Given a discrete-time stable, asymmetric matrix $A$ with maximum eigenvalue strictly less than one in absolute value and a vector $x$, is it necessarily true that $||A^t x|| \leq ||x||$ for all $t\geq 0$?
It is clearly true in the limits, since $A^0 x = x$ and $\lim_{t\rightarrow \infty} A^t = \mathbf 0$ by stability of $A$. But I am not sure if it is true in general for $ 0 < t < \infty$.
If not, is there another bound on the norm of $A^t x$?
On
The solution of discrete linear dynamical system $x_+ = A x$ is given by the discrete series $x_t = A^t x_0$, for $t \in \mathbb{N}_{\geq 0}$. Complying with your conditions, the solution converges to zero due stability. The norm is always bounded by $\lVert x_t \lVert = \lVert A^t \lVert \lVert x_0 \lVert \leq \lVert \lambda_+ \lVert^t \lVert x_0 \lVert$ due triangular inequality .
No. E.g. pick any two negative real numbers $a,b\in(-1,0)$. Then $$ \left\|\pmatrix{a&1\\ 0&b}\pmatrix{0\\ 1}\right\|_2 =\left\|\pmatrix{1\\ b}\right\|_2 >\left\|\pmatrix{0\\ 1}\right\|_2. $$ That the spectral radius of a matrix is smaller than $1$ doesn’t mean that the operator $2$-norm is smaller than $1$.