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Let $q_1,q_2:R^2 \rightarrow R $ be the following quadratic forms $$ \forall x = (x_1,x_2) \in \mathbb{R}^2 \quad q_1(x)=x_1^2+x_1x_2-x_2^2 \qquad \text{and} \qquad q_2(x)=x_1^2-2x_1x_2 \, .$$

Determine whether or not there exists a basis $B$ so that in respect to that basis, both $q_1$ and $q_2$ are diagonal.

My attempt: obviously if one of the quadratic forms was either definitely positive or definitely negative, then we would have the desired simultaneous diagonalization, but they both aren't, so I'm thinking there probably does not exist such a basis. Is there an iff condition for simlutaneous diagonalization of two quadratic forms? If not, what are necessary conditions for such diagonalization, that would help disprove the claim? Any help is appreciated.

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Suppose they are simultaneously diagonalisable. Then $q_2(x)=x_1^2-2x_1x_2$ and $(2q_1+q_2)(x)=3x_1^2-2x_2^2$ are simultaneously diagonalisable too. That is, there exists $P=\pmatrix{a&b\\ c&d}\in GL_2(\mathbb R)$ such that \begin{cases} \pmatrix{a&c\\ b&d}\pmatrix{1&-1\\ -1&0}\pmatrix{a&b\\ c&d}=\pmatrix{\ast&0\\ 0&\ast},\\ \pmatrix{a&c\\ b&d}\pmatrix{3&0\\ 0&-2}\pmatrix{a&b\\ c&d}=\pmatrix{\ast&0\\ 0&\ast}. \end{cases} Since all anti-diagonal entries on the RHS are zero, the above two equations and our assumption on $P$ give \begin{align} &ab-ad-bc=0,\tag{1}\\ &3ab=2cd,\tag{2}\\ &ad-bc\ne0.\tag{3} \end{align} Now the problem boils down to inspecting whether $(1),(2),(3)$ are simultaneously solvable over $\mathbb R$. This shouldn't be difficult to answer.

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