Question

Find a basis for R(A)^⊥, where R(A) denotes the row space of the matrix
A where A is a 3x4 matrix (1 0 4 0 \ 0 1 2 -1 \ 1 -1 2 1)

Answer 1

$R(A)^{\perp}=$ a basis of the orthogonal complement of $R(A).$

Let $R(A)=\{ v_1,v_2,v_3 \},$ where

$v_1=(1,0,4,0),v_2=(0,1,2,-1) \ \text{and} \ v_3=(1,-1,2,1)$.

Since $v_2+v_3-v_1=0,$ the vectors $v_1, v_2, v_3$ are linearly dependent. Thus, $\{v_1,v_2\}$ is the maximal subset of linearly independent vectors in the row space.

If $w_i \in R(A)^{\perp}$, $w_i=(a_i,b_i,c_i,d_i),$ $v_1=(1,0,4,0) \in R(A),$ $v_2=(0,1,2,-1) \in R(A),$ then $w_i\cdot v_1=0,w_i \cdot v_2=0,$ hence $a_i+4c_i=0, b_i+2c_i-d_i=0.$

If $c_1=1, d_1=0,$ then $a_1=-4c_1=-4,$ $b_1=-2c_1+d_1=-2,$ hence $w_1=(a_1,b_1,c_1,d_1)=(-4,-2,1,0).$

If $c_2=0,d_2=1,$ then $a_2=-4c_2=0,$ $b_2=-2c_2+d_2=1,$ hence $w_2=(a_2,b_2,c_2,d_2)=(0,1,0,1).$

Thus, $R(A)^{\perp}=\{w_1,w_2\}=\{ (-4,-2,1,0),(0,1,0,1)\}.$