Characteristic Polynomial of A and A^2

[estro]

Hello,
I'm trying to figure out connection between the characteristic polynomials for real matrices [3x3] and their powers.

Suppose A is a real matrix [3x3] which's c.p is t^3+t^2+t-3, how can i find the c.p. of A^2.
Now suppose p(t)=a_1t^3+a_2t^2+a_3t+a_4

Right away I can know that a_1=1 and a_4=det(A)*det(A)=9.

But what can I do about a_2 and a_3?

 

[thegreenlaser]

I think it's easier to focus on the roots of the polynomial, which should be fine since that tells you pretty much everything you need to know about the polynomial anyway. The characteristic polynomial of A is det(A-xI), correct? Say the roots of that are x=x₁, x₂, and x₃. Now look at the characteristic polynomial of A²:
$$\det (A^2 - \lambda I) = \det \left( \left(A-\sqrt{\lambda}\right) \cdot \left(A+\sqrt{\lambda}\right) \right) = \det \left(A-\sqrt{\lambda}\right) \cdot \det \left(A+\sqrt{\lambda}\right)$$
But, you already know the roots of det(A-xI), so the roots of the above equation just come from the equation
$$\pm \sqrt{\lambda}=x \Longrightarrow \lambda = x^2$$
So in this case, the roots of the characteristic polynomial of A² would be x₁², x₂², and x₃²
So if the characteristic polynomial for A looked like this:
$$p(x) = (x-x_1)(x-x_2)(x-x_3)$$
then the characteristic polynomial for A² would look like:
$$p_2(x) = (x-x_1^2)(x-x_2^2)(x-x_3^2)$$
In fact, the characteristic polynomial for Aⁿ is:
$$p_n(x) = (x-x_1^n)(x-x_2^n)(x-x_3^n)$$

Now, using eigenvalues rather than determinants, this property is actually easier to derive. You know that if x is a root of the characteristic polynomial of A, then x is an eigenvalue, so there exists a non-zero vector v such that:
$$A\vec{v}=x\vec{v}$$
Left multiply both sides by A to get
$$A^2\vec{v}=x(A\vec{v})$$
But you know that Av=xv, so:
$$A^2\vec{v} = x(x\vec{v}) = x^2\vec{v}$$
That tells you that x² is an eigenvalue of A², and you can easily show by induction (just left multiply by A, n times) that xⁿ is an eigenvalue of Aⁿ. Since an eigenvalue is a root of the characteristic polynomial, xⁿ must be a root of the characteristic polynomial of Aⁿ. So once you know the characteristic polynomial of A, you know that the roots of the characteristic of Aⁿ are the roots of the characteristic of A raised to the n^th power.

That may or may not translate to something nice in the p(x) = ax³+bx²+cx+d form (I'll let you try that... ), but it tells you something very nice about the roots which is actually true for any mxm matrix, not just a 3x3 matrix. If you can turn the characteristic of an mxm A into the form:
$$p(x) = (x-x_1)\cdot (x-x_2) \cdot (x-x_3) \cdots (x-x_m)$$
Then you immediately know that the characteristic of Aⁿ is:
$$p(x) = (x-x_1^n)\cdot (x-x_2^n) \cdot (x-x_3^n) \cdots (x-x_m^n)$$

Hope that helps answer your question, even if I didn't answer it directly.

 

[estro]

Thank you for your detailed response, the problem is that this c.p. cano't be shown as a multiplication of linear terms, and there is only 1 real eigenvalue.

I'll try to put more thought on you suggestions in case I missed something in your hint.

[EDIT]
On second thought I don't see any reason why not applying your idea for all the eigenvalues even if the're not real.

 

[lanedance]

nise post greenlaser - should be no problem expressing the multiplication as complex numbers

 

[estro]

Thank you very much guys I was able to comprehend the idea...=)