1
$\begingroup$

Prove that if $f : F^4 → F^2$ is a linear map such that $$\ker (f)= \big\{ (x_1, x_2, x_3, x_4)^T\ :\ x_1 = 3x_2,\ x_3 = 7x_4\big\}$$ then $f$ is surjective.

I know that all $x_1,x_2,x_3,x_4$ that satisfy the above properties are in the kernel, so I am assuming they are mapped to $(0,0)^T$. However, how do I show that all other elements of $F^2$ are getting mapped onto? That is for an arbitary pair $y_1,y_2 \in F^2$ how do I know that there exist $f(x_1,x_2,x_3,x_4)^T=(y_1,y_2)^T$ and not a single element is missed?

$\endgroup$
1
  • 2
    $\begingroup$ You might consider to show that the kernel has dimension $2$ and some theoretical results. $\endgroup$
    – mfl
    Jan 25, 2016 at 21:01

1 Answer 1

2
$\begingroup$

Hint: If $f\colon U \to V$ is linear and $U,V$ are finite dimensional vector spaces, then $$\dim(U)=\dim(\operatorname{Im}(f))+\dim(\ker(f))$$

Show that $\dim(\ker(f))=2$. Then we have $\dim(\operatorname{Im}(f))=2$ and conclude using $\operatorname{Im}(f)\subset F^2$.

$\endgroup$
6
  • $\begingroup$ So using the Rank Nullity I can show that $im(f)=2$, but I'm not able to see why this would imply surjectivity. Certanly I see that $2<4=dim U$ $\endgroup$
    – GRS
    Jan 25, 2016 at 21:07
  • $\begingroup$ @Epsilon $f$ is surjective if $\operatorname{Im}(f)=V=F^2$ $\endgroup$
    – Surb
    Jan 25, 2016 at 21:08
  • $\begingroup$ thanks, I didn't know this fact $\endgroup$
    – GRS
    Jan 25, 2016 at 21:09
  • $\begingroup$ @Epsilon I recommend you to recall the definition of $\operatorname{Im}(f)$ to understand why it is true. $\endgroup$
    – Surb
    Jan 25, 2016 at 21:10
  • $\begingroup$ So since the image has dim $2$, it means that there are $2$ elements from $F^4$ mapping to it at all times. Thus every $x,y$ is mapped on by some elements in $F^4$? $\endgroup$
    – GRS
    Jan 25, 2016 at 21:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .