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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?

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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 $$Tv_1=Tv_2=v_1-v_2,\quad Tv_3=Tv_4=v_3-v_4.$$

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  • $\begingroup$ 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
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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 $Tv_3 = Tv_4 = Te_4 = e_2.$

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  • $\begingroup$ 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

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