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Let $T: V \longrightarrow V, S: V \longrightarrow V$ be two linear operators. Let $P: V \longrightarrow V$ be another linear operator. Suppose $P \circ S=S \circ P$, then prove or disprove that $$ \operatorname{dim}(\operatorname{ker}(T \circ S+P))=\operatorname{dim}(\operatorname{ker}(S \circ T+P)) $$

Is there any isomorphism possible from the space $R_1=\{T(S(x))+P(x):~x \in V\}$ to $R_2=\{S(T(y))+P(y): y \in V\}$ using the fact $P(S(x))=S(P(x))$ so that we can apply Rank-Nullity theorem? Is there any result regarding $\dim \ker (T+P)$ for linear maps $T$ and $P$?

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This is not true. Counterexample: $$ T=\left(\begin{array}{cc|cc}0&0&1&0\\ 0&0&0&1\\ \hline 1&0&0&0\\ 0&1&0&0\end{array}\right), \ S=\left(\begin{array}{cc|cc}0&0&0&0\\ 0&1&0&0\\ \hline 0&0&0&1\\ 0&0&0&0\end{array}\right), \ P=\left(\begin{array}{cc|cc}0&0&0&0\\ 0&0&0&0\\ \hline 0&0&1&0\\ 0&0&0&1\end{array}\right), $$ $$ TS+P=\pmatrix{0&0&0&1\\ 0&0&0&0\\ 0&0&1&0\\ 0&1&0&1}, \ ST+P=\pmatrix{0&0&0&0\\ 0&0&0&1\\ 0&1&1&0\\ 0&0&0&1}. $$ We have $PS=SP$ but $\operatorname{rank}(TS+P)=3\ne2=\operatorname{rank}(ST+P)$.

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Note, however, if $S$ is assumed to be invertible, then the statement is true:

If $S$ is invertible, then $T\circ S+P$ has same rank as $(T\circ S+P)\circ S^{-1}=T+P\circ S^{-1}=T+S^{-1}\circ P$ which has the same rank as $S\circ (T+S^{-1}\circ P) =S\circ T+P$.

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