value of $det(P^2+Q^2)$ [closed]
0
$begingroup$
If $P$ and $Q$ be $3times 3$ matrices and $Pneq Q. $
If $P^3=Q^3$ and $P^2Q=Q^2P.$ Then $det(P^2+Q^2)$ is
Try: From
$P^3-Q^3=ORightarrow (P-Q)(P^2+PQ+Q^2)=O$
So either $P=Q$ or either $P^2+PQ+Q^2=O$
So $P^2+Q^2=-PQ$, Now i did not understand how
i use $P^2Q=Q^2P$ and find $det(P^2+Q^2)$
Could some help me how to find it, thanks
determinant
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
$endgroup$
closed as off-topic by Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy Dec 18 '18 at 21:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
If $P$ and $Q$ be $3times 3$ matrices and $Pneq Q. $
If $P^3=Q^3$ and $P^2Q=Q^2P.$ Then $det(P^2+Q^2)$ is
Try: From
$P^3-Q^3=ORightarrow (P-Q)(P^2+PQ+Q^2)=O$
So either $P=Q$ or either $P^2+PQ+Q^2=O$
So $P^2+Q^2=-PQ$, Now i did not understand how
i use $P^2Q=Q^2P$ and find $det(P^2+Q^2)$
Could some help me how to find it, thanks
determinant
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
$endgroup$
closed as off-topic by Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy Dec 18 '18 at 21:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hint: $det(AB) = det(A)det(B)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:32
1
$begingroup$
@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$
$endgroup$
– DXT
Dec 18 '18 at 15:34
$begingroup$
consider $det(P^2Q) = det(Q^2P)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:36
add a comment |
0
0
0
$begingroup$
If $P$ and $Q$ be $3times 3$ matrices and $Pneq Q. $
If $P^3=Q^3$ and $P^2Q=Q^2P.$ Then $det(P^2+Q^2)$ is
Try: From
$P^3-Q^3=ORightarrow (P-Q)(P^2+PQ+Q^2)=O$
So either $P=Q$ or either $P^2+PQ+Q^2=O$
So $P^2+Q^2=-PQ$, Now i did not understand how
i use $P^2Q=Q^2P$ and find $det(P^2+Q^2)$
Could some help me how to find it, thanks
determinant
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
$endgroup$
If $P$ and $Q$ be $3times 3$ matrices and $Pneq Q. $
If $P^3=Q^3$ and $P^2Q=Q^2P.$ Then $det(P^2+Q^2)$ is
Try: From
$P^3-Q^3=ORightarrow (P-Q)(P^2+PQ+Q^2)=O$
So either $P=Q$ or either $P^2+PQ+Q^2=O$
So $P^2+Q^2=-PQ$, Now i did not understand how
i use $P^2Q=Q^2P$ and find $det(P^2+Q^2)$
Could some help me how to find it, thanks
determinant
determinant
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
asked Dec 18 '18 at 15:30
DXTDXT
5,9022731
5,9022731
closed as off-topic by Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy Dec 18 '18 at 21:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy Dec 18 '18 at 21:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lord_Farin, Don Thousand, Brian Borchers, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hint: $det(AB) = det(A)det(B)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:32
1
$begingroup$
@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$
$endgroup$
– DXT
Dec 18 '18 at 15:34
$begingroup$
consider $det(P^2Q) = det(Q^2P)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:36
add a comment |
1
$begingroup$
Hint: $det(AB) = det(A)det(B)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:32
1
$begingroup$
@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$
$endgroup$
– DXT
Dec 18 '18 at 15:34
$begingroup$
consider $det(P^2Q) = det(Q^2P)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:36
1
1
$begingroup$
Hint: $det(AB) = det(A)det(B)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:32
$begingroup$
Hint: $det(AB) = det(A)det(B)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:32
1
1
$begingroup$
@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$
$endgroup$
– DXT
Dec 18 '18 at 15:34
$begingroup$
@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$
$endgroup$
– DXT
Dec 18 '18 at 15:34
$begingroup$
consider $det(P^2Q) = det(Q^2P)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:36
$begingroup$
consider $det(P^2Q) = det(Q^2P)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:36
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Some hints:
- Establish the equality $left(P^2+Q^2right)left(P-Qright)=0$.
- Using $det(AB)=det(A)det(B)$, we derive that either $detleft(P^2+Q^2right)=0$ or $detleft(P-Qright)=0$.
- Suppose that $detleft(P^2+Q^2right)neq 0$. Then $P^2+Q^2$ is invertible: left-multiply by $left(P^2+Q^2right)^{-1}$ in equality $left(P^2+Q^2right)left(P-Qright)=0$ to reach a contradiction.
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Some hints:
- Establish the equality $left(P^2+Q^2right)left(P-Qright)=0$.
- Using $det(AB)=det(A)det(B)$, we derive that either $detleft(P^2+Q^2right)=0$ or $detleft(P-Qright)=0$.
- Suppose that $detleft(P^2+Q^2right)neq 0$. Then $P^2+Q^2$ is invertible: left-multiply by $left(P^2+Q^2right)^{-1}$ in equality $left(P^2+Q^2right)left(P-Qright)=0$ to reach a contradiction.
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
add a comment |
2
$begingroup$
Some hints:
- Establish the equality $left(P^2+Q^2right)left(P-Qright)=0$.
- Using $det(AB)=det(A)det(B)$, we derive that either $detleft(P^2+Q^2right)=0$ or $detleft(P-Qright)=0$.
- Suppose that $detleft(P^2+Q^2right)neq 0$. Then $P^2+Q^2$ is invertible: left-multiply by $left(P^2+Q^2right)^{-1}$ in equality $left(P^2+Q^2right)left(P-Qright)=0$ to reach a contradiction.
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
add a comment |
2
2
2
$begingroup$
Some hints:
- Establish the equality $left(P^2+Q^2right)left(P-Qright)=0$.
- Using $det(AB)=det(A)det(B)$, we derive that either $detleft(P^2+Q^2right)=0$ or $detleft(P-Qright)=0$.
- Suppose that $detleft(P^2+Q^2right)neq 0$. Then $P^2+Q^2$ is invertible: left-multiply by $left(P^2+Q^2right)^{-1}$ in equality $left(P^2+Q^2right)left(P-Qright)=0$ to reach a contradiction.
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
Some hints:
- Establish the equality $left(P^2+Q^2right)left(P-Qright)=0$.
- Using $det(AB)=det(A)det(B)$, we derive that either $detleft(P^2+Q^2right)=0$ or $detleft(P-Qright)=0$.
- Suppose that $detleft(P^2+Q^2right)neq 0$. Then $P^2+Q^2$ is invertible: left-multiply by $left(P^2+Q^2right)^{-1}$ in equality $left(P^2+Q^2right)left(P-Qright)=0$ to reach a contradiction.
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
edited Dec 18 '18 at 15:51
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
answered Dec 18 '18 at 15:39
Davide GiraudoDavide Giraudo
127k16151265
127k16151265
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
add a comment |
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31
add a comment |
1
$begingroup$
Hint: $det(AB) = det(A)det(B)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:32
1
$begingroup$
@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$
$endgroup$
– DXT
Dec 18 '18 at 15:34
$begingroup$
consider $det(P^2Q) = det(Q^2P)$
$endgroup$
– Stockfish
Dec 18 '18 at 15:36