# Thread: Solution for a particular column vector implies solution for all column vectors.

1. ## Solution for a particular column vector implies solution for all column vectors.

Let $A$ be a square matrix. Show that if the system $AX=B$ has a unique solution for some particular column vector $B$, then it has a unique solution for all $B$.

So I'm not really sure how to go about this since there's no assumption that $A$ is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any connections. Any help would be appreciated, thanks.

2. Originally Posted by Pinkk
Let $A$ be a square matrix. Show that if the system $AX=B$ has a unique solution for some particular column vector $B$, then it has a unique solution for all $B$.

So I'm not really sure how to go about this since there's no assumption that $A$ is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any connections. Any help would be appreciated, thanks.
Here is my thoughts on the question.

Every solution to a linear system can be decomposed into two separate pieces.

$X=X_c+X_p$ where $X_c$ is the complimentary solution (the solution to the homogeneous equation) and $X_p$ is a particular solution.

Since we know the solution is unique for some particular $b$

We have
$Ax=A(x_c+x_p)=b$ Since the solution is unique The complementary solution cannot have any free parameters.

e.g the kernel of this Matrix has only one vector in it and since $\vec{0} \in x_c$ the only solution to the system

$Av=0$ is $v=0$

This will give the desired conclusion.

3. I'm not sure I exactly follow (we haven't covered kernel, etc).

Are you saying that since the solution is unique, it must be the case that $x_{c}$ is the zero column vector?

And I'm not sure I make the connection to how it follows that there must be a unique solution for any $B$.

4. Originally Posted by Pinkk
So I'm not really sure how to go about this since there's no assumption that $A$ is invertible, and I'm assuming I have to utilize row operations or the like, but I just can't make any connections. Any help would be appreciated, thanks.

If $AX=B_0$ has unique solution, then

$E_n\cdot\ldots\cdot E_1AX=E_n\cdot\ldots\cdot E_1B_0\Leftrightarrow UX=E_n\cdot\ldots\cdot E_1B_0$

(where $U$ is a row echelon form) has unique solution. This implies that all leading coefficients of $U$ are different from $0$. Now, you can conclude.

Fernando Revilla

5. Whew, I get it now. Thanks so much!