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

• Feb 1st 2011, 07:31 PM
Pinkk
Solution for a particular column vector implies solution for all column vectors.
Let $\displaystyle A$ be a square matrix. Show that if the system $\displaystyle AX=B$ has a unique solution for some particular column vector $\displaystyle B$, then it has a unique solution for all $\displaystyle B$.

So I'm not really sure how to go about this since there's no assumption that $\displaystyle A$ 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.
• Feb 1st 2011, 08:56 PM
TheEmptySet
Quote:

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

So I'm not really sure how to go about this since there's no assumption that $\displaystyle A$ 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.

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

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

We have
$\displaystyle Ax=A(x_c+x_p)=b$ 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 $\displaystyle \vec{0} \in x_c$ the only solution to the system

$\displaystyle Av=0$ is $\displaystyle v=0$

This will give the desired conclusion.
• Feb 1st 2011, 09:09 PM
Pinkk
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 $\displaystyle x_{c}$ 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 $\displaystyle B$.
• Feb 2nd 2011, 12:34 AM
FernandoRevilla
Quote:

Originally Posted by Pinkk
So I'm not really sure how to go about this since there's no assumption that $\displaystyle A$ 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 $\displaystyle AX=B_0$ has unique solution, then

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

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

Fernando Revilla
• Feb 2nd 2011, 04:16 PM
Pinkk
Whew, I get it now. Thanks so much!