# Characteristic Polynomial of A and A^2

• estro
In summary, the characteristic polynomial of a real matrix [3x3] A can be used to find the characteristic polynomial of A^2 by looking at the roots of the polynomial. This can also be done using eigenvalues, and it applies to any mxm matrix. Even if the polynomial cannot be shown as a multiplication of linear terms, this method still works.
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?

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

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

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.

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

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

Last edited:

## 1. What is the characteristic polynomial of a matrix A?

The characteristic polynomial of a matrix A is a polynomial equation in the variable λ, where A is a square matrix of size n × n. It is defined as det(λI − A), where I is the identity matrix of size n × n. The characteristic polynomial gives important information about the eigenvalues of A.

## 2. How is the characteristic polynomial of A related to its eigenvalues?

The roots of the characteristic polynomial of A are the eigenvalues of A. This means that by solving the characteristic polynomial, we can find the eigenvalues of A. Furthermore, the multiplicity of each eigenvalue in the characteristic polynomial corresponds to the algebraic multiplicity of that eigenvalue.

## 3. What is the significance of the characteristic polynomial of A?

The characteristic polynomial of A is important because it helps us understand the behavior of A as a linear transformation. It also helps in finding the eigenvalues and eigenvectors of A, which are essential in various applications such as solving systems of linear equations, diagonalizing matrices, and understanding the stability of dynamic systems.

## 4. How is the characteristic polynomial of A^2 related to the characteristic polynomial of A?

The characteristic polynomial of A^2 is the square of the characteristic polynomial of A. This means that the roots of the characteristic polynomial of A^2 are the squares of the eigenvalues of A. Additionally, the coefficients of the characteristic polynomial of A^2 can be expressed in terms of the coefficients of the characteristic polynomial of A.

## 5. Can the characteristic polynomial of A^2 be used to find the eigenvalues of A?

Yes, the characteristic polynomial of A^2 can be used to find the eigenvalues of A. This is because the eigenvalues of A are the square roots of the roots of the characteristic polynomial of A^2. However, this method may not be as efficient as directly solving the characteristic polynomial of A. It is mostly used when A is a complex matrix or when the characteristic polynomial of A is difficult to solve.

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