# Finding a cubic polynomial with Cayley-Hamilton Theorem

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

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

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

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

• 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... – notadoctor Nov 15 '18 at 19:54

Hint: The characteristic polynomials share a common root, $$2$$.

• 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 $0$. – notadoctor Nov 15 '18 at 19:52