Question

Show that when you plug a matrix into its own characteristic polynomial you get the zero vector.

Show that the determinant of a matrix exponential is equal the exponential of the trace of the matrix.

Answer 1

For example, \[A=\left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]\] has characteritic polynomial \[\det\left[\begin{matrix}1- \lambda & 2 \\ 3 & 4-\lambda\end{matrix}\right]=\lambda^2-5\lambda -2\] So if we plug in A for lambda we get: \[A^2-5A -2 = \left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]^2-5\left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]-2\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\] \[\left[\begin{matrix}7 & 10 \\ 15 & 22\end{matrix}\right]+\left[\begin{matrix}-5 & -10 \\ -15 & -20\end{matrix}\right]+\left[\begin{matrix}-2 & 0 \\ 0 & -2\end{matrix}\right] = \left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right] \]
Interesting!

Answer 2

http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem

Answer 3

The proof of the theorem is complicated.

Answer 4

I was looking at this but I think I see no reason why any of their justifications make this proof incorrect @Alchemista

\[\Large p(\lambda) = |A-\lambda I| \\ \Large p(A) = |A-AI| \\ \Large p(A)= |0|=0\]