# Prove that two square non-singular matrices are row equivalent

I'm trying to prove that any two square non-singular row matrices are equivalent and don't know if my proof is correct/need more details:

We have two following $$n \times n$$ matrices:

$$A=\begin{pmatrix} a_{11}&a_{12}&\cdots& a_{1n}\\ a_{12}&a_{22}&\cdots &a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots& a_{nn} \end{pmatrix}$$

and $$B = \begin{pmatrix} b_{11}&b_{12}&\cdots& b_{1n}\\ b_{12}&b_{22}&\cdots &b_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ b_{n1}&b_{n2}&\cdots& b_{nn} \end{pmatrix}$$

Because they are both non-singular matrices, then they are the matrices of coefficients of two homogeneous system with unique solutions. If so, then the only solution would be an $$n$$-tuple $$(0,0, \cdots, 0)$$

Therefore, both $$A$$ and $$B$$ are reducible to a $$n \times n$$ identity matrix.

$$\begin{pmatrix} 1&0&\cdots& 0\\ 0&1&\cdots &0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots& 1 \end{pmatrix}$$

So $$A$$ and $$B$$ are row-equivalent.

Thank you all for your help.

• Seems fine to me. It could be improved slightly if you can specifically cite results in your course that you're using (e.g. non-singular matrices are row-equivalent to $I$). Mar 1 '20 at 1:39

There is a result which states "every $$n\times n$$ invertible matrix is row-equivalent to the $$n\times n$$ identity matrix". Since row-equivalence is an equivalence relation, that is to say, it is reflexive, symmetric and transitive, if $$A$$ and $$B$$ are row equivalent to the identity $$I$$, we have that $$A\sim I \sim B$$, thus $$A\sim B$$.
But why does this equivalence relation hold? To begin with, we say the $$m\times n$$ matrices $$A$$ and $$B$$ are row-equivalent if $$B$$ can be obtained from $$A$$ through a finite number of row-operations applied to $$A$$. These operations can be typified into three kinds: multiplication of a row by a non-zero scalar $$c$$; exchange of two rows; or substitution of some row by it plus another one multiplied by a constant. Each of these operations admit an inverse which is also from the same type.
Based on such considerations, the reflixivity is obvious: $$A\sim A$$ because it needs $$0$$ row operations to get it from itself. To prove the symmetric property, let us consider a chain of single row operations which takes $$A$$ into $$B$$: $$A\rightarrow A_{1}\rightarrow\ldots\rightarrow A_{n}\rightarrow B$$. Thus $$B\sim A$$. Since we can obtain $$A$$ from $$B$$ by reversing each row-operation from the right to the left, we conclude that $$A\sim B$$.
Finally, in order to prove the transitivity, it suffices to consider the chains: \begin{align*} A\rightarrow A_{1}\rightarrow\ldots\rightarrow A_{n}\rightarrow B\rightarrow B_{1} \rightarrow \ldots\rightarrow B_{m}\rightarrow C \end{align*} Consequently, if $$B\sim A$$ and $$C\sim B$$, it results that $$C\sim A$$, because there is a chain of elementary row-operations applied to $$A$$ which leads to the matrix $$C$$, proving the transitivity.