Mixing
Finding a matrix $P$ such that $B=P^TAP$ given $A,B$
1
$begingroup$
I have 2 matrices, $A,B$ associated with the same bilinear application, where $A$ is expressed using the canonical basis $(1,0,0),(0,1,0),(0,0,1)$. I need to find the basis $B$ is associated with so I can compute $P$.
$$B=P^TAP$$
I would love a method that doesn't involve solving a big system of equations.
$$
A= begin{bmatrix}
3 & -2 & 0 \
5 & 7 & -8 \
0 & -4 & -1 \
end{bmatrix}$$
$$
B=
begin{bmatrix}
25 & 6 & -8 \
17 & 0 & -9 \
-4 & -5 & -1 \
end{bmatrix}
$$
linear-algebra
edited Jan 31 at 18:57
Nij
2,01611323
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
$endgroup$
|
show 2 more comments
1
$begingroup$
I have 2 matrices, $A,B$ associated with the same bilinear application, where $A$ is expressed using the canonical basis $(1,0,0),(0,1,0),(0,0,1)$. I need to find the basis $B$ is associated with so I can compute $P$.
$$B=P^TAP$$
I would love a method that doesn't involve solving a big system of equations.
$$
A= begin{bmatrix}
3 & -2 & 0 \
5 & 7 & -8 \
0 & -4 & -1 \
end{bmatrix}$$
$$
B=
begin{bmatrix}
25 & 6 & -8 \
17 & 0 & -9 \
-4 & -5 & -1 \
end{bmatrix}
$$
linear-algebra
edited Jan 31 at 18:57
Nij
2,01611323
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
$endgroup$
3
$begingroup$
Can you tell about your method and progress for the question?
$endgroup$
– SNEHIL SANYAL
Jan 31 at 17:39
$begingroup$
Making P a matrix with all its integers variables a,b,c,d,e,f,g,h,i and then making matrix multiplication, then solving the system
$endgroup$
– Marco Villalobos
Jan 31 at 18:01
$begingroup$
If $A,B$ were symmetric, then you could orthogonally diagonalize each of them, getting $A=P^TCP$ and $B=Q^TDQ$. Assuming $C=D$ (else there is no solution), then $B=Q^TPAP^TQ=(P^{-1}Q)^TA(P^{-1}Q)$.
$endgroup$
– vadim123
Jan 31 at 18:08
$begingroup$
Does a method for non-symmetric matrices exist?
$endgroup$
– Marco Villalobos
Jan 31 at 19:05
$begingroup$
How are these matrices “associated with a bilinear application?” If the bilinear form is something like $x^TAx$, then only the symmetric part of $A$ contributes to it, so there’s no a priori reason to believe that there is a solution in the first place. What makes you think that one is possible?
$endgroup$
– amd
Feb 1 at 1:38
|
show 2 more comments
1
1
1
1
$begingroup$
I have 2 matrices, $A,B$ associated with the same bilinear application, where $A$ is expressed using the canonical basis $(1,0,0),(0,1,0),(0,0,1)$. I need to find the basis $B$ is associated with so I can compute $P$.
$$B=P^TAP$$
I would love a method that doesn't involve solving a big system of equations.
$$
A= begin{bmatrix}
3 & -2 & 0 \
5 & 7 & -8 \
0 & -4 & -1 \
end{bmatrix}$$
$$
B=
begin{bmatrix}
25 & 6 & -8 \
17 & 0 & -9 \
-4 & -5 & -1 \
end{bmatrix}
$$
linear-algebra
edited Jan 31 at 18:57
Nij
2,01611323
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
$endgroup$
I have 2 matrices, $A,B$ associated with the same bilinear application, where $A$ is expressed using the canonical basis $(1,0,0),(0,1,0),(0,0,1)$. I need to find the basis $B$ is associated with so I can compute $P$.
$$B=P^TAP$$
I would love a method that doesn't involve solving a big system of equations.
$$
A= begin{bmatrix}
3 & -2 & 0 \
5 & 7 & -8 \
0 & -4 & -1 \
end{bmatrix}$$
$$
B=
begin{bmatrix}
25 & 6 & -8 \
17 & 0 & -9 \
-4 & -5 & -1 \
end{bmatrix}
$$
linear-algebra
linear-algebra
edited Jan 31 at 18:57
Nij
2,01611323
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
edited Jan 31 at 18:57
Nij
2,01611323
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
edited Jan 31 at 18:57
Nij
2,01611323
edited Jan 31 at 18:57
Nij
2,01611323
edited Jan 31 at 18:57
Nij
2,01611323
2,01611323
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
asked Jan 31 at 17:35
Marco VillalobosMarco Villalobos
61
61
3
$begingroup$
Can you tell about your method and progress for the question?
$endgroup$
– SNEHIL SANYAL
Jan 31 at 17:39
$begingroup$
Making P a matrix with all its integers variables a,b,c,d,e,f,g,h,i and then making matrix multiplication, then solving the system
$endgroup$
– Marco Villalobos
Jan 31 at 18:01
$begingroup$
If $A,B$ were symmetric, then you could orthogonally diagonalize each of them, getting $A=P^TCP$ and $B=Q^TDQ$. Assuming $C=D$ (else there is no solution), then $B=Q^TPAP^TQ=(P^{-1}Q)^TA(P^{-1}Q)$.
