Question: Suppose $\alpha,\beta \in V^*$ are linearly independent. Compute the rank and signature of the quadratic form $v \mapsto \alpha(v)\beta(v).$

Some information about $\alpha$ and $\beta$:

$\alpha\cdot\beta:V\times V \rightarrow \mathbb{R}$ is defined by $(\alpha \cdot \beta)(v,w)= \alpha(v) \beta(w)$.

There is a linear map $\Phi:V^*\otimes V^* \rightarrow (V \otimes V)^*$ with $\Phi(\alpha \otimes\beta)(v \otimes w) = \alpha(v) \beta(w) $.

I understand the signature is defined as a pair $(p,q)$ where $p$ is the maximum dimension of a subspace on which $Q$ (the quadratic form) is positive definite and $q$ is the maximum dimension of a subspace on which Q is negative definite. But I am unsure on how to compute these values.


Set $\varphi^1 = \alpha$ and $\varphi^2 = \beta$ and $\dim V = n \geq 2$. Since $(\varphi^1, \varphi^2)$ are linearly independent, we can complete them to a basis $(\varphi^1, \dots, \varphi^n)$ of $V^{*}$. Let $v_1, \dots, v_n$ be the associated dual basis of $V$ (so that $\varphi^i(v_j) = \delta^i_j$). Finally, let $q(v) = \alpha(v)\beta(v)$ be the quadratic form and let

$$\begin{aligned} g(u,v) &= \frac{q(u + v) - q(u) - q(v)}{2} = \frac{\alpha(u + v)\beta(u + v) - \alpha(u)\beta(u) - \alpha(v)\beta(v)}{2} \\ &= \frac{\alpha(u)\beta(v) + \alpha(v)\beta(u)}{2} \end{aligned}$$

be the associated symmetric bilinear form. We need to find the rank and signature of $g$. Note that

$$ g(v_i, v_j) = \frac{\varphi^1(v_i) \varphi^2(v_j) + \varphi^1(v_j) \varphi^2(v_i)}{2} = \frac{\delta^1_i \delta^2_j + \delta^1_j \delta^2_i}{2} = \begin{cases} \frac{1}{2} & i = 1, j = 2 \text{ or } i = 2, j = 1, \\ 0 & \text{otherwise}. \end{cases} $$

which means that $g$ is represented with respect to the basis $(v_1, \dots, v_n)$ by the block matrix

$$ A = \begin{pmatrix} \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix} & 0_{2 \times (n - 2)} \\ 0_{(n - 2) \times 2} & 0_{(n - 2) \times (n - 2)} \end{pmatrix}. $$

The rank of $A$ (and thus, of $g$) is $n - 2$ and the signature of $A$ (and thus, of $g$) is $(1,1)$.


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