let $V_1$ and $V_2$ be subspaces of $V$. then which of the following statements is Corrects?

Question

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 ??

Answer 1

(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).

Answer 2

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.

Answer 3

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 $T(V_1)\cup T(V_2)\subset T(V_1\cup V_2)$.