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Suppose $W_1$, $W_2$ are subspaces of a finite dimensional vector space $V$. Show that the map $T: W_1\times W_2 → V$ defined by $T(w_1, w_2) = w_1 + w_2$ is linear.

What is $\ker(T)$ and $\operatorname{Im}(T)$?

Use the first isomorphism theorem to deduce $\dim(W_1+W_2)=\dim W_1 +\dim W_2 -\dim(W_1∩W_2)$.

$\operatorname{Im}(T)= W_1+W_2$ I get that and kernel is $W_1+W_2=0$? If kernel was $W_1 ∩ W_2$ then I could apply isomorphism theorem to conclude the result. But is the kernel $W_1∩ W_2$ ??

If not how to do it?

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  • $\begingroup$ hey anyone have any ideas? $\endgroup$
    – ramAN
    Oct 6, 2020 at 20:16

3 Answers 3

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Note that $$\ker(T) = \{(w,-w) : w \in W_1 \cap W_2\} \cong W_1 \cap W_2$$ and then, by the rank-nullity theorem we have \begin{align} \dim(W_1) + (W_2) &= \dim(W_1 \times W_2) \\ &= \dim(\ker T) + \dim(\operatorname{im} T) \\ &= \dim(W_1 \cap W_2) + \dim(W_1 + W_2). \end{align}

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  • $\begingroup$ i get this . can u deduce then the dim(W1+w2) part $\endgroup$
    – ramAN
    Oct 6, 2020 at 20:31
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Hint:

Check you have an exact sequence: \begin{alignat}{2} 0\longrightarrow W_1\cap W_2 &\longrightarrow&W_1\oplus W_2&\longrightarrow W_1+W_2 \longrightarrow 0\\ w&\longmapsto& (w,-w)\\ & &(w_1,w_2)&\longmapsto w_1+w_2 \end{alignat}

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As you note, the first isomorphism theorem will work, if $\rm{ker}T=W_1\cap W_2$.

First, note that $W_1\cap W_2$ isn't a subset of $W_1×W_2$. But it's isomorphic to a subset under the isomorphism $v\mapsto(v,v)$. So we make that identification.

For one inclusion, let $T(w_1,w_2)=0$. Then $w_1+w_2=0$. Then $w_1=-w_2\in W_1\cap W_2$. So $(w_1,w_2)=(w_1,-w_1)\in W_1\cap W_2$.

For the other, let $w\in W_1\cap W_2$. Then, under our identification, we have $w=(w,w)$. Thus $T(w)=w-w=0$.

So we see that the kernel of $T$ has the same dimension as $W_1\cap W_2$. The result follows.

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