# Is the set of all invertible linear operators dense in the set of all linear operators?

Exercise 2 after $$\S$$ 91 from Paul R. Halmos's "Finite-Dimensional Vector Spaces" (second edition) invites to prove or disprove the following assertion.

For every (bounded) linear transformation $$A$$ (on an inner product space) there exists a sequence $$(A_n)$$ of invertible linear transformations such that $$A_n \rightarrow A$$.

The inner product space, say $$\mathcal V$$, of the assertion is not specified to be over the complex (or real) field, and is not said to be finite-dimensional or complete either. Also, for reference, $$\S 91$$ (from the book) identifies "$$A_n \rightarrow A$$" in inner product spaces by

$$\Vert A_n - A \Vert \rightarrow 0$$ as $$n \rightarrow \infty$$.

Further, $$\S \ 87$$ has the following definition for the norm $$\Vert \cdot \Vert$$ of a linear operator:

$$\Vert A \Vert = \inf \ \big\{K: \Vert Ax \Vert \leq K \Vert x \Vert \text{ for all vectors } x\big\}.$$

I am able to see why the assertion holds if $$\mathcal V$$ is finite-dimensional; see my "constructive" proof below. I am unable to imagine what happens in the general case however, and would appreciate a proof or a counterexample in infinite dimensional spaces. Would also appreciate an advice if my proof is found to be inaccurate! Thanks.

Proof (in finite dimensions): Let $$\mathcal V$$ be a $$k$$-dimensional inner product space ($$0 \leq k < \infty$$), and let $$A$$ be any linear operator on $$\mathcal V$$. If $$\mathcal N(A)$$ is the null-space of $$A$$, then $$\mathcal N^\perp(A) \oplus \mathcal N(A) = \mathcal V$$. Also, if the rank of $$A$$ is $$m \ (\leq k)$$, and if $$\mathcal R(A)$$ is the range of $$A$$, then $$m =$$ dim $$\mathcal R(A)=$$ dim $$\mathcal N^\perp(A) = k-$$dim $$\mathcal N = k-$$dim $$\mathcal R^\perp(A)$$. It is also true that $$\mathcal R^\perp(A) \oplus \mathcal R(A) = \mathcal V$$.

We now begin constructing a sequence $$(A_n)$$ of invertible linear operators such that $$A_n \rightarrow A$$. Let $$B$$ be the restriction of $$A$$ to the $$m$$-dimensional $$\mathcal N^\perp(A)$$. It is clear that $$B$$ maps $$m$$-dimensional $$\mathcal N^\perp(A)$$ onto $$m$$-dimensional $$\mathcal R(A)$$, and is invertible. Next, write $$C_n x_i = \frac{1}{n}y_i$$ for $$n = 1, 2, \cdots$$ and for $$i = 1, \cdots, k-m$$, where $$\{x_1, \cdots, x_{k-m}\}$$ and $$\{y_1, \cdots, y_{k-m}\}$$ are any bases in the $$(k-m)$$-dimensional subspaces $$\mathcal N(A)$$ and $$\mathcal R^\perp (A)$$ respectively. (It is clear that $$C_n$$ is a linear map with rank $$k-m$$, and is invertible.) Next, if, for every $$z = (z_1+z_2)$$ in $$\mathcal V$$, we write $$A_n z = B z_1 + C_n z_2$$ for $$n = 1, 2, \cdots$$, whenever $$z_1$$ and $$z_2$$ are in $$\mathcal N^\perp(A)$$ and $$\mathcal N(A)$$ respectively, then we find that $$A_n$$ is a linear mapping of $$\mathcal V$$ onto $$\mathcal V$$, and is invertible. Finally, because $$A_n z - Az = (Az + C_nz_2) - Az = C_n z_2$$, which implies that $$(A_n - A)z \rightarrow 0$$, it follows that $$A_n \rightarrow A$$.

The assertion fails in the infinite dimensional case: For example consider the Hilbert space $$V=l^2(\mathbb{N},\mathbb{R})$$ and let $$R,L$$ be the right and left shift operator on $$V$$, that is for $$x=(x_n) \in V$$ $$L(x)=(x_2,x_3, \dots), ~ R(x)=(0,x_1,x_2, \dots).$$ Then $$L$$ is not invertible ($$L$$ is not injective), $$R$$ is a right inverse of $$L$$ ($$LR =I_V$$) and $$\|L\|,\|R\| = 1$$. Now let $$A:V \to V$$ be any linear and continuous operator with $$\|A\|<1$$. Since $$\|RA\| \le \|R\| \|A\| < 1$$ the operator $$(I+RA)$$ is invertible (Neumann's series). Since $$L+A = L(I+RA)$$ we have $$L=(L+A)(I+RA) ^{-1}$$. Thus $$L+A$$ can't be invertible (otherwise $$L$$ would be invertible). So, no sequence of invertible operators can have limit $$L$$.
Just to give a simpler proof in the finite dimensional case. Let $$a_n$$ be a sequence of numbers such that
(1) $$a_n\rightarrow 0$$ as $$n\rightarrow\infty$$
(2) $$a_n$$ is not an eigenvalue of $$A$$ for any $$n$$.
Then $$A-a_n I_n$$ is invertible and converges to $$A$$.
Note that this argument fails in infinite dimensional case for a simple reason: The spectrum of $$A$$ is not necessarily finite, therefore we may not be able to pick $$a_n\rightarrow 0$$ that avoids spec($${A}$$). Indeed, the shift operators have the closed unit disk as its spectrum.