Chapter 4 - Vector Spaces - 4.4 Spanning Sets - Problems - Page 283: 25

See below

Given: $A=\begin{bmatrix} 1 & 4 \end{bmatrix}$ Obtain: $\begin{bmatrix} 1 & 4 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}=0$ We have the system: $x_1+4x_2=0\\ x_1-4x_2=0$ Since $N(A)=(x:Ax=0)\\ \rightarrow N(A)=((x_1,x_2)\in R^2:x_1=-4x_2\\ \rightarrow (x_1,x_2) \in N(A),x_1=-4x_2\\ \rightarrow (-4x_2,x_2)=x_2(-4,1)$ Hence, $N(A)=span (-4,1)$