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If $A,P$ are two matrices of order $n\times n.$ and

$A^2P=AP^2$ and $A^3=P^3.$Then $|A^2+P^2|=$

what i try

from $A^2P=AP^2$

pre multiply both side by $A$

$A^3P=A^2P^2\Rightarrow P^4=A^2P^2\cdots (1)$

from $A^2P=AP^2$

post multiply both side by $P^2$

$A^2P^3=AP^4\Rightarrow A^5=AP^4\cdots (2)$

How do i solve it Help me please

Community
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jacky
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2 Answers2

2

The condition $A^3=P^3$ implies that $A$ and $P$ are both nonsingular or both singular.

If they are nonsingular, then $A^2P=AP^2$ implies that $A=P$ and you may write $\det(A^2+P^2)$ as $\det(2AP),\,2^n\det(A)^2,\,2^n\det(P)^2,\,\det(AP+PA)$ etc..

Now suppose they are singular. If $0$ is not a semisimple eigenvalue of $P$ (i.e. if its algebraic multiplicity and geometric multiplicity are different), then there exists some vector $x$ such that $Px\ne0$ but $P^2x=0$. So, from the two given conditions, we get $0\ne Px\in\ker(A^2)\cap\ker(P^2)$. Hence $\det(A^2+P^2)=0$.

If $0$ is a semisimple eigenvalue of $P$, then $\ker(P^2)=\ker(P^3)$. In turn, we have $$ P^3\left(\ker(P^2)+\ker(A^2)\right) =P^3\ker(A^2) =A^3\ker(A^2)=0, $$ so that $\ker(P^2)+\ker(A^2)\subseteq\ker(P^3)=\ker(P^2)$. Hence $\ker(A^2)\subseteq\ker(P^2)$. As $A$ is singular, we obtain $\ker(A^2)\cap\ker(P^2)=\ker(A^2)\ne0$. Hence $\det(A^2+P^2)=0$.

user1551
  • 139,064
  • It means $\det(2AP),,\det(2A^2),,\det(2P^2),,2^n\det(A)\det(P),,2^n\det(A)^2$ and $2^n\det(P)^2$ are all correct answers. – user1551 Apr 22 '23 at 07:57
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Note that $P^4=PP^3=PA^3$ and $A^2P^2=(A^2P)P=(AP^2)P=A^4$.

Then $(A^2+P^2)^2=A^4+A^2P^2+P^2A^2+P^4=(A^3+A^3+P^2A+PA^2)A$.

Hence $A^2+P^2$ is singular if $A$ is singular. Otherwise, $A=P$.

In both cases we can write $|A^2+P^2|=2^n|A|^2$.

user1172706
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