[Math] Why is $\text{dim ker}(T) = 0$ a requirement for invertibility of a finite dimensional operator, but not for an infinite dimensional operator

functional-analysislinear-transformationsoperator-theory

I have read that Fredholm operators generalize the notion of invertibility, and that all finite dimensional operators are Fredholm with index 0. Also, if you have a Fredholm operator with index zero, then it is surjective if and only if it is injective. So if a Fredholm operator has index 0 and is injective then it will be invertible.

So consider the finite dimensional operator $T: \mathbb{R}^n \to \mathbb{R}^n$ given by the matrix $A$ as follows:

$$
A =
\begin{bmatrix}
1 & 0 & 0 & \dots & \dots & 0 \\
0 & 1 & 0 & \dots & \dots & 0 \\
0 & 0 & 1 & \dots & \dots & 0 \\
0 & 0 & 1 & \dots & \dots & 0 \\
\vdots & \dots & \ddots & \ddots & 0 & 0 \\
\vdots & \dots & \dots & 0 & 1 & 0 \\
0 & 0 & \dots & 0 & 0 & 0 \\
\end{bmatrix}.
$$

That is, the elements of the diagonal of $A$ are $1$, except for the final diagonal element which is $0$.

$A$ has a kernel and cokernel (they are the same) both with dimension $1$ so it is Fredholm with index zero. However it is not injective and therefore not invertible. It seems that for any finite dimensional operator (matrix) I choose with non-trivial kernel, I will never have an injective operator as it can always be reduced to row echelon form in which it will have one or more rows that are all zero.

So the only way an operator $T$ can be invertible in the finite dimensional case is if it has $\text{dim ker}(T) = \text{dim (coker}(T)) = 0$.

On the other hand, from reading about infinite dimensional operators, it seems invertibility is often shown by first showing the operator is Fredholm with index 0, and then showing it is injective.

I don't understand why having $\text{dim ker}(T) \neq 0$ prevents invertibility in the finite dimensional case, yet it doesn't seem to be an issue in the infinite dimensional case? For a finite dimensional operator with a non-trivial kernel we can never have injectivity, so how is it possible that an infinte dimensional operator with non-trivial kernel can be injective?

Best Answer

Because it only shows injectivity in the infinite case. Recall you use dimension arguments in the finite case to show injective iff finite when the dimensions match by showing a basis is mapped to a basis--you show it is mapped to a linearly independent set, and then use counting to show this is a basis, effectively reducing this to the statement "injective functions on finite sets of the same cardinality are bijections". Obviously this is not doable when the dimension is infinite, since there are plenty of injective functions on infinite sets which are not surjections. However, consider the clearly injective map

$$T: \begin{cases} C^{\infty}(\Bbb R)\to C^\infty(\Bbb R)\\ f\mapsto \displaystyle\int_0^x f(t)\,dt \end{cases}$$

By convention (and necessarily) $T(0)=0$. This gives a specific anti-derivative for any given function (all anti-derivatives differ by a constant, so the lower limit of $0$ makes the choice and forces injectivity). However, the map is not surjective as, for example, there is no $f$ so that $T(f) = 1$.

If you don't like the calculus-based example, there's always a shift map, Consider the set of bounded, real sequences $\ell^\infty(\Bbb R)$

$$\sigma: \begin{cases} \ell^\infty(\Bbb R)\to\ell^\infty (\Bbb R) \\ (x_1, x_2,\ldots, x_n,\ldots)\mapsto (0,x_1,x_2,\ldots, x_{n-1},\ldots)\end{cases}$$

is clearly injective and linear, but not surjective as no sequence beginning with $1$ is in the image.

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