# Can we deduce the characteristic polynomial for this matrix?

Given a square $$n \times n$$ matrix $$A$$ that satisfies $$\sum\limits_{k=0}^n a_k A^k = 0$$ for some coefficients $$a_1, a_2, \dots, a_n,$$ can we deduce that its characteristic polynomial is $$\sum\limits_{k=0}^n a_k x^k$$?

• Up to a non-zero constant factor, yes if $a_n\ne 0$, because we have a multiple of the minimal polynomial of $A$ which has the expected degree. – Bernard Jan 12 at 22:48
• I don't believe you can deduce the characteristic polynomial with this information, although you can narrow it down to finitely many possibilities. In my answer below, I have given an example where you can't deduce the minimal polynomial. Please let me know if you have questions. – Andrew Ostergaard Jan 12 at 23:18

You are given a degree $$n$$ polynomial $$p(x)$$ that an $$n\times n$$ matrix $$A$$ satisfies. This is not enough information to find the characteristic polynomial $$c_A(x)$$, although you will be able to narrow it down to finitely many possibilities.

Let's look at an example to see why. Suppose your friend picks the matrix $$A$$. Suppose he doesn't tell you $$A$$, but he does tell you $$p(x)$$ and asks you to guess $$c_A(x)$$. Suppose your friend picked

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

and told you that $$p(x)=x^3-6x^2+11x-6$$.

You could then reason that $$p(x)=(x-1)(x-2)(x-3)$$. Which would mean that the minimal polynomial $$m_A(x)$$ must be one of the following: $$(x-1)$$, $$(x-2)$$, $$(x-3)$$, $$(x-1)(x-2)$$, $$(x-1)(x-3)$$, $$(x-2)(x-3)$$, $$(x-1)(x-2)(x-3)$$.

Hence the characteristic polynomial $$c_A(x)$$ must be one of the following: $$(x-1)^3$$, $$(x-2)^3$$, $$(x-3)^3$$, $$(x-1)^2(x-2)$$, $$(x-1)(x-2)^2$$, $$(x-1)^2(x-3)$$, $$(x-1)(x-3)^2$$, $$(x-2)^2(x-3)$$, $$(x-2)(x-3)^2$$, $$(x-1)(x-2)(x-3)$$.

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

the characteristic polynomial is $$c_A(x)=(x-1)^2(x-2)$$, but you can't prove that, because for all you know he may have picked

$$B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$

which also satisfies $$p(x)=x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$$.

Take matrix $$A = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right)$$ Then $$A^4 - 11 A^3 + 41 A^2 - 61 A + 30 I = 0$$ However, the characteristic polynomial is $$x^4 - 6x^3 + 13x^2 - 12x + 4$$ and the minimal polynomial is $$x^2 - 3 x + 2$$

The zero matrix satisfies every homogeneous polynomial, but clearly not every homogeneous polynomial is the characteristic polynomial of the zero matrix.

Let $$m$$ be the minimal polynomial and $$p$$ be the characteristic polynomial of $$A$$. Then $$m(x)\mid p(x)$$ and $$m(x)\mid q(x)$$, where $$q(x)=\sum a_kx^k$$. The question is whether or not this actually implies that $$p(x)=kq(x)$$ for a constant (nonzero) factor $$k$$, which is clearly untrue. In particular, the polynomial $$p(x)/m(x)$$ can be switched out for anything else and multiplying back by $$m(x)$$ will give you another multiple of $$m(x)$$ with the same degree but is different from a constant multiple of $$p(x)$$.

The result however is true if we assume that $$m(x)$$ has degree $$n$$, since then $$p(x)$$ and $$q(x)$$ will both just be a constant multiple of $$m(x)$$.