Showing invertibility of bounded operators. 
Let $S$ and $T$ be two bounded linear operators on a complex separable Hilbert space $\mathcal H$ satisfying the following relations $:$
$(1)$ $SS^{\ast} + TT^{\ast} = 1,$
$(2)$ $S^{\ast} S + T^{\ast} T = 1,$
$(3)$ $ST + TS = 0,$
$(4)$ $T^{\ast} T - TT^{\ast} = 0,$
$(5)$ $ST^{\ast} + T^{\ast} S = 0.$
Let $A$ be the $C^{\ast}$-subalgebra of $\mathcal L (\mathcal H)$ generated by $S$ and $T.$
Assume
$(a)$ $\dim \mathcal H \gt 1,$ and
$(b)$ there is no non trivial proper closed subspace $\mathcal H_{0}$ of $\mathcal H$ such that $A \mathcal H_{0} \subseteq \mathcal H_{0}.$
Show that
$(a)$ $S$ and $T$ are invertible,
$(b)$ there exist a real $t \in (0,1)$ and two unitaries $U$ and $V$ such that $S = t U,$ $T = \sqrt {1-t} V$ and $UV + VU = 0.$
$(c)$ $\dim \mathcal H = 2.$

The above question appeared in an entrance exam. I can see that both $S$ and $T$ are necessarily normal. But I have no idea about how to show the invertibility of $S$ and $T.$ Any suggestion would be greatly appreciated.
 A: Since $A$ acts irreducible on $H$ by (b), we have
$$
A'=\{R\in L(H)\mid RX=XR\text{ for all }X\in A\}=\mathbb C I
$$
by Schur's lemma. Direct calculations using (3)-(5) show that $S^\ast S,T^\ast T\in A'$. Combined with (2) this implies $S^\ast S=tI$, $T^\ast T=(1-t)I$ for some $t\in[0,1]$. If $t=0$ or $t=1$, then $A$ would be commutative, but all irreducible representations of commutative $C^\ast$-algebras are $1$-dimensional, in contradiction to (a). Thus $t\in (0,1)$.
Moreover, $SS^\ast=S^\ast S=t I$ for $t\neq 0$ implies $\ker S=\{0\}$ and $\operatorname{ran}S=H$. Hence $S$ is invertible, and the same holds for $T$. Furthermore, the operators $U=t^{-1/2}S$ and $V=(1-t)^{-1/2}T$ are surjective isometries, hence unitaries, and anticommute because $S$ and $T$ do.
I don't see how to prove (c) at the moment. I'll come back to that later or maybe someone else can help out.
A: Too long for a comment, but building on MaoWao's answer it is not hard to show that $\dim H=2$. The spirit of the argument is to notice that necessarily
$$
S=\sqrt t\,\begin{bmatrix} 1&0\\0&-1\end{bmatrix},\qquad T=\tfrac{\sqrt{1-t}}2\begin{bmatrix} 0&1\\1&0\end{bmatrix}
$$
(properly, some unitary conjugation of those). I expect that the argument can be simpler, I didn't spend time on it and just went with what came naturally first.
With all the properties established by MaoWao, we are now given unitaries $U$ and $V$ such that $UV=-VU$ and the C$^*$-algebra $A$ they generate is irreducible.
From $UV=-VU$ we get that $U^2V=VU^2$. From Flugede-Putnam we also get $U^2V^*=V^*U^2$. So $U^2\in A'=\mathbb CI$. This implies that after multiplying by an appropriate scalar we may assume that $U^2=I$ and the spectrum of $U$ is $\{1,-1\}$ (it cannot be just $\{1\}$ because $U\ne I$, and multiplying $U$ by a scalar will still generate the same algebra and satisfy the same relations). Similarly, we can assume $\sigma(V)=\{1,-1\}$.
We can now write $U=2P-I$, $V=2Q-I$ for certain projections $P$ and $Q$ (these are the respective spectral projections corresponding to $1$, so $U=P-(I-P)$, $V=Q-(I-Q)$). The equality $UV=-VU$ now reads
$$
4PQ-2Q-2P+I=-(4QP-2P-2Q+I).
$$
This we may write as
$$\tag1
4PQ+4QP-4Q-4P+2I=0.
$$
If we multiply by $I-P$ on the left and right,
$$
-4(I-P)Q(I-P)+2(I-P)=0.
$$
That is, $$(I-P)Q(I-P)=\tfrac12\,(I-P).$$ Conjugating $(1)$ with $P$ we get
$$
PQP=\tfrac12\,P. 
$$
If we take $W=\sqrt2\,PQ(I-P)$, it is easy to check from the above formulas (or variations of them using the symmetry of the roles) that $W^*W=I-P$, $WW^*=P$. Thus
$$\tag2
P\simeq I-P.
$$
The relation $(2)$ allow us to assume below that $PH$ and $(I-P)H$ are the same space.
Now we express everything as $2\times2$ matrices in terms of the decomposition $H=PH\oplus (I-P)H$. So
$$
P=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad Q=\begin{bmatrix} 1/2&T\\ T^*&1/2\end{bmatrix}. 
$$
From $Q^2=Q$ we get that $T=\operatorname{Re} T$, and $T^2=\tfrac14\,I$. So $T$ is selfadjoint and $\sigma(T)=\{1/2,-1/2\}$. This means that there exists a projection $R$ with $T=R-\tfrac12\,I$. The element
$$
\begin{bmatrix} R&0\\0&R\end{bmatrix}
$$
commutes with both $P$ and $Q$, and so it is in $A'$. This means that $R=0$ or $R=I$, as it is a projection. Thus $T=\pm\tfrac12$.
It is now trivial to check via matrix computations that $A'$ (that is the set of operators that commute with both $P$ and $Q$) consists of all operators of the form
$$
\begin{bmatrix} Z&0\\0&Z\end{bmatrix}.
$$
As $A'=\mathbb C I$, such $Z$ has to be a scalar multiple of the identity. This is only possible if $P$ has rank 1. Thus $I-P$ has rank 1 by $(2)$, and $\dim H=2$.
