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Let $T$ be a linear transformation on a finite dimensional vector space $V$ and let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects ?

$(a)$ $T(V_1\cap V_2) = T(V_1) \cap T(V_2).$

$(b)$ $T(V_1 \cup V_2) = T(V_1) \cup T(V_2).$

My attempts : i know that intersection two subspaces is subspaces and union of two subspace need not to be subspaces

so option $a)$ is correct

option $b)$ is not corrects

Is it true ??

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  • $\begingroup$ Have you tried to write down a standard double-inclusion argument, or perhaps a counterexample, in either case? That's where I'd start. $\endgroup$ – The Count Aug 6 '18 at 21:37
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    $\begingroup$ Try approaching the question via definition and element chasing. Suppose that $x\in T(V_1\cap V_2)$. Then there must be some $y\in V_1\cap V_2$ such that $T(y)=x$. Does it follow then that $x\in T(V_1)$? Does it also follow that $x\in T(V_2)$? How about whether or not $x\in T(V_1)\cap T(V_2)$? (The element $y$ that I point out the existence of plays a critical role in these last few steps). Now... there is one more direction left for part (a) and two directions to do for part (b). Can you continue? $\endgroup$ – JMoravitz Aug 6 '18 at 21:37
  • $\begingroup$ Your remarks about intersection and union are correct, nevertheless, that does not completely verify nor refute the equalities. Even if you have established that both sides are subspaces, you still have to show them to be equal. $\endgroup$ – blub Aug 6 '18 at 21:39
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(a) is false. Take $T\colon\mathbb{R}^2\longrightarrow\mathbb{R}^2$ defined by $f(x,y)=(x+y,0)$. If$$V_1=\{(x,0)\,|\,x\in\mathbb{R}\}\text{ and }V_2=\{(0,x)\,|\,x\in\mathbb{R}\},$$then $T(V_1\cap V_2)=T(\{0\})=\{0\}$, whereas $T(V_1)\cap T(V_2)=\{(x,0)\,|\,x\in\mathbb{R}\}$.

(b) this holds for every funtion (linear or otherwise).

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  • $\begingroup$ for b) if i will take ur counter example as u have taken for option a) then it will not True ..can elaborate option b) more ...@Jose sir $\endgroup$ – Messi fifa Aug 6 '18 at 21:45
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    $\begingroup$ @Messififa $T(V_1\cup V_2)=T(V_1)\cup T(V_2)=\{(x,0)\,|\,x\in\mathbb{R}\}$. $\endgroup$ – José Carlos Santos Aug 6 '18 at 21:48
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No, in contrary $$T(V_1 \cup V_2) = T(V_1) \cup T(V_2)$$is the true choice. Note that $$ V_1 \cup V_2$$ does not have to be a subspace for the above statement to be true.

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Your reasoning is wrong - the question didn't ask whether any of those sets were subspaces.

And your conclusions are both wrong as well. Another answer already shows that (a) is false. In fact (b) is true.

If $y\in T(V_1\cup V_2)$ then there exists $x\in V_1\cup V_2$ with $y=T(x)$. But $x\in V_1\cup V_2$ says $x\in V_1$ or $x\in V_2$; hence $y=T(x)$ is either in $T(V_1)$ or $T(V_2)$, so $y\in T(V_1)\cup T(V_2)$.

So $T(V_1\cup V_2)\subset T(V_1)\cup T(V_2)$. Conversely, $V_1\subset V_1\cup V_2$ shows that $T(V_1)\subset T(V_1\cup V_2)$. Similarly $T(V_2)\subset T(V_1\cup V_2)$, hence $T(V_1)\cup T(V_2)\subset T(V_1\cup V_2)$.

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