I believe that there are counterexamples for the following statements:

Let $V$ be a vector space and let $V_1$ and $V_2$ be subspaces.

  1. $V\cong V_1\times V_2 \Rightarrow V=V_1\oplus V_2$


  1. $\phi:V\rightarrow V$ linear operator $\Rightarrow V\cong \ker \phi\times I m \phi$

I am not so sure about the first statement. I know that $ V=V_1\oplus V_2$ if and only if the map $\psi:V_1\times V_2\rightarrow V_1\oplus V_2$ defined by $(v_1,v_2)\mapsto v_1+v_2$ is an isomorphism, but the statement for (1) above is not exactly this last statement I wrote.

I see that (2) holds if $\phi$ is invective, but I can't find a counterexample for the general statement. Finding counterexamples has never been my strong suit. Thand you for your help.

  • $\begingroup$ Finite direct product and finite direct sum are the same thing ($\prod_{i = 1}^n V_i \cong \bigoplus_{i=1}^n V_i$), maybe the first statement should say something like $V_1,V_2 \leq V$. $\endgroup$ – Gilberto López Jul 17 '17 at 17:32
  • $\begingroup$ For the second statement, remember the First Isomorphism Theorem for vector spaces. $\endgroup$ – Gilberto López Jul 17 '17 at 17:35

For your first statement, there is indeed a counterexample (with $V_1,V_2 \subseteq V$, moreover). Take $$ V_1 = V_2 = \{(t,0): t \in \Bbb R\} \subset \Bbb R^2 = V $$ We have $V \cong V_1 \times V_2$, but $V \neq V_1 \oplus V_2$. Look at the definition of $\oplus$ to verify that we indeed have $V \neq V_1 \oplus V_2$.

For your second statement: there is no counterexample. If you are only working on finite dimensional vector spaces, the rank-nullity theorem is sufficient here. Otherwise, recall that with the first isomorphism theorem, we have $$ V/ \ker \phi \cong \operatorname{Im} \phi $$ and $V = (V/\ker \phi) \oplus (\ker \phi)$

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