# Proving a statement about invertible square matrices

$$RTP:$$ Show, that if a homogeneous system of equations $$Ax=0$$ has only a trivial solution, then $$A$$ is non-singular/invertible.

My idea is to prove this by contrapositive. Lets assume, that $$A$$ is singular. This means the inverse matrix $$A^{-1}$$ doesn't exist.

A linear system $$Ax=0$$ has a unique solution ( e.g only a trivial solution ), given by $$x=A^{-1}0$$. However, since a singular matrix has no inverse matrix $$A^{-1}$$ the system has does not have a unique solution. Thus we have proven our original statement by contrapositive. Is this correct?

No, it is not correct. There is an unjustifed jump from the non-existence of $$A^{-1}$$ to “the system has does not have a unique solution”.

You can do it as follows: since $$A$$ is singular, there is a vector $$v=(x_1,\ldots,x_n)\neq(0,\ldots,0)$$ such that $$A.v=0$$. So either the system has no solution or, if it has a solution $$(y_1,\ldots,y_n)$$, then$$(x_1+y_1,\ldots,x_n+y_n)$$shall be another distinct solution.

As your work implies, the existence of an inverse implies the uniqueness of the trivial solution. But you have merely assumed that the negation of this statement holds as well, rather than having proved it. That is, you have assumed that the non-existence of an inverse implies the non-uniqueness of the solution. This is the very contrapositive you set out to prove.

What to do instead? I will assume you define an inverse of a square matrix $$A$$ to be a matrix $$B$$ such that $$BA=AB=I$$.

You can apply elementary row operations to $$A$$ to put it in reduced row-echelon form. There is a matrix $$C$$ that achieves this by left multiplication, that is, $$CA$$ is the reduced row-echelon form of $$A$$. If $$CA$$ is not the identity, then the system has an arbitrary parameter or parameters, and hence non-trivial solutions. So we conclude that $$CA$$ must be the identity.

Now we can solve the equations $$Ax=e_i$$, where $$e_i$$ are the standard basis vectors, that is, the columns of $$I$$. We do this by multiplying the equations on the left by $$C$$ to conclude $$x=IX=CAx=Ce_i=c_i$$, where $$c_i$$ is the $$i^\text{th}$$ column of $$C$$. But $$Ac_i=e_i$$ for all $$i$$ implies $$AC=I$$. Hence $$CA=AC=I$$, and $$C$$ is an inverse of $$A$$.

By the way, such an inverse is unique: $$AB_1=AB_2=I$$ and $$B_1A=B_2A=I$$ imply that $$A(B_1-B_2)=0$$ and, after multiplying on the left by $$B_1$$ (or $$B_2$$), that $$B_1-B_2=0$$.