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Let $V$ a vector space and $W$ be its linear subspace. Give an example of a linear map that satisfies $\mathrm{im}(f)=W$ and $\ker(f) \oplus \mathrm{im}(f)=V$, but $f^2 \neq f$.

Would $f(v)=2v$ be the right example? Since the kernel of it is $0$ and the map itself is surjective so the condition $\ker(f)\oplus \mathrm{im}(f)=V$ satisfied, also $W=V$ in this case and $f^2 \neq f$ is also satisfied.

A next question is once I restrics $f$ to $W$ then will it be always isomorohism. My approach was that the map will be surely surjective, since $f:W\rightarrow W$ and $W=\mathrm{im}(f)$. But the injectivity I check by using the given $\ker(f)\oplus \mathrm{im}(f)=V$. Can someone help me on that. I am stuck. Please.

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Your map $v\mapsto 2v$ works only for the special case $W=V$ and characteristic $\ne 2$. I guess you are supposed to find such $f$ for an arbitrary subspace $W$ of $V$. Note that with $f^2=f$, we would have a projection from $V$ to $W$. Starting from such a projection, yuo can find a map as requested by composing with something like $v\mapsto v$.

Then again, some conditions seem to be misisng from the problem statekemnt:

  • If $W=0$, then $f^2=f$ cannot be avoided.
  • If $\dim W=1$ and the ground field is $\mathbb F_2$, then $f^2=f$ cannot be avoided.

But if your task is really merely to find some example, then you are fine.

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For the second question, there are 3 things to check: (i) $f(W) \subset W$; (ii) $f(W) \supset W$ (surjectivity); and (iii) $f|_W$ is injective. Item (i) is immediate from the hypothesis that $W = \operatorname{Im}(f)$. For item (ii), let $w\in W$. We need to find an $x\in W$ with $f(x) = w$. Since $W = \operatorname{Im}(f)$, we can find a $y\in V$ with $f(y) = w$. Since $V = W\oplus \operatorname{ker}(f)$, we can write $y = x+z$, with $x\in W$ and $f(z) = 0$. Therefore $w = f(x) + f(z) = f(x)$. Finally, to prove injectivity, let $x\in W$ and suppose $f(x) = 0$. Then $x\in \operatorname{ker}(f)$. By definition of a direct sum, the intersection of $\operatorname{ker}(f)$ and $W$ is 0, so $x = 0$, which proves injectivity.

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