Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 6 / c

Gaussian elimination gives the row operations $R_2\leftarrow R_2-3R_1$, $R_3\leftarrow R_3-R_1$, $R_4\leftarrow R_4-2R_1$, producing the lower three rows $(0,1,4,1)$, $(0,1,4,1)$, $(0,-3,2,-1)$. Subtract the new row two from row three and add three times row two to row four; the remaining nonzero row is $(0,0,14,2)$. Move it above the zero row and finish elimination to obtainThe pivot columns are one, two and three. Row operations preserve the linear dependence relations among columns, so the corresponding columns of the original matrix form a basis for the column space, which is the image of the linear map:In particular, the matrix rank is three. Columns of the reduced matrix itself would generally give a different subspace, so the original columns must be used.

The reduced row echelon form gives $u_1=2u_4/21$, $u_2=-3u_4/7$, and $u_3=-u_4/7$. Taking $u_4=21t$ gives every vector in the null space as $t(2,-9,-3,21{)}^{\top }$. Thus a basis for the kernel of the linear map isDirect multiplication gives $A(2,-9,-3,21{)}^{\top }=0$. The rank-nullity theorem also checks the count: matrix rank three plus nullity one equals the domain dimension four.

No. The nonzero vector $v=(2,-9,-3,21{)}^{\top }$ is in the kernel of the linear map, so $T(v)=T(0)=0$ although $v\ne 0$. A linear map is injective exactly when its kernel consists only of zero.