Row Equivalent Homogenous Matrix Systems have the same Solution Set

Last updated Nov 1, 2022

Let $F$ be a Field, let $m,n \in \mathbb{N}$ and let $A \mathbf{x} = \mathbf{0}$ and $B \mathbf{x} = \mathbf{0}$ be two homogenous Matrix Equation Systems. If $A \sim_{R} B$, then $${\mathbf{x} \in F^{n} : A \mathbf{x} = \mathbf{0}} = {\mathbf{y} \in F^{n} : B \mathbf{y} = \mathbf{0}}.$$ That is, the two Matrix Equation Systems have the same Solution Set.

Since $A \sim_{R} B$, there exists $e_{1}, \dots, e_{k} \in \text{Row}F^{m \times n}$ so that $$e_{1} \circ \cdots \circ e_{k}(A) = B.$$ Sincee row operations can be represented as matrices, we can rewrite this chain of Function Composition as $$E_{1}\cdots E_{k} A = B,$$ Matrix Multiplication is a Linear Function and Linear Operators preserve 0 so $E_{1} \cdots E_{k} \mathbf{0} = \mathbf{0}$. Thus, since Matrix Multiplication Distributes over Columns, we have that $$\begin{pmatrix}A &\Big| &\mathbf{0}\end{pmatrix} \sim_{R} \begin{pmatrix}B &\Big| &\mathbf{0}\end{pmatrix}.$$ Recalling that Row Equivalent Matrix Systems have the same Solution Set completes the proof. $\blacksquare$

• A Linear Equation System is Homogenous iff its constants are 0
• Matrix Equation Systems are Linear Systems
• Row Equivalent Matrix Systems
• Row Equivalent Matrices