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- #1

- Feb 5, 2012

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Take a look at the following question.

**Problem:**

Determine which of the following quadratic functions \(q_1,\,q_2:\,V\rightarrow\Re\) is positive definite and find a basis of \(V\) where one of \(q_1,\,q_2\) has normal form and the other canonical: \(q_1(x_1,\,x_2)=x_{1}^{2}+2x_1 x_2+2x_{2}^{2}\) and \(q_{2}(x_1,\,x_2)=x_{1}^{2}-2x_1 x_2-4x_{2}^{2}\).

**My Ideas:**

The first part of finding the positive definiteness is easy and by writing these quadratic functions as \(x^{t}Ax\) and calculating the eigenvalues of \(A\) we can find out that \(q_{1}\) is positive definite and \(q_{2}\) is not.

However then to find a basis where one of these has normal form (having only squared terms with coefficients \(\pm 1\) or \(0\)) and the other having the canonical form (having only squared terms), in my opinion is impossible.

\(q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2\)

\[q_2(x_1,\,x_2)=(x_1-x_2)^2-5x_{2}^2\]

Now if we choose \(y=x_1+x_2\) and \(z=x_1-x_2\) then \(q_1\) transforms to it's normal form but \(q_{2}\) does not transform to it's canonical form. If \(q_{2}\) was instead given as \(q_{2}(x_1,\,x_2)=x_{1}^{2}\color{red}{+}2x_1 x_2-4x_{2}^{2}\) then the above transform works since substituting \(y=x_1+x_2\) and \(z=x_1-x_2\) we get,

\[q_1(x_1,\,x_2)=(x_1+x_2)^2+x_{2}^2=y^2+z^2\]

and

\[q_2(x_1,\,x_2)=(x_1+x_2)^2-5x_{2}^2=y^2-5z^2\]

So I am assuming that this problem has a typo. Can anybody confirm my assumption or tell me where I went wrong.