 Solution

To find the kernel of a linear map, perform the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-(1+\alpha )R_1$, followed by $R_3\leftarrow R_3-R_2$. The equivalent homogeneous equations areThe resulting determinant is ${\alpha }^2(3-\alpha )$. If $\alpha \notin \{0,3\}$ all coordinates vanish. At $\alpha =0$ only $x-y+z=0$ remains; at $\alpha =3$ the equations give $y=2z$, $x=-5z$. Therefore kernel bases areThe empty family is the basis of the zero vector subspace.

For the change of basis, let $B_0,C_0$ be the matrices whose columns are the two ordered bases in standard coordinates. Then $v=B_0[v{]}_B=C_0[v{]}_C$, so $[v{]}_B=S[v{]}_C$ with $S=B_0^{-1}{C}_0$. If $A$ and $A^{\prime }$ represent the same linear map in the two bases, $[\Phi (v){]}_B=A[v{]}_B$ and also $[\Phi (v){]}_B=S{A}^{\prime }[v{]}_C$. Therefore $S{A}^{\prime }=AS$, andFor the concrete bases, their vectors satisfy $c_1=b_1+b_2$, $c_2=b_2+b_3$, $c_3=b_1+b_2+b_3$. These coordinate columns giveAt $\alpha =0$, every row of $A_0$ is $(1,-1,1)$, so every row of $A_{0}S$ is $(0,0,1)$. Multiplying on the left by the displayed inverse givesThe original PDF asks for this full product; its final factor $S$ is missing from the converted TeX.