Similarity classification of idempotent linear maps
= Similarity classification of idempotent linear maps
If $T^2=T$, every <vector> splits as $v=(v-Tv)+Tv$ in $\ker T\oplus\operatorname{im}T$. On these summands $T$ is respectively zero and the identity. Its <matrix> in an adapted <basis> is therefore $\operatorname{diag}(0,I_r)$, where $r$ is its <rank of a linear map>. Two <idempotent linear maps> on the same finite-dimensional <vector space> are <similar matrices> exactly when their ranks agree.