Let V be a vector space over $\mathbb{R}$ and let $B=\{v_1,v_2,v_3,v_4\}$ be its basis.
Let $T : V \to V$ be a linear transformation which satisfies the following condition:

  1. $T(v_1) = T(v_2)$
  2. $T(v_3) = T(v_4)$
  3. $kerT \subseteq ImT$

Can you find an example for a basis which satisfies the above conditions?
(The question actually asks to prove that $kerT = ImT$, but the conditions which are presented here seem unreasonable.)

Take for example the standard basis:
$$ T(e_1) = T(e_2) = e_1 \\ T(e_3) = T(e_4) = e_3 $$ Thus the image is spanned by $\{e_1,e_3\}$, but if $kerT = ImT$, then $T(e_1) = 0$ !
We conclude that $ImT = \{0\}$ which isn't true according to the data of the question.
What am I missing?


Let $w_1=v_1-v_2,w_2=v_3-v_4$. Then $w_1,w_2\in\ker T$ and they are linearly independent. Hence $\dim\ker T\ge2$. By the rank-nullity theorem, $\dim\text{im}T=4-\dim\ker T\le2$. However $\ker T\subseteq\text{im}T$ implies $2\le\dim\ker T\le\dim\text{im}T\le2$ so we must have $\ker T=\text{im}T$.

To find an example of such an operator, note that you cannot have $Tv=v$ for any $v\ne0$, since this would imply $v\in\text{im}T=\ker T$ so $v=Tv=0$, a contradiction. Try instead $$Tv_1=Tv_2=v_1-v_2,\quad Tv_3=Tv_4=v_3-v_4.$$

  • $\begingroup$ In the last example - Amongst $v1,v2,v3,v4$, who is in $kerT$ and who is in $ImT$? $\endgroup$ – Dor Feb 9 '15 at 7:39
  • $\begingroup$ None. We have $\ker T=\text{im}T=\text{span}\{v_1-v_2,v_3-v_4\}$. $\endgroup$ – Jason Feb 9 '15 at 8:03

what about a transformation $T$ defined by $Te_1 = Te_2 = 0, Te_3 = e_1, Te_4 = e_2$ then a basis for both $ker T$ and $im T$ is $\{e_1, e_2\}$

we can fix $v$'s by $v_1 = e_1 + e_3, v_2 = -e_1 + e_3, v_3 = e_2+e_4, v_4=-e_2 + e_4.$

showing $\{v_1, v_2, v_3, v_4\}$ is linearly independent. suppose $av_1+ bv_2 + cv_3 + dv_4 = 0.$ then $a - b= 0, c+d = 0, a + b= 0, c-d = 0$. only solution is $a = b = c = d = 0.$ that shows the independence of the $v$'s.

we can verify that $Tv_1 = Tv_2 = Te_3 = e_1$ and $Tv_3 = Tv_4 = Te_4 = e_2.$

  • $\begingroup$ I'm not sure that I understand the example with the $v$'s. Why does that "fix" them? (I understood why they are linearly independent). $\endgroup$ – Dor Feb 9 '15 at 7:41
  • $\begingroup$ $e$'s satisfy your third condition. you need to satisfy the first two. $v_1, v_2$ have one element from the $ker$ and one from $im$ so that $Tv_1 = Tv_2$. same for $v_3, v_4.$ $\endgroup$ – abel Feb 9 '15 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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