$\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ has an inverse $\implies a,c$ have inverses?

It is possible that $a$ and $c$ are not invertible.

HINT:

Consider a invertible linear transformation $\phi$ with an invariant subspace $W$ so that the restriction $\phi_{| W}$ is not invertible. (Clearly $W$ has to have an infinite dimension.) For this, take a bijective map $f$ from $\mathbb{Z}$ to $\mathbb{Z}$ with an invariant subset $M$ so that $f_{|M}$ is not surjective. (Eg: $f(z)=z+1$, $M = \mathbb{N}$).Now linearize everything.

In our example $R$ will be the ring of square matrices $(a_{ij})_{ij \in \mathbb{N}\times \mathbb{N}}$ with entries in $\mathbb{R}$ such that in every column there are finitely many non-zero entries. Therefore, $R$ is isomorphic to ring of linear transformations of $\mathbb{R}^{(\mathbb{N})} $.

A more explicit form of the above counter example: Consider a space of bounded one-sided sequences $X=\{(x_0,x_1,x_2,...)\}$ and let $R$ be the ring of bounded linear operators on $X$. Now let $$ a(x_0,x_1,...)=(0,x_0,x_1,...),$$ $$ b(y_0,y_1,...)=(y_0,0,...),$$ $$c(y_0,y_1,...)=(y_1,y_2,...) $$

You may then verify that $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}^{-1} = \begin{pmatrix} c & 0 \\ b & a \end{pmatrix}$. It is a way of decomposing a full shift into one sided shifts.