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Let $x_p$ be the particular solution of $Ax = b$ and $x_h$ be the solution to the homogeneous system $Ax = O$. All the solutions of $Ax = b$ are of the form $x_p + x_h$

Proof:

Let $x$ be the solution of $Ax = b$, then $A(x − x_p) = Ax − Ax_p = b − b = 0 \to x_h = x − x_p \to x = x_p + x_h$

We need to show all the solutions are of this format $x_p + x_h$. Let $x'$ be a solution of $Ax = 0$, then $A(x + x') = Ax + Ax' = Ax + 0 = b + 0 = b$. Hence $x + x'$ is a solution of $Ax = b$.

I understand the technical details of this proof, but I am not sure about the intent of the arguments.

I think the first part is saying if $x $ is the solution of $Ax = b$, then $x = x_p + x_h.$

Second part is saying that $x_p + x_h$ is a solution to every system $Ax= b.$

Does that make sense?

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    Is this a proof in the book? It doesn't look very logical to me. – 2017-01-04
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    @ KittyL, yes. Capture from my book: https://s30.postimg.org/es1og65b5/capt.png – 2017-01-04
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    I still think this proof is very confusing. To get the best of it, the first part proves that if $x$ is a solution of $Ax=b$, it must be of the form $x_p+x_h$. The second part is the most confusing. It seems the author wants to prove that, if $x$ is a particular solution (I think he should have used $x_p$), and $x'$ (should have been $x_h$) is a solution of the homogeneous part, then $x+x'$ is a solution of $Ax=b$. Although his statement is opposite. – 2017-01-04
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    So in summary, the first part proves that, any solution is of the assumed format. The second part proves that, an $x$ of this format is indeed a solution. – 2017-01-04

1 Answers 1

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Yes, presented more clearly

$\quad $ if $\,\ ax_p = b\,\ $ then $\,\ ax=b \!\iff\! a(x\!-\!x_p) = 0\!\iff\! \ x-x_p = x_h$ and $\, ax_h = 0$

Remark $\ $ More generally this hold for any operator $A$ that is $\,\rm\color{#c00}{L}inear,\,$ i.e.

$\quad $ if $\,\ \color{#0a0}{Ax_p = b}\,\ $ then $\,\ Ax=b \!\iff\! A(x\!-\!x_p) = 0\!\iff\! \ x-x_p = x_h$ and $\, Ax_h = 0$

because $\ \ \ \smash[t]{ A(x-x_p)\overset{\rm\color{#c00}{\large L}} = Ax - \color{#0a0}{Ax_p} = Ax -\color{#0a0}b} $

therefore $\ A(x-x_p)\, =\, 0 \ \iff\ Ax - b = 0$

Thus this relationship between general, particular and homogeneous solutions also holds for linear differential and difference equations (recurrences)