PropositionNull space and orthogonal complement

The null space of an \((m,n)\)-matrix \(\Mtrx{A}\) is the orthogonal complement of the row vectors of \(\Mtrx{A}\) in \(\RNrSpc{n}\). Therefore \(\NullSpc{A}\) is a subvector space of \(\RNrSpc{n}\).

Proof

By definition, the null space of \(\Mtrx{A}\) consists of all solutions of the matrix equation

\(\Mtrx{A}\cdot\Vect{x}\)\(=\)\(\Vect{0}\)

Now express \(\Mtrx{A}\) in terms of its row vectors:

\[ A = \left[ \begin{array}{c} R_1 \\ \vdots \\ R_m \end{array} \right] \]

Then we find:

\[ \Mtrx{A}\cdot\Vect{x} = \left[ \begin{array}{c} R_1 \\ \vdots \\ R_m \end{array} \right] \Vect{x} = \left[ \begin{array}{c} \DotPr{ R_1 }{ \Vect{x} } \\ \vdots \\ \DotPr{ R_m }{ \Vect{x} } \end{array} \right] \]

and so we see that \(\Vect{x}\) is a solution of \(\Mtrx{A}\cdot \Vect{x}=\Vect{0}\) exactly when \(\Vect{x}\) is perpendicular to each of the row vectors \(R_1\), ... , \(R_m\). But this happens exactly when \(\Vect{x}\) belongs to the null space of \(\Mtrx{A}\).