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From Serge Lang's Linear Algebra:

Let $V$ be a finite dimensional space over $\mathbb{R}$, and let $\langle \, , \, \rangle$ be a scalar product on $V$. Show that $V$ admits a direct sum decomposition:

$$V = V^{+1} \oplus V^{-1} \oplus V^0$$

Show that the dimensions of the spaces $V^+$, $V^-$ are the same in all such decompositions.

Considering that $V$ is finite dimensional vector space, assume that $V=\mathbb{R}^n$ and consider some basis $\{v_1, ... , v_n\}$ of $V$. Then:

$$\{\langle v_i, v_i \rangle > 0 \mid v_i \in \{v_1, ... , v_n\} \} \subset V^{+1}$$ $$\{\langle v_i, v_i \rangle < 0 \mid v_i \in \{v_1, ... , v_n\} \} \subset V^{-1}$$ $$\{\langle v_i, v_i \rangle = 0 \mid v_i \in \{v_1, ... , v_n\} \} \subset V^{0}$$

In terms, $V^{+1}$, $V^{-1}$ and $V^{0}$ are all subspaces respectively having vectors such that they are positive definite, negative definite and zero.


My "proof":

In order to show that $V$ is composed of subspaces $V^{+1}$, $V^{-1}$ and $V^{0}$, it must be shown that there exists unique elements $v^{+1} \in V^{+1}$, $v^{-1} \in V^{-1}$ and $v^{0} \in V^{0}$ such that $\forall v \in V, v=v^{+1} + v^{-1} + v^{0}$. Also, we know that these elements are unique iff $V^{+1} \cap V^{-1} \cap V^{0} = \{0\}$.

Scalar product is a bilinear form $\langle \, , \, \rangle: V \times V \rightarrow \mathbb{R}$, thus obviously $(\langle v, v \rangle > 0) \, \lor \, (\langle v, v \rangle < 0) \, \lor \, (\langle v, v \rangle = 0), \forall v \in V$. Equivalently:

$$(v=v^{1} + O + O) \, \lor \, (v=O + v^{-1} + O) \, \lor \, (v=O + O + v^{0}), \forall v \in V$$

This is because all subspaces have zero element $O$ in common, thus every for $v \in V$, $v = v^{+1} + v^{-1} + v^{0}$.

Now it must be shown that $V^{+1} \cap V^{-1} \cap V^{0} = \{0\}$. If there exists $v^{+1}, v^{-1}, v^{0}$ such that $v^{+1} = v^{-1} = v^{0}$, then $\langle v^{+1}, v^{+1} \rangle = \langle v^{-1}, v^{-1} \rangle = \langle v^{0}, v^{0} \rangle$, which is obviously false.

Hence:

$$V = V^{+1} \oplus V^{-1} \oplus V^0$$

Note: As for last question, I assume author implies Sylvester's theorem (that index of positivity is same for all bases in the vector space), which is already proven in the textbook and can be generalized for index of negativity.

Is my "proof" correct? If not, what's the mistake (is it rigorous enough)?

Thank you!

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    $\begingroup$ It might be more complete it you first show your sets are indeed subspaces of $V$. Also, I'm not sure I follow your last bit at the end. You want to show if you pick 1 vector in the intersection of all 3 subspaces, then it must be the zero vector. You should show that any combination of subspaces being intersected is the zero vector. $\endgroup$ – user23793 Aug 2 '19 at 19:01
  • $\begingroup$ @user23793 I thought about proving that sets were indeed subspaces as well, but they were already mentioned in the whole exercise (I only took some part), thus I thought it would be obsolete. As for the end, I apologize that it is little implicit: I just showed that if there is any element between the subspaces (besides zero) that they have in common, then quadratic scalar products of these elements must be common as well (which is not the case). Thank you for the response! $\endgroup$ – ShellRox Aug 2 '19 at 19:09
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    $\begingroup$ Ah, I see what you're trying to say. I think it's obvious that they are even pair-wise disjoint. The strict inequality directly gives this result. $\endgroup$ – user23793 Aug 2 '19 at 19:12
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You need to be clear what you want to prove. For such a decomposition to be unique, you need to require the bilinear form to be positive definite on $V^+$, negative definite on $V^-$ and $V_0$ to be the kernel of the form (i.e. the set of $x$ such that $\langle x,y \rangle=0$ for all $y\in V$).

It is not enough to consider any basis, but it works to consider an orthogonal basis $(v_1\ldots v_n)$ for your form (if you know that they exist). Then: $$\{\langle v_i, v_i \rangle > 0 \mid v_i \in \{v_1, ... , v_n\} \}$$ is not a subspace, it is just a set of elements: so you want to consider the subspace spanned by these elements: $$V^+=\text{Span}(\{\langle v_i, v_i \rangle > 0 \mid v_i \in \{v_1, ... , v_n\} \})$$ and similarly for $V^-$ and $V_0$.

Then using the orthogonality of $(v_1\ldots v_n)$, you can prove that the bilinear form is positive definite on $V^+$: for all $x=\sum \alpha_i v_i \in V^+\setminus\{0\}$, $$\langle x,x \rangle = \sum \alpha_i^2 \langle v_i, v_i \rangle > 0$$ and similarly for $V^-$.

For $V_0$, you need to show that it is actually equal to the kernel of the form, so you have to proceed by double inclusion:

  • if $x$ is in the kernel of the form, then for all $v_i$ $\langle x,v_i\rangle = 0$ so its coordinates in the elements of $V^+$ and $V^-$ must be zero, which proves that $x\in V_0$.

  • if $x$ is in $V_0$, use the orthogonality of your basis to show that $x$ is in the kernel of the form.

The fact that the sum is direct is absolutely not equivalent to $V^+\cap V^- \cap V_0=\{0\}$. What you need to show is that the decomposition $v=v^++v^-+v_0$ exists and is unique for all elements $v\in V$. This is pretty obvious given that these sets are generated by a partition of a basis (orthogonality is not relevant here).

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  • $\begingroup$ Hello, thank you for the response. Few remarks: 1. I completely agree that existence of orthogonal basis must be specified to show the existence of such subspaces (I was just confused about author's assertion that indexes of positivity/negativity/nullity do not depend on choice of orthogonal basis). 2. In my initial definition, I've mentioned that sets of elements are proper subsets of subspaces, not the subspaces themselves, I also initially assumed that $V^0, V^{+}, V^{-1}$ are indeed subspaces since they are trivial to prove. $\endgroup$ – ShellRox Aug 3 '19 at 15:53
  • $\begingroup$ 3: As for showing uniqueness of elements in sum of subspaces, I think intersection is a fair measure (at least that's what my textbook introduced me to long time ago). Hence in order to show that direct sum exists, two axioms must be satisfied: 1. $V = V^0 + V^{+} + V^{-}$ and 2. $V^{+} \cap V^{-} \cap V^{0} = \{0\}$. Thank you again! $\endgroup$ – ShellRox Aug 3 '19 at 16:00
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    $\begingroup$ @ShellRox This characterization of a direct sum is only valid for two elements. In the plane $\mathbb R^2$, the three lines $\mathbb R(1,0)$, $\mathbb R(0,1)$ and $\mathbb R(1,1)$ have an empty intersection but are not in direct sum! $\endgroup$ – FXV Aug 3 '19 at 16:21
  • $\begingroup$ I apologize, I didn't really think about that before making such generalization, and I didn't consider the fact that these subspaces are spanned from partitions of same base either! That seems like much better proof, thanks! $\endgroup$ – ShellRox Aug 3 '19 at 18:45

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