P is a 2*4 matrix, which has rank (P) = 2, L: M 4*4 -> M 2*2 is a linear mapping, defined by L(A) = P A P^T, ---(PAP transpose). I can see that L is not one-to-one, as A must be in the null-space of P and thus the Ker(L) has to be non-zero. But how to prove or disprove L is onto?
Thanks!
Since $\operatorname{rank}(P)=2$, $P$ is the matrix of a surjective linear map from $K^4$ to $K^2$. This linear map has a linear section, from $K^2$ to $K^4$, with matrix $S$, i.e. $PS=I_2$. Now let $B$ be a 2×2 matrix, and set $A= SB \,^\mathrm{t}\!S $: it is a 4×4 matrix and $$PA \,^\mathrm{t}\!P=PSB \,^\mathrm{t}\!S \,^\mathrm{t}\!P=(PS)B \,^\mathrm{t}(PS )=I_2BI_2=B.$$ Thus, the mapping is surjective.