Let $T$ be the right shift operator, that is, $T(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$, suppose $E$ is a compact operator from $\ell^{\infty}\rightarrow \ell^{\infty}$, can $T+E$ be an invertible operator from $\ell^{\infty}\rightarrow \ell^{\infty}$? It seems that $T+E$ can not be invertible, but I could not prove this.

We know that $T$ and $E$ are not surjective operators, I tried to show that $T+E$ is not surjective so that it can not be invertible but failed. Could some give any comments?

  • $\begingroup$ If $E=0$ then $T+E$ is certainly not surjective. $\endgroup$ Sep 21, 2020 at 15:18
  • $\begingroup$ Here E is not 0. Then what is the case? $\endgroup$
    – Rebeca J
    Sep 21, 2020 at 15:19

1 Answer 1


$T+E$ will be a Fredholm operator with index -1, so it will not be invertible.

  • $\begingroup$ So you mean a Fredholm operator with index -1 is not invertible? Could you please give an explanation? Is a Fredholm operator with a negative index not invertible? $\endgroup$
    – Rebeca J
    Sep 30, 2020 at 15:47
  • $\begingroup$ Every invertible operator has Fredholm index zero because both the dimension of the kernel and the codimension of the range are zero. $\endgroup$
    – Ruy
    Sep 30, 2020 at 16:22
  • $\begingroup$ I got it, thank you for your help. $\endgroup$
    – Rebeca J
    Oct 1, 2020 at 1:05

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