0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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 $A\sim B$.

EXPLANATION

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.