Null Space of Reduced Echelon Form
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Theorem
Let $\mathbf A$ be a matrix in the matrix space $\map {\MM_\R} {m, n}$ such that:
- $\mathbf A \mathbf x = \mathbf 0$
represents a homogeneous system of linear equations.
The null space of $\mathbf A$ is the same as that of the null space of the reduced row echelon form of $\mathbf A$:
- $\map {\mathrm N} {\mathbf A} = \map {\mathrm N} {\map {\mathrm {rref} } {\mathbf A} }$
Proof
By the definition of null space:
- $\mathbf x \in \map {\mathrm N} {\mathbf A} \iff \mathbf A \mathbf x = \mathbf 0$
From the corollary to Row Equivalent Matrix for Homogeneous System has same Solutions:
- $\mathbf A \mathbf x = \mathbf 0 \iff \map {\mathrm {rref} } {\mathbf A} \mathbf x = \mathbf 0$
Hence the result, by the definition of set equality.
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.