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Consider the following system of Fredholm integral equations with constant kernel matrix $$ f(x)=K(x)\int_{0}^{1}f(s)ds $$ where $K(x)\in C([0,1];M_{2\times 2}(% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion )).$ For the scalar case, it is known by the Fredholm alternative that the above equation has a unique solution equals to zero if and only if $K(x)$ is different from $1$ for any $x\in [0,1]$.

My question is: What happens in the system case?. How I can check the spectrum condtion?. Thank you in advance.

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    $\begingroup$ Your claim for the scalar case is not correct. In the scalar case, $1$ is an eigenvalue of your operator on the right if and only if $K$ has integral $1$. (By the way, this has nothing to do with the Fredholm alternative.) $\endgroup$ Mar 30, 2021 at 20:28
  • $\begingroup$ Thank you sir I'm very thankful. $\endgroup$
    – Gustave
    Mar 30, 2021 at 22:49

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The operator $f\to K(x) \int_0^1 f(s) ds$ is a compact operator from $C([0,1];\mathbb R ^n)$ equipped with the sub norm into itself, if $K$ is an $n\times n$ matrix valued function for any finite $n\geq1$, simply because bounded intervals in $\mathbb R$ are compact. An even simpler argument is that the range is finite dimensional.

So indeed, $1$ has finite multiplicity (at most $n$).

Now integrating over $(0,1)$, you find that if $1$ is an eigenvalue for $K$ with eigenvector $f$ then $1$ is an eigenvalue of $M=\int_0^1 K(x) dx \in \mathbb R ^{n\times n}$ with eigenvector $\int_0^1 f(x) dx$, and the multiplicity of $1$ for $K$ is smaller or equal to that of $1$ for $M$. If $L$ is an eigenvector for $M$ with eigenvalue $1$, then $KL$ is the corresponding eigenvector for $K$. So, just like in the one dimensional case, all is decided by the constant matrix $M$.

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  • $\begingroup$ Thank you sir for the answer. Is there any changes if I work in $L^2$ instead of the space of continuous functions?. $\endgroup$
    – Gustave
    Mar 30, 2021 at 22:18
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    $\begingroup$ No difference. $L^1$ works as well. It is a finite dimensional problem.. $\endgroup$
    – username
    Mar 31, 2021 at 5:21

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