I have two matrices:
$A = \begin{bmatrix} 1 &2 \\ -1 &4 \end{bmatrix} $
and
$B = \begin{bmatrix} 0 &-1 \\ 2 &3 \end{bmatrix} $
I need to find a monic cubic polynomial $g$ such that $g(A)=g(B)=0$, the zero matrix.
I understand through the Cayley-Hamilton Theorem that both $A$ and $B$ satisfy their own characteristic equations. That is,
$f_A(A)=0\,$ for $\,f_A(t)=t^2-5t+6$
and
$f_B(B)=0\,$ for $\,f_B(t)=t^2-3t+2$
I can verify this computationally or simply by citing Cayley-Hamilton, so that's fine. However, I'm not sure how to combine these and find the cubic polynomial that's been requested so that both matrices evaluate the polynomial to $0$.
Am I missing something deeper about the Cayley-Hamilton Theorem, or am I just forgetting some basic algebra tricks?
the least common multiple of $(t-3)(t-2)$ and $(t-1)(t-2)$ is $$ (t-1)(t-2)(t-3) $$
As with natural numbers, the LCM of two polynomials is their product divided by their gcd. If you did not notice the common root, the euclidean algorithm can find the gcd
$$ \left( x^{2} - 5 x + 6 \right) $$
$$ \left( x^{2} - 3 x + 2 \right) $$
$$ \left( x^{2} - 5 x + 6 \right) = \left( x^{2} - 3 x + 2 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( - 2 x + 4 \right) $$ $$ \left( x^{2} - 3 x + 2 \right) = \left( - 2 x + 4 \right) \cdot \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x + 3 }{ 2 } \right) }{ \left( \frac{ - x + 1 }{ 2 } \right) } $$ $$ \left( x - 3 \right) \left( \frac{ 1}{2 } \right) - \left( x - 1 \right) \left( \frac{ 1}{2 } \right) = \left( -1 \right) $$ $$ \left( x^{2} - 5 x + 6 \right) = \left( x - 3 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 0 \right) $$ $$ \left( x^{2} - 3 x + 2 \right) = \left( x - 1 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x - 2 \right) } $$ $$ \left( x^{2} - 5 x + 6 \right) \left( \frac{ 1}{2 } \right) - \left( x^{2} - 3 x + 2 \right) \left( \frac{ 1}{2 } \right) = \left( - x + 2 \right) $$