$endgroup$
– vadim123
Jan 31 at 18:08
$begingroup$
Does a method for non-symmetric matrices exist?
$endgroup$
– Marco Villalobos
Jan 31 at 19:05
$begingroup$
How are these matrices “associated with a bilinear application?” If the bilinear form is something like $x^TAx$, then only the symmetric part of $A$ contributes to it, so there’s no a priori reason to believe that there is a solution in the first place. What makes you think that one is possible?
$endgroup$
– amd
Feb 1 at 1:38
|
show 2 more comments
3
$begingroup$
Can you tell about your method and progress for the question?
$endgroup$
– SNEHIL SANYAL
Jan 31 at 17:39
$begingroup$
Making P a matrix with all its integers variables a,b,c,d,e,f,g,h,i and then making matrix multiplication, then solving the system
$endgroup$
– Marco Villalobos
Jan 31 at 18:01
$begingroup$
If $A,B$ were symmetric, then you could orthogonally diagonalize each of them, getting $A=P^TCP$ and $B=Q^TDQ$. Assuming $C=D$ (else there is no solution), then $B=Q^TPAP^TQ=(P^{-1}Q)^TA(P^{-1}Q)$.
$endgroup$
– vadim123
Jan 31 at 18:08
$begingroup$
Does a method for non-symmetric matrices exist?
$endgroup$
– Marco Villalobos
Jan 31 at 19:05
$begingroup$
How are these matrices “associated with a bilinear application?” If the bilinear form is something like $x^TAx$, then only the symmetric part of $A$ contributes to it, so there’s no a priori reason to believe that there is a solution in the first place. What makes you think that one is possible?
$endgroup$
– amd
Feb 1 at 1:38
3
3
$begingroup$
Can you tell about your method and progress for the question?
$endgroup$
– SNEHIL SANYAL
Jan 31 at 17:39
$begingroup$
Can you tell about your method and progress for the question?
$endgroup$
– SNEHIL SANYAL
Jan 31 at 17:39
$begingroup$
Making P a matrix with all its integers variables a,b,c,d,e,f,g,h,i and then making matrix multiplication, then solving the system
$endgroup$
– Marco Villalobos
Jan 31 at 18:01
$begingroup$
Making P a matrix with all its integers variables a,b,c,d,e,f,g,h,i and then making matrix multiplication, then solving the system
$endgroup$
– Marco Villalobos
Jan 31 at 18:01
$begingroup$
If $A,B$ were symmetric, then you could orthogonally diagonalize each of them, getting $A=P^TCP$ and $B=Q^TDQ$. Assuming $C=D$ (else there is no solution), then $B=Q^TPAP^TQ=(P^{-1}Q)^TA(P^{-1}Q)$.
$endgroup$
– vadim123
Jan 31 at 18:08
$begingroup$
If $A,B$ were symmetric, then you could orthogonally diagonalize each of them, getting $A=P^TCP$ and $B=Q^TDQ$. Assuming $C=D$ (else there is no solution), then $B=Q^TPAP^TQ=(P^{-1}Q)^TA(P^{-1}Q)$.
$endgroup$
– vadim123
Jan 31 at 18:08
$begingroup$
Does a method for non-symmetric matrices exist?
$endgroup$
– Marco Villalobos
Jan 31 at 19:05
$begingroup$
Does a method for non-symmetric matrices exist?
$endgroup$
– Marco Villalobos
Jan 31 at 19:05
$begingroup$
How are these matrices “associated with a bilinear application?” If the bilinear form is something like $x^TAx$, then only the symmetric part of $A$ contributes to it, so there’s no a priori reason to believe that there is a solution in the first place. What makes you think that one is possible?
$endgroup$
– amd
Feb 1 at 1:38
$begingroup$
How are these matrices “associated with a bilinear application?” If the bilinear form is something like $x^TAx$, then only the symmetric part of $A$ contributes to it, so there’s no a priori reason to believe that there is a solution in the first place. What makes you think that one is possible?
$endgroup$
– amd
Feb 1 at 1:38
|
show 2 more comments
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3
$begingroup$
Can you tell about your method and progress for the question?
$endgroup$
– SNEHIL SANYAL
Jan 31 at 17:39
$begingroup$
Making P a matrix with all its integers variables a,b,c,d,e,f,g,h,i and then making matrix multiplication, then solving the system
$endgroup$
– Marco Villalobos
Jan 31 at 18:01
$begingroup$
If $A,B$ were symmetric, then you could orthogonally diagonalize each of them, getting $A=P^TCP$ and $B=Q^TDQ$. Assuming $C=D$ (else there is no solution), then $B=Q^TPAP^TQ=(P^{-1}Q)^TA(P^{-1}Q)$.
$endgroup$
– vadim123
Jan 31 at 18:08
$begingroup$
Does a method for non-symmetric matrices exist?
$endgroup$
– Marco Villalobos
Jan 31 at 19:05
$begingroup$
How are these matrices “associated with a bilinear application?” If the bilinear form is something like $x^TAx$, then only the symmetric part of $A$ contributes to it, so there’s no a priori reason to believe that there is a solution in the first place. What makes you think that one is possible?
$endgroup$
– amd
Feb 1 at 1:38