Characteristic Polynomial of A and A^2

1. Jul 11, 2011

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?

2. Jul 11, 2011

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 nth 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:
$$p(x) = (x-x_1^n)\cdot (x-x_2^n) \cdot (x-x_3^n) \cdots (x-x_m^n)$$

3. Jul 11, 2011

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.

4. Jul 11, 2011

lanedance

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

5. Jul 11, 2011

estro

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

Last edited: Jul 11, 2011