Let $V$ be a vector space , $T:V \to V$ be a linear operator , then is it true that
$(\ker(T) \cap R(T) ) \times R(T^2) \cong R(T)$ ?
(note that the direct product is well-defined as both the spaces are subspaces of $V$ so have same ground field )
See $V$ be a vector space , $T:V \to V$ be a linear operator , then is $\ker (T) \cap R(T) \cong R(T)/R(T^2) $? , note that the isomorphism holds when $V$ is finite dimensional as it is known that if $U,W$ are finite dimensional vector spaces over a same field , then $\dim (U \times W)=\dim U + \dim W$ . Also note that the example given in the linked question doesn't violate the stated isomorphism here . Please help . Thanks in advance .
EDIT : One general thing that can be noted is that $R(T)/R(T)\cap \ker (T) \cong R(T^2)$ holds , but I don't know whether it can be used to prove or disprove the claim in question
NOTE : If $V,W$ are vector spaces over a field $F$ then $V \times W$ is a vector space with addition and multiplication defined as $(v,w)+(v',w'):=(v+v',w+w') $ ; $ k.(v,w):=(kv,kw)$ . It can be seen that if $B_V , B_W$ are bases for $V , W$ repectively , then $(B_V \times \{O_W\})\cup (\{O_V\}\times B_W)$ is a basis for $V \times W$ , so that if $V,W$ are finite dimensional then so is $V \times W$ and
$\dim (V \times W)=|(B_V \times \{O_W\}) \cup (\{O_V\}\times B_W)|=|B_V \times \{O_W\}|+|\{O_V\}\times B_W|=|B_V|+|B_W|=\dim V + \dim W$
So this "direct product" is substantially different from "direct sum"