# Find the $\ker T$ and deduce $\dim(W_1+W_2)=\dim W_1 +\dim W_2 -\dim(W_1∩W_2)$

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?

• hey anyone have any ideas? Oct 6, 2020 at 20:16

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}

• i get this . can u deduce then the dim(W1+w2) part Oct 6, 2020 at 20:31

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}

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.