# Linear Algebra Proof - Prove that when A+D=0, D is unique.

• Sep 3rd 2007, 10:14 AM
Fourier
Linear Algebra Proof - Prove that when A+D=0, D is unique.
I need to prove that for each $\displaystyle m \times n$ matrix A, there exists a unique $\displaystyle m \times n$ matrix D such that A+D=O. Where O is the $\displaystyle m \times n$ zero matrix.

I have a textbook that states the proof simply as:

$\displaystyle \textrm{For } A=[a_{ij}] \textrm{ Let } D=[-a_{ij}].$

I'm not sure how to construct a a better proof for this result. Could anyone offer a starting point?
• Sep 3rd 2007, 11:22 AM
red_dog
Let $\displaystyle A=(a_{ij}), \ D=(x_{ij})$
Then $\displaystyle A+D=(a_{ij}+x_{ij})$.
$\displaystyle A+D=O_{mn}\Rightarrow a_{ij}+x_{ij}=0\Rightarrow x_{ij}=-a_{ij}\Rightarrow D=(-a_{ij})$

Suppose $\displaystyle D'$ is another matrix such as $\displaystyle A+D'=D'+A=O_{mn}$.
Then $\displaystyle D'=D'+O_{mn}=D'+(A+D)=(D'+A)+D=O_{mn}+D=D$
• Sep 4th 2007, 04:28 PM
Fourier
Thanks for giving the alternate proof.