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 $2.5 +/- \sqrt({21}/2$

I named the matrix to be solved $C$,

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

$trace(C) =$ 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.

  • 2
    Characteristic polynomial is invariant under similarity, so you made a computational mistake. – RghtHndSd Oct 12 at 0:21
  • 2
    it is pretty easy to guess a 2 by 2 matrix of integers with trace 5 and determinant 1 – Will Jagy Oct 12 at 0:40
up vote 6 down vote accepted

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?

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.