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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?

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2 Answers 2

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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

$$ \left( x^{2} - 5 x + 6 \right) $$

$$ \left( x^{2} - 3 x + 2 \right) $$

$$ \left( x^{2} - 5 x + 6 \right) = \left( x^{2} - 3 x + 2 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( - 2 x + 4 \right) $$ $$ \left( x^{2} - 3 x + 2 \right) = \left( - 2 x + 4 \right) \cdot \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x + 3 }{ 2 } \right) }{ \left( \frac{ - x + 1 }{ 2 } \right) } $$ $$ \left( x - 3 \right) \left( \frac{ 1}{2 } \right) - \left( x - 1 \right) \left( \frac{ 1}{2 } \right) = \left( -1 \right) $$ $$ \left( x^{2} - 5 x + 6 \right) = \left( x - 3 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 0 \right) $$ $$ \left( x^{2} - 3 x + 2 \right) = \left( x - 1 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x - 2 \right) } $$ $$ \left( x^{2} - 5 x + 6 \right) \left( \frac{ 1}{2 } \right) - \left( x^{2} - 3 x + 2 \right) \left( \frac{ 1}{2 } \right) = \left( - x + 2 \right) $$

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  • $\begingroup$ Of course it is. Thank you! I saw the shared root but didn't remember that through basic algebra with polynomials I could convince myself that the LCM would also share roots. Whoops... $\endgroup$
    – notadoctor
    Nov 15, 2018 at 19:54
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Hint: The characteristic polynomials share a common root, $2$.

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  • $\begingroup$ Thanks! I agree. I noticed this but I think it was basic algebra tricks I was forgetting (I am studying linear algebra after a 7-year hiatus from mathematics!). I didn't know how to convince myself that the shared root meant a cubic polynomial would evaluate to $0$. $\endgroup$
    – notadoctor
    Nov 15, 2018 at 19:52

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