I'm given this representation matrix of a linear operator in $\mathbb{R^4}$ , i'm trying to find the values of $a,b,c,d,e,f$. Now to my understanding every column vector of the matrix is $[Tv_1]_B,[Tv_2]_B,[Tv_3]_B,[Tv_4]_B$. i try to find the third column by $[Tv_3]_B=[T]_B*v3=(3,3,3,3)$ and then i try to find [Tv4]_B in the same way but i'm not comfortable with my answer because i get $[Tv_4]_B=[T]_B*v_4=(1,1,1,1)$ but the given vector $[Tv_4]_B$ starts with $4$ am i going at this wrong?
Ok, after clarification in the chat, I know that $$ \dim \ker T =3 $$ the Rank–nullity theorem implies that $$ \dim \operatorname{Im} T = 4- 3 = 1$$ Meaning that for every matrix representation, $$ \dim \operatorname{rank} [T]_B = 1$$ And so there is only one vector in the basis of the space generated by the matrix's columns. One can see it is the vector $(1,1,1,1)^T$ Now you can understand that the missing columns are $(3,3,3,3) , (4,4,4,4)$