6) Over the field of real numbers, find the rational canonical form of the matrix
\[ A =
\begin{pmatrix}
-1 & -9 & 0 \\
1 & 5 & 0 \\
1 & 3 & 2
\end{pmatrix}.
\]
Proof. First find the eigenvalues as the roots of the characteristic polynomial \( c_A(x) = \) \({\text{det}(xI - A)} \).
\[ c_A(x) = {\text{det}(xI - A)} = \begin{vmatrix}
x+1 & 9 & 0 \\
-1 & x-5 & 0 \\
-1 & -3 & x-2
\end{vmatrix} = {(x-2) \cdot \begin{vmatrix}
x+1 & 9 \\
-1 & x-5
\end{vmatrix}}.
\]
\[ \therefore c_A(x) = {(x-2) \big((x+1)(x-5) + 9 \big)} = {(x-2)(x^2-4x + 4)} = {(x-2)^3}. \]
So \( \lambda = 2 \) is the only eigenvalue. We'll need to know the minimal polynomial \( m_A(x) \) as well, which, as a divisor of
\( c_A(x), \) must be \( x - 2 \) or \( (x - 2)^2 \) or \( (x - 2)^3 \). The minimal polynomial is the polynomial of least degree
such that \( m_A(x) = 0 \). Obviously \( m_A(x) \neq x - 2, \) since otherwise \( A = 2I. \) A direct calculation shows that
\( {(A - 2I)^2} = 0, \) so \( m_A(x) = {(x - 2)^2}. \)
\( m_A(x) \) is the largest invariant factor and \( c_A(x) \) is the product of all the invariant factors; it follows that the
invariant factors are exactly \( a_1(x) = {x - 2} \) and \( a_2(x) = {(x - 2)^2} = {x^2 - 4x + 4}. \) The companion matrices for these two invariant factors
are:
\[ \mathcal{C}_{a_1(x)} = (2), \hspace{20pt} \mathcal{C}_{a_2(x)} = {\begin{pmatrix} 0 & -4 \\ 1 & 4 \end{pmatrix}}. \]
Run these two companion matrices down the diagonal, filling with zeroes elsewhere, to produce the rational canonical form:
\[ \text{RCF}(A) = {\left( \begin{array} {c|cc}
2 & 0 & 0 \\ \hline
0 & 0 & -4 \\
0 & 1 & 4
\end{array} \right)}
\]
QED.