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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that $\eta: V \to V/W$ defined as $\eta(x)=x+W$ is a surjective linear transformation, and $\ker(\eta)=W$.

My proof: Let $y \in V/W$. Then let $x=y+W$. So $\eta(x)=\eta(y+W)=(y+W)+W=y+W$

I don't think that last step is right at all. I'm not too sure how to find the kernel either.

Thanks in advance.

share|cite|improve this question
up vote 2 down vote accepted

The elements of $V/W$ are the sets of the form $x+W$ for $x\in V$, so if $y\in V/W$, then $y$ already is $x+W$ for some $x\in V$. That means that $y=\eta(x)$, so we’ve just proved that $\eta$ is surjective. You still have to prove that $\eta$ is linear and find $\ker\eta$.

To show that $\eta$ is linear, you have to show that for every $x,y\in V$ and scalars $\alpha,\beta$, $$\eta(\alpha x+\beta y)=\alpha\eta(x)+\beta\eta(y)\;.$$ To do that, use the definition of $\eta$ to write down what $\eta(\alpha x+\beta y)$ and $\alpha\eta(x)+\beta\eta(y)$ are, and see if you can see (and explain!) why they must be equal.

The kernel of $\eta$ is the set of vectors $x\in V$ such that $\eta(x)$ is the zero vector in $V/W$. What is that zero vector? Once you figure out what it is, it’s not to hard to figure out which elements $x+W$ of $V/W$ are equal to it.

share|cite|improve this answer

Hint: What is the zero vector of $V/W$?

share|cite|improve this answer
Is it just $0+W=W$? – tk2 Sep 24 '12 at 22:56
@tkrm Mhm...knowing that, what must the kernel be? – AsinglePANCAKE Sep 24 '12 at 22:57
The kernal must be W then. – tk2 Sep 24 '12 at 22:59

Your proof needs slight modification. Let $y \in V/ W$. By definition of what it means to be an element of $V / W$ we may choose $x \in V$ such that $y = x + W$. Then $\eta(x) = x+W = y$ which shows $\eta$ is surjective.

Then we need to show that $\ker(\eta) = W$. Firstly, we must consider what the zero vector in $V / W$ is. It is the equivalence class of $0\in V$. Hence, it is $W$. Let any $x \in W$ be given, then we certainly have $\eta(x) = x+W=W$ so $\ker(\eta) \supseteq W$. Alternatively, suppose $x \in V$ but $x \notin W$, then $\eta(x) = x+W \neq W$, so $x \notin \ker(\eta)$. It follows that $\ker(\eta) = W$.

Note: In the previous proof we used the fact that $x +W = W \iff x \in W$ since $W$ is a subspace.

share|cite|improve this answer

Your Answer


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.