Diagonalize the quadratic form by finding an orthogonal matrix $Q $ such that the change of variable $x = Qy $ transforms the given form into one with no cross-product terms. Give $Q $ and the new quadratic form, $f(y)$.

Assume $y = \begin{bmatrix} y_1\\y_2\end{bmatrix}$

$$x^2 + 14xy + y^2$$

Current (wrong) answer: Matrix of the quadratic equation is $\begin{bmatrix} 1 & 7\\ 7 & 1\end{bmatrix}$, eigenvalues of matrix are $8$ and $-6$. Eigenvectors of matrix are $\begin{bmatrix} 1 \\1 \end{bmatrix}$ and $\begin{bmatrix} -1\\1 \end{bmatrix}.$

Thus the orthogonal matrix $Q$ is: $\begin{bmatrix} 8 & 8\\ 6 & -6\end{bmatrix}. $

And the new quadratic form is: $$8y_1^2 - 6y_2^2 + 2y_1y_2$$

Not too sure where I went wrong, help would be much appreciated, thanks.


You seem to have some of the right ideas organized incorrectly.

Here's a hint:

Let $$ A = \begin{bmatrix}1&7\\7&1\end{bmatrix} $$ and note that $$ A = QJQ^\top $$ where \begin{align*} Q &= \frac{1}{\sqrt2}\begin{bmatrix}-1& 1\\1&1\end{bmatrix} & J &= \begin{bmatrix}-6&0\\0&8\end{bmatrix} \end{align*} Also note that $Q^\top=Q^{-1}$ so that $Q$ is orthogonal.

This tells you that your "new" quadratic form is $-6\,x^2+8\,y^2$. Can you fill in the details?


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