at 22:15
KevinKevin
16211
$endgroup$
I have the quadric $3X^2-Y^2-Z^2=0$ expressed in the canonical form, and the matrix of change of basis from a basis B to the canonical form is
$$P=begin{bmatrix}frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & 0\ frac{1}{sqrt{2}} & -frac{1}{sqrt{2}} &0\ 0 & 0 &-1end{bmatrix}$$
Now, I would like to express the quadric in respect to the basis B. How do I do that?
linear-algebra quadrics
linear-algebra quadrics
asked Jan 20 at 22:15
KevinKevin
16211
asked Jan 20 at 22:15
KevinKevin
16211
asked Jan 20 at 22:15
KevinKevin
16211
asked Jan 20 at 22:15
KevinKevin
16211
asked Jan 20 at 22:15
KevinKevin
16211
16211
$begingroup$
If $Px_B=x_c$ (the subscripts B and c meaning that the object is considered with basis B or canonical) then, being $A$ the matrix associated to the quadratic form, $(Px_B)^tA_c(Px_B)=x_B^t(P^tA_cP)x_B=x_B^t(A_B)x_B$
$endgroup$
– gabriele cassese
Jan 20 at 22:21
$begingroup$
@gabrielecassese You should write that up as an answer instead of a comment.
$endgroup$
– amd
Jan 20 at 23:45
add a comment |
$begingroup$
If $Px_B=x_c$ (the subscripts B and c meaning that the object is considered with basis B or canonical) then, being $A$ the matrix associated to the quadratic form, $(Px_B)^tA_c(Px_B)=x_B^t(P^tA_cP)x_B=x_B^t(A_B)x_B$
$endgroup$
– gabriele cassese
Jan 20 at 22:21
$begingroup$
@gabrielecassese You should write that up as an answer instead of a comment.
$endgroup$
– amd
Jan 20 at 23:45
$begingroup$
If $Px_B=x_c$ (the subscripts B and c meaning that the object is considered with basis B or canonical) then, being $A$ the matrix associated to the quadratic form, $(Px_B)^tA_c(Px_B)=x_B^t(P^tA_cP)x_B=x_B^t(A_B)x_B$
$endgroup$
– gabriele cassese
Jan 20 at 22:21
$begingroup$
If $Px_B=x_c$ (the subscripts B and c meaning that the object is considered with basis B or canonical) then, being $A$ the matrix associated to the quadratic form, $(Px_B)^tA_c(Px_B)=x_B^t(P^tA_cP)x_B=x_B^t(A_B)x_B$
$endgroup$
– gabriele cassese
Jan 20 at 22:21
$begingroup$
@gabrielecassese You should write that up as an answer instead of a comment.
$endgroup$
– amd
Jan 20 at 23:45
$begingroup$
@gabrielecassese You should write that up as an answer instead of a comment.
$endgroup$
– amd
Jan 20 at 23:45
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Notation: I will use the subscript c meaning that the object is considered in the canonical basis, while b if considered in basis b, and $x_c=(X,Y,Z)$.
The quadratic form is indipendent from basis, so we can write
$x_c^tA_cx_c=x_b^tA_bx_b$.
Using the relation $Px_c=x_b$, we get:
$(Px_c)^tA_b(Px_c)=x_c^t left(P^t A_b Pright)x_c=x_c^tA_cx_c$
Thus, $A_b=left(P^{-1}right)^tA_cP^{-1}$
Note: while in this case your change of basis matrix was symmetric and orthogonal,so there was no difference in $P, P^{-1}, P^t$, you have in general to pay close attention to this differences. Those difficulties lead to the concept of contravariant and covariant transformation
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notation: I will use the subscript c meaning that the object is considered in the canonical basis, while b if considered in basis b, and $x_c=(X,Y,Z)$.
The quadratic form is indipendent from basis, so we can write
$x_c^tA_cx_c=x_b^tA_bx_b$.
Using the relation $Px_c=x_b$, we get:
$(Px_c)^tA_b(Px_c)=x_c^t left(P^t A_b Pright)x_c=x_c^tA_cx_c$
Thus, $A_b=left(P^{-1}right)^tA_cP^{-1}$
Note: while in this case your change of basis matrix was symmetric and orthogonal,so there was no difference in $P, P^{-1}, P^t$, you have in general to pay close attention to this differences. Those difficulties lead to the concept of contravariant and covariant transformation
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
$endgroup$
add a comment |
$begingroup$
Notation: I will use the subscript c meaning that the object is considered in the canonical basis, while b if considered in basis b, and $x_c=(X,Y,Z)$.
The quadratic form is indipendent from basis, so we can write
$x_c^tA_cx_c=x_b^tA_bx_b$.
Using the relation $Px_c=x_b$, we get:
$(Px_c)^tA_b(Px_c)=x_c^t left(P^t A_b Pright)x_c=x_c^tA_cx_c$
Thus, $A_b=left(P^{-1}right)^tA_cP^{-1}$
Note: while in this case your change of basis matrix was symmetric and orthogonal,so there was no difference in $P, P^{-1}, P^t$, you have in general to pay close attention to this differences. Those difficulties lead to the concept of contravariant and covariant transformation
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
$endgroup$
add a comment |
$begingroup$
Notation: I will use the subscript c meaning that the object is considered in the canonical basis, while b if considered in basis b, and $x_c=(X,Y,Z)$.
The quadratic form is indipendent from basis, so we can write
$x_c^tA_cx_c=x_b^tA_bx_b$.
Using the relation $Px_c=x_b$, we get:
$(Px_c)^tA_b(Px_c)=x_c^t left(P^t A_b Pright)x_c=x_c^tA_cx_c$
Thus, $A_b=left(P^{-1}right)^tA_cP^{-1}$
Note: while in this case your change of basis matrix was symmetric and orthogonal,so there was no difference in $P, P^{-1}, P^t$, you have in general to pay close attention to this differences. Those difficulties lead to the concept of contravariant and covariant transformation
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
$endgroup$
Notation: I will use the subscript c meaning that the object is considered in the canonical basis, while b if considered in basis b, and $x_c=(X,Y,Z)$.
The quadratic form is indipendent from basis, so we can write
$x_c^tA_cx_c=x_b^tA_bx_b$.
Using the relation $Px_c=x_b$, we get:
$(Px_c)^tA_b(Px_c)=x_c^t left(P^t A_b Pright)x_c=x_c^tA_cx_c$
Thus, $A_b=left(P^{-1}right)^tA_cP^{-1}$
Note: while in this case your change of basis matrix was symmetric and orthogonal,so there was no difference in $P, P^{-1}, P^t$, you have in general to pay close attention to this differences. Those difficulties lead to the concept of contravariant and covariant transformation
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
edited Jan 21 at 20:23
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
answered Jan 21 at 7:12
gabriele cassesegabriele cassese
870314
870314
add a comment |
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$begingroup$
If $Px_B=x_c$ (the subscripts B and c meaning that the object is considered with basis B or canonical) then, being $A$ the matrix associated to the quadratic form, $(Px_B)^tA_c(Px_B)=x_B^t(P^tA_cP)x_B=x_B^t(A_B)x_B$
$endgroup$
– gabriele cassese
Jan 20 at 22:21
$begingroup$
@gabrielecassese You should write that up as an answer instead of a comment.
$endgroup$
– amd
Jan 20 at 23:45