91Ó°ÊÓ

Consider an n ×n matrix A and a scalar λ. Then λ is an eigenvalue5 of A.

$\mathbf{det}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{λ\pm ô}\mathbf{)}\mathbf{=}\mathbf{0}$

This is called the characteristic equation (or the secular equation) of matrix A.

We solve:

$$\begin{gathered} \det (\mathrm{A}-\mathrm{λ\pm ô})=0 \\ \left|\begin{array}{ccc}-\mathrm{λ}& 0& \mathrm{a}\\ 1& -\mathrm{λ}& 0\\ 0& 1& -\mathrm{λ}\end{array}\right|=0 \\ -{\mathrm{λ}}^3+\mathrm{a}=0 \end{gathered}$$

Let$\mathrm{g}\left(\mathrm{λ}\right)=-{\mathrm{λ}}^3+3\mathrm{λ}$. The function has local extremes at g(-1) = - 2 and g(1) = 2.

So, for$a∉\left[-2,2\right]$, we will only have one real eigenvalue (and other two complex), and it will only be of algebraic multiplicity 1, thuswill not be diagonalizable.

For $\mathrm{a}=Â\pm 2$, we will have two distinct real eigenvalues, one of which will have an algebraic multiplicity of 2. That one will be $∓1$, and in either case, it will apply (A-l) = {0} , so will not be diagonalizable.

For , we will have three distinct real eigenvalues on a $3×3$matrix, therefore will be diagonalizable

Here, the final result is $3×3$matrix, therefore will be diagonalizable.