# Prove that the zero set of a quadratic function has measure zero

Let $f:R^n \rightarrow R$ be a function of the form $f(x) = x'Ax$, where $A$ is a non-zero symmetric matrix from $R^{n \times n}$.

Under which conditions does $Z = \{x \in R^n | f(x) = 0\}$ has Lebesgue measure zero?

Wrong assumptions of original question: Matrix A was not necessarily symmetric. Under this assumption, I assumed that the zero set $Z = \{x \in R^n | f(x) = 0\}$ has Lebesgue measure zero.

• Since $A$ is not symmetric, $f$ is not quadratic. Commented Feb 26, 2018 at 9:58
• Thanks! I modified the question accordingly. Commented Feb 26, 2018 at 10:17

Let $A=\begin{pmatrix}0&1\\ -1&0& \end{pmatrix}$, then $f(x)=0$ for all $x \in \mathbb R^2$ !

• Thank you! I modified the question thanks to your example. Commented Feb 26, 2018 at 9:48

In the case where $A$ is non-zero and symmetric, $Z$ always has zero as Lebesgue measure.

Since $A$ is symmetric, you can diagonalize it into an orthonormal basis, which is equivalent to say that it exists $a_i\in \mathbb{R}$ such that $$f(x)=\sum_{i=1}^n a_i|x_i|^2$$ (with at least one $a_i\neq 0$ since $A$ is non-zero).

Then $$Z= \{ x ; \sum_{i=1}^n a_i|x_i|^2 =0 \}$$ which has a zero as Lebesgue measure.

To see that last point, without loss of generality, let us assume that $a_n=1$, then (with $\lambda_n$ the Lebesgue measure),

\begin{align*} \lambda_n(Z) &= \int_{\mathbb{R}^n}1_{\sum_{i=1}^n a_i|x_i|^2 =0}(x) dx_1\dots d x_n \\ & = \int_{\mathbb{R}^{n-1}}\bigg(\underbrace{\int_{\mathbb{R}} 1_{x_n^2=-\sum_{i=1}^{n-1} a_i|x_i|^2}(x_1,\dots,x_{n-1},x_n) d x_n}_{=0}\bigg)dx_1\dots d x_{n-1} \\ &=0 \end{align*}

• Thanks! Suppose $A = Q^TDQ$. The set $Z$ consist of all zeros $x$ transformed by the orthogonal map $Q$. Why is the set $Q^TZ$ also a set of measure zero? Commented Feb 26, 2018 at 11:57