The isotropic cone of a positive semi-definite bilinear is a subspace of $\Bbb R^n$.

Question

Let $A \in \Bbb R^n \times \Bbb R^n$ be a symmetric and positive semidefinite matrix. Let $q(x) =x^\top Ax$ and let the isotropic cone be defined as

$$W= \{x ; q(x) = 0 \}$$

I want to show that $W$ is a subspace of $\Bbb R^n$.

My closest attempt:

Take $x$ and $y$ in $W$ (the zero vector always belongs to $W$ and is a subspace of $\Bbb R^n$). let $\alpha, \beta \in\Bbb R $ It is known that $$q(\alpha x- \beta y) = -2 \alpha \beta x^\top Ay + \alpha^2 q(x) + \beta^2 q(y) = -2 \alpha \beta x^\top Ay$$

So ignoring constants we are done if $x^\top Ay = 0$ for $x,y \in W$. By the spectral theorem we have that $A = UDU^\top$ where $D$ is diagonal and $U$ is unitary. This means that $$ x^\top Ay = (U^\top x)D (U^\top y) $$ but I don't see where I can go from here.

Answer 1

If $A$ is positive semi-definite, then the Cauchy-Schwarz inequality holds for the (possibly degenerate) scalar product $\langle x,y\rangle_A\doteq x^\top Ay$, but this time you can have equality without $x$ and $y$ being proportional. And $\|x\|_A\doteq \sqrt{\langle x,x\rangle_A}$ is a semi-norm (with a similar remark for the triangle inequality).

Clearly $0\in W$ and $x\in W$, $\lambda \in \Bbb R$ implies $\lambda x\in W$. Finally, if $x,y\in W$ we have $$\begin{align} 0\leq \langle x+y,x+y\rangle_A &=\langle x,x\rangle_A+2\langle x,y\rangle_A+\langle y,y\rangle_A \\ &= 2\langle x,y\rangle_A \\ &\leq 2\|x\|_A\|y\|_A = 0,\end{align} $$as wanted.

Note that semi-definiteness is essential. In $\Bbb R^2$, the matrix $A={\rm diag}(1,-1)$ gives a counter-example ($W$ is the union of the lines $y=x$ and $y=-x$).

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Note that this argument holds for positive semi-definite bilinear forms in any vector space (including the infinite dimensional case).

Answer 2

Sylvester's Law of Inertia says that after a change of basis, we may assume that $A$ is a diagonal matrix with only $1$'s, $-1$'s, and $0$'s on the diagonal. Since the form is positive semidefinite, this means there are no $-1$'s. So we may assume that

$$ A = \left( \begin{array}{cccccc} 1 & & & & & \\ & \ddots & & & & \\ & & 1 & & & \\ & & & 0 & & \\ & & & & \ddots & \\ & & & & & 0 \end{array} \right) $$

Looking at it this way, $W$ is the span of the basis vectors representing the right-most columns, and the form $q$ in this basis looks like

$$ q\left( \sum_{i=1}^n \alpha_i e_i \right) = \sum_{i=1}^k \alpha_i^2, $$ where $k$ is the number of $1$'s on the diagonal.

Answer 3

Let $\beta(x,y) := x^T A y$ be the bilinear form associated to $A$. By the spectral theorem there exists a orthogonal matrix $U \in \operatorname{O}(n)$ such that $A = U D U^T$ for a diagonal matrix $D$ with non-negative entries, say $$ D = \operatorname{diag}(c_1, \dotsc, c_r, 0, \dotsc, 0) $$ with $c_1, \dotsc, c_r > 0$. Then $\mathcal{B} := (v_1, \dotsc, v_n)$ given by $v_i := U e_i$ is a basis of $\mathbb{R}^n$ which is orthogonal with respect to $\beta$; more precisely, we have that $$ \beta(v_i, v_j) = \begin{cases} c_i \delta_{ij} & \text{if $1 \leq i,j \leq n$}, \\ 0 & \text{otherwise}. \end{cases} $$ (In other words: With respect to the basis $\mathcal{B}$ the bilinear form $\beta$ is represented by the diagonal matrix $D$.) It follows that for every vector $x \in \mathbb{R}^n$, $x = \lambda_1 v_1 + \dotsb + \lambda_n v_n$ we have that $$ q(x) = \beta(x,x) = c_1 \lambda_1^2 + \dotsb + c_r \lambda_r^2. $$ It follows from $c_1, \dotsc, c_r > 0$ that $q(x) = 0$ if and only if $\lambda_1 = \dotsb = \lambda_r = 0$. Thus we have that $W$ is precisely the span of the basis vectors $v_{r+1}, \dotsc, v_n$. It is in particular a linear subspace of $\mathbb{R}^n$.