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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

$$beginbmatrix 1 & 1 \ 0 & 1 endbmatrix.$$

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.

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

Guide:

Let $C=beginbmatrixa & b \ c & d endbmatrix$.

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?

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

$A = beginbmatrix 0 & 1 \ -a & -b endbmatrix: tag 1$

$det (A - lambda I) = det left (beginbmatrix -lambda & 1 \ -a & -b - lambda endbmatrix right ) = lambda^2 + blambda + a; tag 2$

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

$P = beginbmatrix 0 & 1 \ -1 & -5 endbmatrix, tag 3$

which may be easily checked:

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

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 $S^-1XS$ 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

$C(q(lambda)) = beginbmatrix 0 & 1 & 0 & ldots & 0 \ 0 & 0 & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 endbmatrix; tag 6$

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

$C(q(lambda) - lambda I = beginbmatrix -lambda & 1 & 0 & ldots & 0 \ 0 & -lambda & 1 & ldots & 0 \
vdots & vdots & ldots & vdots & vdots \ -q_0 & -q_1 & -q_2 & ldots & -q_n - 1 - lambda endbmatrix; tag 7$

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.