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.

  • $\begingroup$ 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 $I$). $\endgroup$
    – user754697
    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.


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