# Finding a matrix given its characteristic polynomial

I am asked to find a $$2 \times 2$$ matrix with real and whole entries given it's characteristic polynomial:

$$p^2 -5p +1.$$

This is what I have done thus far:

I equated the polynomial to zero, and the roots (eigenvalues) were found to be $$2.5 \pm \sqrt{21}/2$$

I named the matrix to be solved $$C$$,

so $$\det(C) =$$ product of eigenvalues $$= 1$$

$$\text{trace}(C) =$$ sum of eigenvalues $$=5$$

I then tried to find C by solving $$T^{-1} \times D \times T$$, where $$D$$ is a matrix whose diagonal entries are the eigenvalues solved above, and $$T$$ is any matrix who's determinant is non zero.

I used $$T$$ as a $$2 \times 2$$ being

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.$$

I solved $$T^{-1} \times D \times T$$, and threw it in a calculator to make sure I made no algebraic mistakes, but the answer I received is wrong.

I appreciate any help, thank you.

• Characteristic polynomial is invariant under similarity, so you made a computational mistake. Commented Oct 12, 2018 at 0:21
• it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1 Commented Oct 12, 2018 at 0:40

Let's start off by looking at the characteristic polynomial of the $$2 \times 2$$ matrix

$$A = \begin{bmatrix} 0 & 1 \\ -a & -b \end{bmatrix}: \tag 1$$

$$\det (A - \lambda I) = \det \left (\begin{bmatrix} -\lambda & 1 \\ -a & -b - \lambda \end{bmatrix} \right ) = \lambda^2 + b\lambda + a; \tag 2$$

we see from (2) that we may always present a $$2 \times 2$$ matrix with given characteristic polynomial $$\lambda^2 + b\lambda + a$$ in the form $$A$$; for example, if he quadratic is $$\lambda^2 + 5\lambda + 1$$, as in the present problem (tho' I have replaced $$p$$ with $$\lambda$$), we may take

$$P = \begin{bmatrix} 0 & 1 \\ -1 & -5 \end{bmatrix}, \tag 3$$

which may be easily checked:

$$\det(P - \lambda I) = -\lambda(-5 - \lambda) - 1 ( -1) = \lambda^2 + 5\lambda + 1. \tag 4$$

The matrices $$A$$ and $$P$$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $$X$$ and $$S^{-1}XS$$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If

$$q(\lambda) = \displaystyle \sum_1^n q_i \lambda^i, \; q_n = 1, \tag5$$

we define $$C(q(\lambda))$$ to be the $$n \times n$$ matrix

$$C(q(\lambda)) = \begin{bmatrix} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ldots & \vdots & \vdots \\ -q_0 & -q_1 & -q_2 & \ldots & -q_{n - 1} \end{bmatrix}; \tag 6$$

that is, $$C(q(\lambda))$$ has all $$1$$s on the superdiagonal, the negatives of the coefficients of $$q(\lambda)$$ on the $$n$$-th row, and $$0$$s everywhere else. We have

$$C(q(\lambda) - \lambda I = \begin{bmatrix} -\lambda & 1 & 0 & \ldots & 0 \\ 0 & -\lambda & 1 & \ldots & 0 \\ \vdots & \vdots & \ldots & \vdots & \vdots \\ -q_0 & -q_1 & -q_2 & \ldots & -q_{n - 1} - \lambda \end{bmatrix}; \tag 7$$

it is easy to see, by expanding in minors along the $$n$$-th row, that

$$\det(C(q(\lambda)) - \lambda I) = q(\lambda); \tag 8$$

also, since a matrix and its transpose have equal determinants, the transposed form $$C(q(\lambda))$$, $$C^T(q(\lambda))$$, also gives rise to the same characteristic polynomial. The matrices $$C(q(\lambda))$$ and $$C^T(q(\lambda))$$ are known as companion matrices for the polynomial $$q(\lambda)$$; the linked article has more of the story.

Your choice of $$T$$ would not lead to whole entries.

Guide:

Let $$C=\begin{bmatrix}a & b \\ c & d \end{bmatrix}$$.

We need $$a+d=5$$ and $$ad-bc=1$$

You can let $$a=0$$, then we end up having $$-bc=1$$. Can you choose some possible values of $$b$$ and $$c$$ to let it satisfy the equation?