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`.

1 Answer | Add Yours

mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

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.

 

 

 

 

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question