I'm interested in solving an equation of the form

$$ Ax = b $$

for some bounded linear operator $A: H_1 \mapsto H_2$ where $H_1, H_2$ are some Hilbert spaces.

I've seen in this math.SE post in particular, the claim that

the solution set should always be the kernel plus a particular solution.

Where does this claim come from?

I am also interested to learn in general about how operators in Hilbert spaces are introduced. Is any book on functional analysis good for this purpose?

  • $\begingroup$ I think the particular solution and the compliment are orthogonal and span the solution set. $\endgroup$ – marshal craft Jul 28 '15 at 9:11

Let's prove it. Suppose $k$ is in the kernel of $A$ and $x_0$ is a specific solution, so that $Ax_0=b$. We have $$A(x_0+k)=Ax_0+Ak=Ax_0=b.$$ Conversely, suppose that $x_0$ and $y_0$ are two solutions. Then $$A(x_0-y_0)=0$$ implying that $x_0-y_0=:k$ is in the kernel. Therefore any two solutions differ by an element of the kernel.


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