# Take a matrix `A = ([r_1],[r_2],[.],[.],[.],[r_m])` Let the vector `x` be in the null space of A. Show that the vector `x` is perpendicular to all of the row vectors `r_k`.

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### 1 Answer

The null space or kernel of `A` (m x n) is the set of all vectors `x = (x_1,x_2,...x_n)` (length n) such that

`Ax = overline(0)`

where `overline(0)` is the zero vector and has length n.

Now if ` ` `A = ([r_1],[r_2],[.],[.],[.],[r_m])`

where `r_k` are the row vectors `k in [1,m]` then, necessarily,

`r_k .x = r_(k1)x_1 + r_(k2)x_2 + ... + r_(kn)x_n =0` ` ``forall k in [1,m]`.

In words, the dot product or inner product of the vector `r_k` and the vector `x`.

This implies that each of the `r_k` row vectors are orthogonal to `x`, ie are perpendicular to `x`.

**The dot product of each row vector and the null space vector x is 0. Therefore the row vectors are all perpendicular to x.**