Nontrivial solution

• Mar 6th 2009, 02:26 PM
antman
Nontrivial solution
How do I prove / disprove "If a Matrix A is nonsingular, then the homogeneous system Ax=0 has a nontrivial solution?"

I know that if A is nonsingular, then \$\displaystyle A^{-1} \$exists and multiplying both sides of Ax=0 by \$\displaystyle A^{-1}\$ gives \$\displaystyle A^{-1}(Ax)=A^{-1}(0)\$ so \$\displaystyle (AA^{-1})(x)=0\$ then \$\displaystyle (I_n)x=0\$ and x=0. Therefore, the only solution to Ax=0 is x=0 so a nonsingular matrix never has a trivial solution to Ax=0.

Is this enough to disprove the statement?
• Mar 6th 2009, 04:44 PM
o_O
Quote:

Originally Posted by antman
How do I prove / disprove "If a Matrix A is nonsingular, then the homogeneous system Ax=0 has a nontrivial solution?"

I know that if A is nonsingular, then \$\displaystyle A^{-1} \$exists and multiplying both sides of Ax=0 by \$\displaystyle A^{-1}\$ gives \$\displaystyle A^{-1}(Ax)=A^{-1}(0)\$ so \$\displaystyle ({\color{red}A^{-1}A})(x)=0\$ then \$\displaystyle (I_n)x=0\$ and x=0. Therefore, the only solution to Ax=0 is x=0 so a nonsingular matrix never has a nontrivial solution to Ax=0.

Is this enough to disprove the statement?

Just a minor change. One, you're multiplying both sides by \$\displaystyle A^{-1}\$ from the left so you should stay consistent (even though \$\displaystyle AA^{-1} = A^{-1}A = I\$).

And the second red is probably a typo.
• Mar 7th 2009, 06:44 PM
antman
Thank you! That does make more sense and that definitely was a typo.