# Banach-Steinhaus theorem for finite dimensional space

I am wondering if there is a theorem in matrix theory as an analog to Banach-Steinhaus theorem. Here are some attempts. Suppose $$D_n\in\mathbb{R^{d\times d}}$$, $$n=1,2,\ldots$$ is a sequence of matrix and for every $$x\in\mathbb{R}^d$$, there exists some constant $$c_x>0$$, such that $$\|D_n x\|\leq c_x$$. From Banach-Steinhaus theorem, there exists a constant $$C>0$$ such that $$\|D_n\|\leq C$$.

1. Is the condition “$$\sup\limits_n\|D_n x\|<\infty$$” equivalent to “the eigenvalues of $$D_n$$ has a uniform bound”?
2. For finite dimensional case, is there any simple proof for Banach-Steinhaus theorem?

Thanks.

In any case, having a uniform bound on the eigenvalues of the $$D_n$$ is weaker than being pointwise bounded. For instance, consider the $$2\times 2$$ matrices $$D_n=\begin{pmatrix}1 & n \\ 0 & 0\end{pmatrix}.$$ These matrices are all projections so their eigenvalues are only $$0$$ and $$1$$ but $$D_nx$$ is unbounded for $$x=(0,1)$$.
In finite dimensions, the Banach-Steinhaus theorem is essentially trivial. For instance, a uniform bound on $$\|D_n e_i\|$$ where $$e_i$$ is the $$i$$th standard basis vector gives a uniform bound on the entries of the $$i$$th column of $$D_n$$. Taking this for $$i=1,\dots,d$$, you get a uniform bound on all the entries of $$D_n$$ and hence on its norm (using whatever norm you want since all norms are equivalent in finite dimensions).