We have quadratic forms $Q = \lambda x^2 +4y^2+16z^2$ and $R=2xy+2yz$. For what real values of $\lambda$ does there exist a real linear operator $T$ on $\mathbb{R}^3$ such that $R=Q \circ T$?
From what I know about quadratic forms, this problem simplifies to finding $\lambda$ such that there exists $T$ with $$T^t \begin{pmatrix} \lambda & 0 & 0 \\\ 0 & 4 & 0 \\\ 0 & 0 & 16 \end{pmatrix} T = \begin{pmatrix} 0 & 1 & 0 \\\ 1 & 0 & 1 \\\ 0 & 1 & 0 \end{pmatrix}$$.
However, I don't know how to proceed from here. Just by writing out the elements of T and doing the multiplication longhand I've determined that $\lambda < 0$, but I don't think that approach will yield anything more helpful.
I also saw this: How to solve quadratic matrix equations of the form $A^T B A=C$?
Unfortunately, the solution to that question seemed somewhat specific to that exact problem, and that approach won't exactly work here because his answer for $A$ was the inverse of some matrix, and my $T$ must not be invertible.
Finally, I've considered diagonalizing the matrix for $R$ or writing it alternatively as $\begin{pmatrix} 0 & 2 & 0 \\\ 0 & 0 & 2 \\\ 0 & 0 & 0 \end{pmatrix}$, but I don't see how those will help. Thanks!
You are not asked to construct $T$. You are asked to determine the values of $\lambda$ such that $T$ exists. It suffices to consider the inertia of $Q$ and $R$.
The matrix representation of $R$ has rank two and trace zero. Therefore it has exactly one positive eigenvalue, one negative eigenvalue and one zero eigenvalue. Since the matrix of $Q$ has at least two positive eigenvalues ($4$ and $16$) and the matrix of $Q\circ T$ cannot possess more negative eigenvalues than the matrix of $Q$, in order that $R=Q\circ T$, the matrix of $Q$ must have at least one negative eigenvalue, i.e. $\lambda$ must be negative.
Conversely, if $\lambda$ is negative, define a linear operator $L$ by $Le_1=e_1,\,Le_2=e_2$ and $Le_3=0$. Then $Q\circ L$ has the same inertia as $R$. Therefore, by Sylvester's law of inertia, there exists a linear operator $P$ such that $(Q\circ L)\circ P=R$. Now take $T=L\circ R$ and we are done.