- What is the companion matrix \(M\) of the polynomial \(f=x^{2}-x+2\) over \(C\) ? Prove that \(f\) is the minimal polynomial of \(M\).
Solution
1. For a monic quadratic\[ f(x)=x^2+a_1x+a_0, \]we use the companion-matrix convention\[ C_f=\begin{pmatrix}0&-a_0\\1&-a_1\end{pmatrix}. \]Hence for \(f(x)=x^2-x+2\),\[ M=\begin{pmatrix}0&-2\\1&1\end{pmatrix}. \]
Here \(a_1=-1\) and \(a_0=2\), so substituting into the standard companion form gives the displayed matrix.
2. The characteristic polynomial of \(M\) is \(f\).
We compute \[ \det(xI-M) =\det\begin{pmatrix}x&2\\-1&x-1\end{pmatrix} =x(x-1)+2 =x^2-x+2=f(x). \]
3. The minimal polynomial of \(M\) has degree \(2\).
The vector \(e_1=(1,0)^T\) satisfies \[ Me_1=e_2. \] Thus \(e_1,Me_1\) are linearly independent, so no nonzero polynomial of degree at most \(1\) can annihilate \(M\). Hence the minimal polynomial has degree at least \(2\). Since \(M\) is \(2\times2\), Cayley-Hamilton implies that its minimal polynomial has degree at most \(2\).
4. Therefore the minimal polynomial of \(M\) is \(f(x)=x^2-x+2\).
By 2, \(f(M)=0\). By 3, the monic minimal polynomial has degree \(2\), and it divides the characteristic polynomial \(f\), also monic of degree \(2\). Therefore they are equal.