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Let V be a vector space over the field $k$, and suppose that $V$ is a direct sum, $V=W_1\oplus W_2$, of subspaces $W_1$, $W_2$. Let $g_1$ be a symmetric bilinear form on $W_1$, and let $g_2$ be a symmetric bilinear form on $W_2$. Show that there exists a unique bilinear form $g$ on $V$,such that, if $v=v_1+v_2$ and $w=w_1+w_2$ are elements of $V$, with $v_1,w_1\in W_1$ and $v_2,w_2\in W_2$, then $g(v,w)=g_1(v_1,w_1)+g_2(v_2,w_2)$.

I guess I have to use the bilinear form properties. But how can I prove it? It seems to me intuitive that $g$ is a bilinear form of the whole of $V$. After looking for some theorems, they prove of no use. I cannot even start the proof. I do not know how to prove this statement.

Questions:

Can someone provide me a proof?

Thanks in advance!

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  • $\begingroup$ The only thing you need is that the decomposition $v=v_1+v_2$ as a sum of elements of $W_1$ and $W_2$ is unique. Then $g(v,w)=g_1(v_1,w_1)+g_2(v_2,w_2)$ unambiguously defines a map, and proving it bilinear is then routine. $\endgroup$ – Angina Seng Jul 22 '17 at 14:44
  • $\begingroup$ @LordSharktheUnknown What do you mean by unique? $\endgroup$ – Pedro Gomes Jul 22 '17 at 14:46
  • $\begingroup$ @LordSharktheUnknown Could you give me more hints? $\endgroup$ – Pedro Gomes Jul 22 '17 at 15:02
  • $\begingroup$ @PedroGomes Here, uniqueness of the bilinear form $g$ means essentially the same thing as does the uniqueness of a direct sum decomposition. You should postulate there exists another $\tilde{g}$ satisfying the same properties and then show $g = \tilde{g}$. $\endgroup$ – Benighted Jul 22 '17 at 16:39
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You need to use the definition of a direct sum (i.e. of $\oplus$) here. The definition you use (or a relevant theorem regarding direct sums), you should be able to immediately say

Every $v \in V$ can be uniquely written in the form $v = v_1 + v_2$ with $v_1 \in W_1$ and $v_2 \in W_2$. (In other words: if $v_1 + v_2 = w_1 + w_2$ with $v_j,w_j \in W_j$, then $v_1 = w_1$ and $v_2 = w_2$)

Now, we can define a bilinear form $g$ by $$ g(v,w) = g(v_1,w_1) + g(v_2,w_2) $$ where the $v_j,w_j \in W_j$ are chosen so that $v_1 + v_2 = v, w_1 + w_2 = w$, as is guaranteed by the definition of $\oplus$. Because the decomposition is unique, this function is unambiguous. That is, there is only one function that fits this description. It remains to be shown that $g$ is bilinear.

To that end, note that for $v,v' \in V$ and $\alpha \in K$, note that $$ g(v,w) + \alpha\,g(v',w) = g_1(v_1,w_1) + \alpha g_1(v_1',w_1) + g_2(v_2,w_2) + \alpha\, g_2(v_2',w_2) \\ = g_1(v_1 + \alpha v_1',w_1) + g_2(v_2 + \alpha v_2',w_2) $$ Now, here's a tricky point: take $\tilde v = v + \alpha v'$. We see that $\tilde v_1 = v_1 + \alpha v_1'$ and $\tilde v_2 = v_2 + \alpha v_2'$ is such that $\tilde v_1 \in W_1$, $\tilde v_2 \in W_2$, and $\tilde v = \tilde v_1 + \tilde v_2$. It follows from this and the above that $$ g(v + \alpha v', w) = g(\tilde v,w) = g_1(\tilde v_1 ,w_2) + g_2(\tilde v_2, w_2) \\ = g_1(v_1 + \alpha v_1',w_2) + g_2(v_2 + \alpha v_2',w_2) \\ = g(v,w) + \alpha\,g(v',w) $$ So, we see that $g$ is indeed linear in the first argument. Similarly, we may argue that $g$ is linear in its second argument, and hence bilinear.

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  • $\begingroup$ I have just understood your answer! Thanks a lot! You have been very helpful on this forum! $\endgroup$ – Pedro Gomes Jul 22 '17 at 17:46

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