I'm considering a stochastic multivariate process, the stability of which implies that
- all eigenvalues $\lambda_i$, $i = \overline{1,n}$ of a certain square real-valued matrix $A$ lie within the unit circle.
Besides that we know nothing about $A$. But I also need the condition
- $\| A \|_F < 1$
to hold true and I wonder if there's a connection between these 2 conditions. How much more strict would the latter one be?
Best Answer
Let $\lambda$ be an eigenvalue of $A$ and $v$ an associated eigenvector. We may suppose without loss of generalities that $\|v\|_2 =1$. Thus we have $|v_i|\leq 1$ for every $i$ and there exists $j \in \{1,\ldots,n\}$ such that $|v_j|\geq n^{-1/2}$ since $$ 1 = \left(\sum_{k=1}^n |v_k|^2\right)^{1/2} \leq \left(n\max_{k=1,\ldots,n}|v_k|^2 \right)^{1/2}$$ Then we have $$ |\lambda |= |\lambda n^{1/2}||n^{-1/2}|\leq |\lambda n^{1/2}| |v_j| = | n^{1/2}||(Av)_j|=n^{1/2}\left|\sum_{k=1}^nA_{j,k}v_k\right|\leq n^{1/2}\sum_{k=1}^n|A_{j,k}||v_k|\leq n^{1/2}\sum_{k=1}^n|A_{j,k}|\leq n^{1/2}\max_{j=1,\ldots,n}\sum_{k=1}^n|A_{j,k}|.$$
Now, note that by equivalence between matrix norms we get $$n^{1/2}\max_{j=1,\ldots,n}\sum_{k=1}^n|A_{j,k}| = n^{1/2}\|A\|_{\infty} \leq \|A\|_2 \leq \|A\|_F.$$
It follows that your condition $\|A\|_F < 1$ implies that the eigenvalue of $A$ lie in the unit circle. Now, note that the matrix $\frac{2}{3}\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ has one eigenvalue equal to $2/3$. However, its Frobenius norm is $\sqrt{3\left(\frac{2}{3}\right)^2}=\sqrt{\frac{4}{3}}>1$.
EDIT: In the comment, OP asked about a relation of type $\|A\|_F \leq C\rho(A)$ where $\rho(A)$ is the spectral radius of $A$ and $C >0$ may depend on the dimension $n$, i.e. $A \in \Bbb R^{n\times n}$. Here's a proof that such a bound doesn't exist. Let $n =2 $ and consider the matrix $$ A = \begin{pmatrix} 1 & b \\ 1 & 1 \end{pmatrix},$$ Then for $b >1$ we have $\rho(A) = b^{1/2}+1$ and $\|A\|_F = \sqrt{3+b^2}$, suppose such a $C$ exists then we must have $$ C \geq\frac{\sqrt{3+b^2}}{\sqrt{b}+1}\geq \frac{\sqrt{3+b^2}}{2\sqrt{b}}\geq \frac{\sqrt{b^2}}{2\sqrt{b}}=\frac{\sqrt{b}}{2} \qquad \forall b \geq 1.$$ But note that $\lim_{b\to \infty}\frac{\sqrt{b}}{2}=\infty,$ a contradiction.