Non real roots

spix

Let $A\in\mathcal M_2(R)$ and $P\in R[X]$ with no real roots. If $\det(P(A))=0$ then $P(A)=O_2$ .

Can this problem be generalised?

**Posted:** Wed Mar 08, 2006 4:19 am

MoHaMeD

Nope .Example:for $A \in M_n(R)$ that has its minimal $\pi=(X^2+X+1)*(X-1)$ consider $P=X^2+X+1$

**Posted:** Wed Mar 08, 2006 5:29 am

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jmerry

You can't do that- we're restricting to $2\times 2$ matrices.

The generalized version:
Let $F$ be a field, and $p$ be a polynomial over $F$ such that all irreducible factors of $p$ have degree at least $n$ . Then if $M$ is an $n\times n$ matrix over $F$ and $p(M)$ is singular, $p(M)=0$ .

We have to change fields in the generalization because there are no irreducible polynomials with degree greater than 2 over $\mathbb{R}$ .

**Posted:** Wed Mar 08, 2006 12:04 pm

perfect_radio

For the original problem:

Let $\lambda_i, \overline{\lambda_i} \in \mathbb C \setminus \mathbb R$ be the roots of $P$ .

Then $\det P(A) = \prod \det \left( A - \lambda_i I_2 \right) \det \left( A - \overline{\lambda_i} I_2 \right) = 0$ .

Thus, $A$ has a non-real eigenvalue. Since the sum and the product of the eigenvalues are real, it follows that the conjugate of the previous number is also an eigenvalue.

Hence, $P(A) = 0$ , right?

**Posted:** Thu Mar 09, 2006 3:00 pm

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spix

I don`t quite understand the last part of your solution... here`s mine:
Let $\alpha_1,\alpha_2$ be the eigenvalues of $A$ . Then the eigenvalues of $P(A)$ are $P(\alpha_1)$ and $P(\alpha_2)$ .
So $\det(P(A))=P(\alpha_1)*P(\alpha_2)=0$ wich implies $\alpha_1,\alpha_2$ are roots of $P$ and they are conjugated. But $\alpha_1,\alpha_2$ are roots of $P_A$ so $P_A/P$ equivalent to $P(X)=P_A(X)Q(X)$ and for $X=A$ we get $P(A)=O_2$ .

**Posted:** Fri Mar 10, 2006 12:08 am

perfect_radio

**Posted:** Fri Mar 10, 2006 1:29 am

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

spix

$P(A)$ is a $2x2$ matrix so it has two eigenvalues. There is nothing wrong it`s the same ideea... I did`nt understand how you jumped to that conclusion in the end... but now it`s clear to me... you just said a little more straightforward than me. Basically the solutions are the same... Smile

**Posted:** Fri Mar 10, 2006 1:34 am

Moubinool

**Posted:** Fri Mar 10, 2006 2:15 pm