# Understanding of the term $Ax=b$

I believe i have some difficulty understanding the actual meaning of $Ax=b$

Following are two questions, which i would appreciate if you could guide me through what they actually meant:

Suppose the homogeneous system $Ax=0$ has non-trivial solution. Show that the linear system $Ax=b$ has either no solution or infinitely many solutions.

Answer: If $Ax = b$ has a solution $x = u$, then $u + v$ is also a solution to $Ax = b$ for all solutions $x = v$ to $Ax = 0$. Hence $Ax = b$ has either no solutions or infinitely many solutions.

1. How does assuming $Ax = b$ having a solution $x = u$, leads to $u + v$ also a solution?
2. I don't understand this "for all solutions $x = v$ to $Ax = 0$".
3. How does it lead to the conclusion that $Ax = b$ has either no solutions or infinitely many solutions, when the explanation did not show the "no solution" aspect.

Suppose the homogeneous linear system $Bx=0$ has infinitely many solutions. How many solutions does the system $ABx=0$ have?

Answer: Let $x = u$ be any solution to the system $Bx = 0$. Then $ABu = A0 = 0$. The system $ABx = 0$ has at least as many solutions as the system $Bx = 0$. Thus it has infinitely many solutions.

1. What does this "The system $ABx = 0$ has at least as many solutions as the system $Bx = 0$" mean?

Thanks.

• I highly recommend you go through this playlist by 3Blue1Brown. Sep 30, 2017 at 9:32
• Is English your native language? If not, what is? [if you don't mind, that is] Sep 30, 2017 at 9:32
• @KennyLau I'm Korean, but i'm okay with English. Been using English for the past decade. :) Sep 30, 2017 at 9:42

1. If $Ax=b$, then $A(u+v)=Au+Av=b+0=b$.
2. What I wrote above holds for every solution $v$ of the equation $Av=0$, not just to one or some of them.
3. If $Ax=b$ has a solution $u$, then, since the equation $Ax=0$ has, at least, one solution $v\neq0$, then every $\lambda v$ ($\lambda\in\mathbb R$) is also a solution of $Ax=0$, and therefore every $u+\lambda v$ is a solution of $Ax=b$.

1. If $u$ is a solution of $Bx=0$, then $(AB)u=A(Bu)=A0=0$. Therefore, every solution of $Bx=0$ is also a solution of $ABx=0$.
• Ah thanks a lot! A question: How does $(AB)u=A(Bu)=A0=0$ shows that every solution of $Bx=0$ is also a solution of $ABx=0$? Is it because of the fact that since $Bx=0$ has infinite solutions, the term $ABx$ being $= 0$ also implies the same? Sep 30, 2017 at 9:50
• @idolo If $u$ is such that $Bu=0$ (that is, if $u$ is a solution of $Bx=0$), then $(AB)u=A(Bu)=A0=0$ (that is, $u$ is a solution of $(AB)x=0$). This has nothing to to with the number of solutions. Sep 30, 2017 at 9:53
• How does every solution of $Bx=0$ is also a solution of $ABx=0$? Thanks. Sep 30, 2017 at 9:56
• @idolo I have already proved that twice, first in my answer and then in my previous comment. What is the passage that you don't understand? Sep 30, 2017 at 9:58
• I do not understand how does the above leads to the conclusion that every solution of $Bx=0$ is also a solution of $ABx=0$. Sep 30, 2017 at 11:22

If $Ax = b$ has a solution $x = u$, then $u + v$ is also a solution to $Ax = b$ for all solutions $x = v$ to $Ax = 0$.

This sentence may be a little bit difficult to understand. Allow me to rephrase it:

If $Ax = b$ has a solution $x = u$, then let $x=v$ be a solution to $Ax = 0$: $x = u+v$ would also be a solution to $Ax = b$.

1. $Ax = b$ either has no solution or has a solution.
3. Therefore, $Ax = b$ either has no solution or has infinitely many solutions.