# Upper-triangular matrix is invertible iff its diagonal is invertible: C*-algebra case

Exercise 1.14 of the book Rordam, Larsen and Laustsen "An introduction to K-theory for C*-algebras" asks to prove, that upper triangular matrix with elements from some C*-algebra $A$ is invertible in $M_n(A)$ iff all diagonal entries are invertible in $A$.

Trying to solve this I've found that if $a$ is invertible and $\delta$ is such that $(a^{-1}\delta)^n=0$ then $a+\delta$ is invertible too and its inverse is given by $(a+\delta)^{-1}=\sum_{k=0}^{n} (-a^{-1}\delta)^ka^{-1}$. Using this fact I can show that if diagonal is invertible, then upper-triangular matrix with this diagonal is invertible too, and also that if upper-triangular matrix has upper-triangular inverse, then its diagonal is invertible. So all I need to prove is that if upper-triangular matrix invertible, then its inverse is upper-triangular. I've failed to prove this.

Also there is a hint for this exercise: "Solve the equation $ab=1$ where $a$ is as above [i.e. upper-triangular matrix] and where $b$ is unknown upper triangular matrix". Solution of this equation follows from my reasoning above, but this doesn't help.

Update (counterexample attempt): I've made one more attempt and it looks for me like I have found a counterexample. However I think there is a mistake in it (because otherwise there is a mistake in the book). Here it is. Let $A=B(l^2(\mathbb{N}))$ --- algebra of bounded operators on sequences $x=\{x_i\}_ {i=1}^ \infty:\|x\|^2=\sum_{x=1}^{\infty}|x_i|^2<\infty$. Let $z\in A$ be defined by $(zx)_ {2n-1}=0$, $(zx)_{2n}=x_n$, and $t\in A$ be defined by $(tx)_{2n-1}=x_n$, $(tx)_ {2n}=0$. Then we have $t^*t=z^ * z=tt^* +zz^* =1$, $t^* z=z^* t=0$. From these we have that $$\begin{pmatrix}z&tz^* \\\ 0&t^* \end{pmatrix}\begin{pmatrix}z^* &0\\\ zt^* &t\end{pmatrix}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$$ and $$\begin{pmatrix}z^* &0\\\ zt^* &t\end{pmatrix}\begin{pmatrix}z&tz^* \\\ 0&t^* \end{pmatrix}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}.$$ So now my question should say "Where am I wrong?".

• I hardly know anything about C*-algebras. Is there anything which makes this problem different from the ordinary linear algebra situation? (It doesn't sound so to me.) – Hans Lundmark Oct 25 '10 at 7:34
• @Hans: Yes, solution for linear algebra doesn't work here and I don't know, how to modify it to make it applicable to C*-case. In ordinary linear algebra matrix elements commute. If a,b,z are numbers, then from ab=1, az=0 we can conclude that z=0. In C*-case we cannot: there is an example of a,b,z,t such that ab=1, az=0, tz=1. – Fiktor Oct 25 '10 at 8:18
• OK, but I don't see that noncommutativity should cause much trouble here. Consider for example $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, where $a$, $b$, $c$ are given. This holds if $aA=1$, $aB+bC=0$, $cC=1$. Thus $A=a^{-1}$, $C=c^{-1}$, and $B=-a^{-1}bC$. – Hans Lundmark Oct 25 '10 at 8:30
• Oh, I see. Then I'd better keep quiet and let someone who actually knows something do the talking! ;) – Hans Lundmark Oct 25 '10 at 9:09
• Nice example! It shows that the exercise in the book is incorrect. Based on the hint, perhaps it was intended that you show that if the matrix has upper triangular inverse, then the diagonal entries are invertible. Perhaps it's true if you assume the algebra contains no nonunitary isometries, but I have nothing of substance to add. – Jonas Meyer Oct 26 '10 at 2:43

Let $H$ be a separable Hilbert space with orthonormal basis $\{e_0,e_1,e_2,\ldots\}$. Let $S \in B(H)$ be the unilateral shift determined by $S(e_i) = e_{i+1}$ for all $i$. Let $P_0 \in B(H)$ be rank-1 projection onto the span of $e_0$. Then, $U = \begin{pmatrix} S & P_0 \\ 0 & S^* \end{pmatrix}$ is a unitary in $M_2(B(H))$, despite being nondiagonal, upper-triangular, and having both diagonal entries noninvertible.
It's easy to check $U^* U = UU^* = 1$ directly, but the "real" reason this works is that, under the isomorphism $M_2(B(H)) \cong B(H^2)$, $U$ works on the natural basis of $H^2$ by $$\cdots \mapsto (0,e_2) \mapsto (0,e_1) \mapsto (0,e_0) \mapsto (e_0,0) \mapsto (e_1,0) \mapsto (e_2,0) \mapsto (e_3,0) \mapsto \cdots$$ so $U$ is actually (up to a unitary conjugacy) the bilateral shift.