Prove that if positivity of one bilinear form implies positivity of second bilinear form then they are scalar multiples.

Question

Let $M_2, M_2 \in \mathbb{R}^{d\times d} \setminus \{ 0 \}$. Prove that if for all $x,y \in \mathbb{R}^d$ $$x^T M_1 y > 0 \implies x^T M_2 y > 0$$ then $M_2 = \lambda M_1$.

Answer 1

Step 1: For all $z\in \mathbb{R}^d$ there exists $\lambda_z \in \mathbb{R}$ such that $M_1 z=\lambda_z M_2 z$.

Assume not. Then there exists $w\in\mathbb{R}^d$ such that $M_1w \neq \lambda M_2w$ for all $\lambda\in \mathbb{R}$. Then $M_1w\neq 0$. If $M_2w\neq 0$ we can extend $\{M_1w,M_2w\}$ to a basis $B$ of $\mathbb{R}^d$, else we extend $\{M_1w\}$ to a basis $B$ of $\mathbb{R}^d$. Now we define a linear map $X:\mathbb{R}^d \rightarrow \mathbb{R}$ by letting $X(M_1w)=1$ and $X(b)=0$ for all $b\in B\setminus \{M_1w\}$ and then extend linearly. Since every such linear map is given by a matrix, we can find $x\in \mathbb{R}^d$ such that $X(y)=x^\top y$ for all $y \in \mathbb{R}^d$.
Then we have $x^\top M_1w=X(M_1w)=1$ and $x^\top M_2 w=X(M_2w)=0$, which is a contradiction.

Step 2: Since $M_1\neq0$ we can find $q\in \mathbb{R}^d$ such that $M_1q\neq0$. Now let $\lambda=\lambda_q$ from Step 1. Our goal is to show $M_1v=\lambda M_2v$ for all $v\in \mathbb{R}^d$, which will imply $M_1=\lambda M_2$.

Let $v\in \mathbb{R}^d$. We have \begin{gather} (\lambda_q-\lambda_{q+v})M_2q+(\lambda_v-\lambda_{q+v})M_2v=\lambda_qM_2q+\lambda_vM_2v-\lambda_{q+v}M_2(q+v)\\=M_1q+M_1v-M_1(q+v)=0 \end{gather} So if $M_2q$ and $M_2v$ are idependent $\lambda_v=\lambda_{q+v}=\lambda_q=\lambda$ and hence $M_1v=\lambda_vM_2v=\lambda_qM_2v=\lambda M_2v$ and we are done.

If $M_2q$ and $M_2v$ are not idependent we can write $M_2v=rM_2q$ with $r\in\mathbb{R}$ (since by Step 1 and $M_1q\neq0$ we also have $M_2q\neq0$). Then $M_1v=\lambda_vM_2v=\lambda_vrM_2q$, so if $r=0$ we have $ M_1v=0=\lambda M_2v$ and again we are done.

Finally if $r\neq0$ we let $y=q-\frac1rv$ and $x=M_1y$. Then $x^\top M_2y=x^\top M_2(q-\frac1rv)=x^\top0=0$, so by the contraposition of the assumption we must have $x^\top M_1y\le0$.
Since $x^\top M_1y=(M_1y)^\top (M_1y)=\|M_1y\|^2$ this means $\|M_1y\|=0$, so $M_1y=M_1(q-\frac1rv)=0$ and we see $M_1v=rM_1q=r\lambda_qM_2q=\lambda_qrM_2q=\lambda_qM_2v=\lambda M_2v$, which completes Step 2.

Step 3: Since $M_1=\lambda M_2$ and $M_1\neq0$, also $\lambda\neq0$, so by dividing by $\lambda$ we are done.