Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 6C (Vectors and Matrices)

Write a vector in the domain as $(u,v,w,t{)}^T$. The first and third equations defining the kernel of a linear map imply $v=2\alpha u-2t$. Substitution into the second equation then givesFor $\alpha \ne 1$, this forces $t=(\alpha +1)u$, $v=-2u$ and $w=3u$. HenceThe kernel of a linear map has dimension one, so the rank-nullity theorem gives rank three and proves the assertion about the image of a linear map. Alternatively, the minor formed from columns $2,3,4$ has determinant $2(\alpha -1)$, including a nonzero value at $\alpha =-1$.

For $\alpha =1$, the parameters $u,t$ are free and $v=2u-2t$, $w=-3u+3t$. ThusThe third row is the first row minus the second. The image of a linear map is therefore contained in $y_1-y_2-y_3=0$. The first and third columns, $(1,2,-1{)}^T$ and $(1,0,1{)}^T$, are linearly independent, so the matrix rank is two and this containment is equality:This is the sole kernel jump in a parameter-dependent linear map; there is no further exceptional case at $\alpha =-1$.

Let $e_1,\dots ,e_n$ be the standard basis and define $a_j=f(e_j)$. By linearity,Thus every linear functional has this form. Its representing vector is unique because evaluating $a\cdot x=b\cdot x$ at each $e_j$ gives $a_j=b_j$.

For the prescribed kernel of a linear map, the third listed vector is the first minus the second, so it adds no independent constraint. The first two requireTaking $a_1=s$ and $a_4=t$ gives $a_2=2s-2t$, $a_3=-3s+3t$. Consequently all possible representing vectors areConversely, every such vector is perpendicular to each of the three prescribed vectors, so its linear functional vanishes on them. This also includes the zero linear functional.