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Prove that two homogeneous systems of linear equations are equivalent iff they have the same solution set.

My definition of equivalent of linear equations is that each equation of system $A$ is a linear combination of the equations of system $B$ and converse.

So i have a bit trouble proving this statement in general as $m\times n$ form. I worked with their coefficient matrices and if i show that these two are row equivalent the problem is solved. Is there any method to use for this? If not what's the general prove.

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This cannot be done without the rank theorem or some other theorem of similar power.

We are given an $(m_A\times n)$-matrix $A$ and an $(m_B\times n)$-matrix $B$. Denote by ${\rm ker}(A)\subset{\mathbb R}^n$ the solution space of $Ax=0$, by $A_{i\!-}\in{\mathbb R}^n$ the $i^{\rm th}$ row vector of $A$, and by $${\rm row}(A):=\langle A_{1\!-}\>,\ldots, A_{m_A\!-}\rangle\subset{\mathbb R}^n$$ the row space of $A$. Similarly for $B$. We have to prove the following claim: $${\rm row}(A)={\rm row}(B)\qquad\Leftrightarrow\qquad{\rm ker}(A)={\rm ker}(B)\ . $$ For this it is sufficent to prove $${\rm row}(B)\subset {\rm row}(A)\qquad\Leftrightarrow\qquad{\rm ker}(A)\subset{\rm ker}(B)\ .$$ Proof of $\Rightarrow\>: \quad$ Assume ${\rm row}(B)\subset {\rm row}(A)$, and consider a vector $x\in{\rm ker}(A)$. Let $B_{i\!-}\> x=\sum_{k=1}^n B_{ik}x_k=0$ be an equation of the $B$-system. As $B_{i\!-}=\sum_{j=1}^{m_A} \lambda_j A_{j\!-}$ for certain $\lambda_j\in{\mathbb R}$ we obtain $$B_{i\!-}\>x=\sum_{j=1}^{m_A}\lambda_j A_{j\!-}\>x=0\ .$$ Since this is true for all $i\in[m_B]$ it follows that $x\in{\rm ker}(B)$. (This was the easy part.)

Proof of $\Leftarrow\>: \quad$ If ${\rm row}(B)\not\subset {\rm row}(A)$ then there is a row $B_{i\!-}$ of $B$ that does not belong to ${\rm row}(A)$. Denote by $A'$ the matrix obtained from $A$ by adding the row $B_{i\!-}$ at the bottom. Then $A'$ has rank one larger than $A$, hence ${\rm ker}(A')\subset{\rm ker}(A)$ has dimension one less than ${\rm ker}(A)$. It follows that there are vectors $x\in{\rm ker}(A)\setminus{\rm ker}(A')$. These vectors do not satisfy the last $A'$-equation $B_{i\!-}\>x=0$, hence do not belong to ${\rm ker}(B)$. This proves ${\rm ker}(A)\not\subset{\rm ker}(B)$.

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