value of $det(P^2+Q^2)$ [closed]

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

asked Dec 18 '18 at 15:30

DXT

5,7062731

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 |

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

asked Dec 18 '18 at 15:30

DXT

5,7062731

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 |

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

asked Dec 18 '18 at 15:30

DXT

5,7062731

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

DXT

5,7062731

asked Dec 18 '18 at 15:30

DXT

5,7062731

asked Dec 18 '18 at 15:30

DXT

5,7062731

asked Dec 18 '18 at 15:30

DXT

5,7062731

asked Dec 18 '18 at 15:30

DXT

5,7062731

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

Hint: $det(AB) = det(A)det(B)$

– Stockfish
Dec 18 '18 at 15:32

@Stockfish how i find $det(P)det(Q)$ from $P^2Q=Q^2P$

– DXT
Dec 18 '18 at 15:34

consider $det(P^2Q) = det(Q^2P)$

– Stockfish
Dec 18 '18 at 15:36

add a comment |

1 Answer
1

active

oldest

votes

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.

edited Dec 18 '18 at 15:51

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

$begingroup$
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

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.

edited Dec 18 '18 at 15:51

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31

add a comment |

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.

edited Dec 18 '18 at 15:51

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

$begingroup$
You don't need the second bullet point.
$endgroup$
– Robert Israel
Dec 18 '18 at 16:31

add a comment |

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.

edited Dec 18 '18 at 15:51

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

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.

edited Dec 18 '18 at 15:51

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

edited Dec 18 '18 at 15:51

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

answered Dec 18 '18 at 15:39

Davide Giraudo

127k16151264

$begingroup$
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

You don't need the second bullet point.

– Robert Israel
Dec 18 '18 at 16:31

add a comment |

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