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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 $A^t x$?

phil
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2 Answers2

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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 .

Bruno Lobo
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  • Thank you. I just was not sure if this was true when $A$ is asymmetric – phil Aug 25 '23 at 23:15
  • This is not a necessary sufficient. – Bruno Lobo Aug 25 '23 at 23:16
  • Would you please approve the answer? – Bruno Lobo Aug 25 '23 at 23:19
  • The matrix norm is an upper bound for the spectral radius, not the other way around. Eigenvalues with multiplicity provide counter-examples. – Lutz Lehmann Aug 26 '23 at 04:26
  • @LutzLehmann Does this explain why the answer is correct? Because if $\rho$ is spectral radius, then this would mean that $||Ax||\leq\rho (A) ||x|| \leq ||A|| ||x||$? Or am I understanding your comment incorrectly – phil Aug 26 '23 at 06:16
  • Never mind, there is a counterexample in the other answer – phil Aug 26 '23 at 06:17
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    I said that $\rho(A)\le|A|$. There is a theorem that says that for any $\epsilon$ one can find a vector norm so that in the induced operator norm also $|A|\le\rho(A)+\epsilon$. But that is not helpful for the claim, as that is wrong if the norms are fixed, not adapted to $A$. – Lutz Lehmann Aug 26 '23 at 07:21
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    That said, this is sad. :-( Thanks, @LutzLehmann – Bruno Lobo Aug 26 '23 at 11:15
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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 $1$.

user1551
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  • There must be some convenient bound in terms of $||x||$ and $||A||$, no? – phil Aug 26 '23 at 06:19
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    @phil Yes. E.g. there is some constant $C>0$ and some $r$ such that $\rho(A)\le r<1$ and $|A^kx|\le Cr^k|x|$ for every integer $k\ge0$. However, the constant $C$ can be very large. If you want a practical upper bound to control the size of $|A^kx|$, you need to find a more refined one. – user1551 Aug 26 '23 at 06:58
  • Thank you, this is very helpful. Could you point me toward a source for this bound or some materials that might help me understand how to go about refining it? I'm very interested in how something like this would be obtained – phil Aug 27 '23 at 05:53
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    @phil The bound in my previous comment or similar ones can be found in this answer and this recent answer. I am not sure which books have intensive discussions on this kind of bounds, but you may try those books on matrix analysis, such as Horn and Johnson’s Matrix Analysis, the book with the same title by Bhatia, or Meyer’s Matrix Analysis and Applied Linear Algebra books. Books on numerical linear algebra may also cover this kind of bounds too. – user1551 Aug 27 '23 at 08:38
  • thank you! I will check out these resources – phil Aug 28 '23 at 17:42