# Expressing a matrix operator as a sum of an identity and a compact operator

My problem concerns with the unique solvability of a linear system of integral equations. In my problem, as I was able to write the system in matrix form: M \begin{align} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} f \\ g \end{bmatrix} \end{align} where M = \begin{align} \begin{bmatrix} I - K_1 & S_1 \\ K_2 & S_2 \end{bmatrix} \end{align} is an operator define on $H^{\frac12}(\Gamma_1)\times H^{-\frac12}(\Gamma_2)$ ($\Gamma_1 \cap \Gamma_2 = \emptyset$, $\Gamma_1, \Gamma_2$ have certain smoothness).

To show the well-posedness of the above equation, I just need to show that $M$ is invertible. I doing so, I followed the ideas in Chapter 4-4 of Taylor's book on Pseudo Differential Operators. I understand that the steps are as follows:

1. Show that the adjoint $M^*$ of $M$ is injective.
2. Show that $M^*$ has a closed-range.
3. Show that $M$ is an isomorphism.

I was able to do the first step but I am stuck in showing the last two.

Here are my questions.

First Question I am not sure, but, do I really need to do the second step? I think that if could show that $M$ is an isomorphism, then the condition that the range of $M^*$ is closed is already satisfied (kindly correct me if my argument is wrong and please point out what might be my mistake).

Second Question Now, I believe I can apply Fredholm alternative to show that $M$ is in fact bijective. Knowing that, I need to show that I can express $M$ as a sum of the identity operator and a compact operator, i.e., $M=I+C$, where $C$ is a compact operator. For this one, I believe I can use the condition that $K_1, K_2, S_1, S_2$ are compact, but, I don't know how to do this. Is my argument correct? and how to do I show that $M$ is a sum of the identity operator and a compact operator?

Thank you.

This is more of a comment for your second question. In order for $M$ to be the sum of the identity operator and a compact operator, you would have $$M = \left[\begin{matrix} I - K_1 & S_1 \\ K_2 & S_2 \end{matrix}\right] = \left[\begin{matrix} I & 0 \\ 0 & I \end{matrix}\right] + \left[ \begin{matrix} -K_1 & S_1 \\ \ K_2 & S_2-I\end{matrix}\right].$$ So you would need $$\left[\begin{matrix} -K_1 & S_1 \\ \ K_2 & S_2-I\end{matrix}\right]$$ to be compact. That would require $S_2 - I$ to be compact (since the compacts form an ideal), and since $S_2$ is compact already, this would only be true if your space is finite dimensional. So unless I'm missing something, it looks like you may need something different to show what you want.
• Then according to my answer above, $M \neq I + C$. Commented May 9, 2018 at 16:19
• What happens if $S_2$ is just bounded (but still my space is infinite dimensional)? Commented May 11, 2018 at 4:58
• For $M-I$ to be compact, each of its entries must be compact. So you would need $S_2 = I + T$ where $T$ is compact. Commented May 11, 2018 at 18:20