Proof of global asymptotic stability for dynamical system

Question

I have a question regarding a proof for the following question: Given a dynamical system with the mapping $x_{t+1} = Ax_t$ where A = \begin{bmatrix} \lambda && 1 \newline 0 && \lambda \end{bmatrix}. Prove that if |$\lambda$|<1, Then the fixed point $\tilde{x}$ = 0 is globally asymptotically stable. In order to prove this I know that I need to prove that the fixed point is locally asymptotically stable and that $x_t$ goes to $\tilde{x}$ for all initial points $x_0$. However I encounter the problem that the eigenvalues of the matrix A have algebraic multiplicity of 2 and geometric multiplicity of 1 thus the matrix is not diagonolisable. Can someone help with a proof for this theorem?

Answer 1

If $-1<\lambda < 0$, proof is simple. Assuming $x$ is 1 by 2 vector, vector $x_t$ equals to $$x_t = \{x_{1}\lambda^t, (1 + \lambda)^t(x_1 + \lambda x_2)\}$$ for $t>=0$. Taking $t \to \infty$ gives $$x_t \to \{0,0\}$$ regardless of initial values $x_1, x_2$. However, I don't know how to prove the result for $0 < \lambda < 1$. It seems to me stability should fail because $(1+\lambda) > 1$ implies $\infty$ as the limit. In addition, $\lambda = 0$ will lead to stable system with the limit point $\{0, x_1\}$; global stability here fails because initial condition $x_1$ matters.

Answer 2

The solution to the equation is $x_t=A^tx_0$. We can also write $A=\lambda I+N$ where $N$ is the remaining nilpotent matrix, so $N^2=0$. You can easily verify that $A^t=\lambda^{t} I + t \lambda^{t-1} N$. Since the exponential term converges much faster than $t$ we can conclude that $A^t \to 0$ when $|\lambda|<1$.