3

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?

Bill Dubuque
  • 272,048
Chimp
  • 77
  • 7
  • Is this a proof in the book? It doesn't look very logical to me. – KittyL Jan 04 '17 at 16:40
  • @ KittyL, yes. Capture from my book: https://s30.postimg.org/es1og65b5/capt.png – Chimp Jan 04 '17 at 16:43
  • 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. – KittyL Jan 04 '17 at 16:53
  • 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. – KittyL Jan 04 '17 at 16:55

1 Answers1

1

Yes, presented more clearly
$$\overbrace{{\rm if}\ \ ax_p = b}^{{\rm particular\ solution\ }{\large x_p}}\!\! {\rm then}\ \ ax=b \!\iff\! a(x\!-\!x_p) = 0\!\iff\! \ \underbrace{x-x_p = x_h\ \ {\rm and} \overbrace{ a\, x_h = 0}^{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\rm homogeneous\ solution}\ \large x_h}}_{\large \bbox[5px,border:1px solid #c00]{\begin{align}x\ &=\ \text{general solution }\\ &=\ \rm particular + homogeneous\end{align}}}\qquad$$

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

Thus this relationship between general, particular and homogeneous solutions also holds for linear differential and difference equations (recurrences), linear Diophantine equations (Bezout scalings), and many other common linear equations. In linear algbra such solutions spaces of inhomogenous equations are abstracted in the study of affine spaces.

Bill Dubuque
  • 272,048