Characteristic Polynomial

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March 29th 2010, 06:15 PM
joe909

Characteristic Polynomial

Given that the characteristic polynomial of $A$ is $t^4 + t + 1$ . How would I go about calculating the characteristic polynomial of $A^2$ ?

I originally thought of using the fact that the eigenvalues of $A^2$ would simply be the square of the eigen values of $A$ . But unfortunately I couldnt calculate the eigenvalues of A.

Any help would be great, thanks!
March 30th 2010, 03:02 AM
tonio

Quote:

Originally Posted by joe909

Given that the characteristic polynomial of $A$ is $t^4 + t + 1$ . How would I go about calculating the characteristic polynomial of $A^2$ ?

I originally thought of using the fact that the eigenvalues of $A^2$ would simply be the square of the eigen values of $A$ . But unfortunately I couldnt calculate the eigenvalues of A.

Any help would be great, thanks!

Let $\lambda_i\,,\,i=1,2,3,4$ the eigenvalues of $A\Longrightarrow \lambda_i^2$ are the eigenvalues of $A^2$ . Now, applying Viete formulae we get:

Coefficient of $t^3=-\sum^4_{i=1}\lambda_i=0$

Coefficient of $t^2=\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j=0$

Coefficient of $t=-\!\!\!\!\!\sum^4_{i,j,k=1\,,\,i>j>k}\lambda_i\lamb da_j\lambda_k=1$

Free coefficient $=\prod^4_{i=1}\lambda_i=1$ .

Well, use the above relations for $\lambda_i^2$ and a little algebra to find out the coefficients of the char. pol. of $A^2$ . For example, it may interest you to use

$\sum^4_{i=1}\lambda_i^2=\left(\sum^4_{i=1}\lambda_ i\right)^2-2\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j$ ...
March 30th 2010, 07:03 AM
Opalg

I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, $A^4+A+I=0$ . Therefore $A = -(A^4+I)$ and so $A^2 = (A^4+I)^2 = A^8+2A^4+I$ . Hence $A^8 +2A^4-A^2+I = 0$ , which says that $A^2$ satisfies the polynomial equation $t^4+2t^2-t+1=0$ . So I assume that $t^4+2t^2-t+1$ is the characteristic polynomial of $A^2$ . But I don't know whether that conclusion is justified. (Wondering)
March 30th 2010, 08:21 AM
tonio

Quote:

Originally Posted by Opalg

I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, $A^4+A+I=0$ . Therefore $A = -(A^4+I)$ and so $A^2 = (A^4+I)^2 = A^8+2A^4+I$ . Hence $A^8 +2A^4-A^2+I = 0$ , which says that $A^2$ satisfies the polynomial equation $t^4+2t^2-t+1=0$ . So I assume that $t^4+2t^2-t+1$ is the characteristic polynomial of $A^2$ . But I don't know whether that conclusion is justified. (Wondering)

It looks fine to me: it is of the correct degree, monic and vanishes at $A^2$ ...good!

Tonio