Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.