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Lay P402 : A change of variable is an equation of the form $x=Py$, where $P$ is an invertible matrix and $y$ is the (neW) coordinate vector of $x$ relative to the basis of $\mathbb{R}^{n}$ determined by the columns of $P$. (See Section 4.4.)

If this change of variable is made in a quadratic form $x^{T}Ax$, then $ x^{T}Ax=(Py)^{T}A(Py)=y^{T}(P^{T}AP)y\ $ and the new matrix of the quadratic form is $P^{T}AP$. Since $A$ is symmetric, Theorem 2 guarantees there's an orthogonal matrix $P$ such that $P^{T}AP =$ a diagonal matrix $D$. This is the strategy of the next example.

My Question $1.$ By virtue of Theorem 2, $P$ is orthogonal, but why must $P$ in fact be orthonormal? Lay doesn't reveal this, but the solution to the question below veraciously normalises the columns of P. If I kept $P$ as is (without normalising its columns), I wouldn't obtain the right answer?

Given Question : Determine the standard form of the quadric surface $x^2 + y^2 -2z^2 + 2xy + 8xz + 8yz+ 3x +z = 0 \; (*)$.

Rewrite this as $x^T \begin{bmatrix} 1 & 1 & 4 \\ 1 & 1 & 4 \\ 4 & 4 & -2 \\ \end{bmatrix} x + \begin{bmatrix} 3 & 0 & 1 \\ \end{bmatrix} x = 0.$

Regarding the 3 by 3 symmetric A: $\lambda_1 = -6$ associates with $v_1 = (-1, 1, 2)^T$. $\lambda_2 = 0$ associates with $v_2 = (-1, 1, 0)^T$. $\lambda_3 = 6$ associates with $v_1 = (1, 1, 1)^T$.

P with normalised eigenvectors $ = \begin{bmatrix} \dfrac{1}{ \sqrt{6} } \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} & \dfrac{1}{ \sqrt{2} } \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} & \dfrac{1}{ \sqrt{3} } \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \\ \end{bmatrix} $.

Define $y = (x_2, y_2, z_2)$. Apply the change of variable $x = Py$ to $(*)$:

$ y^T \begin{bmatrix} -6 & & ~ \\ ~ & 0 & ~ \\ ~ & ~ & 6 \\ \end{bmatrix} y + \begin{bmatrix} 3 & 0 & 1 \\ \end{bmatrix} Py = 0 $. Now I complete the square.

$\implies -\color{orangered}{6}(x_2^2 + \dfrac{ -3 + \sqrt{2} }{\color{orangered}{6} \sqrt{6} }x_2 + \color{green}{\dfrac{ (2 - \sqrt{3} )^2 }{ 864 }}) + \color{orangered}{6}(z_2^2 + \dfrac{ 1 + \sqrt{18} }{\color{orangered}{6} \sqrt{3} }z_2 + \color{green}{\dfrac{ (1 + 3\sqrt{2} ) }{ 432 }} ) + \dfrac{ -3 + \sqrt{2} }{ 6 } y_2 = -6 \color{green}{\dfrac{ (2 - \sqrt{3} )^2 }{ 864 }} + 6\color{green}{\dfrac{ (1 + 3\sqrt{2} ) }{ 432 }} $.

This doesn't look right. The answer's $\dfrac{ x_3^2 }{ 1/2} - \dfrac{ y_3^2 }{ 1/2} = z_3$, a hyperboloic paraboloid.

Supplementary: I only lighted upon P11 of this afterwards; did I misconstrue 'orthogonal'?

The terminology is very confusing. The deØnition of an orthogonal matrix requires that the columns be mutually perpendicular and also that they be unit vectors. Unfortunately, the terminology reminds us of the former condition but not of the latter condition. It would have been better if such matrices had been named orthonormal' matrices rather thanorthogonal' matrices, but that is not how it happened, and we don't have the option of changing the terminology at this late date.

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I didn't study all details, but whether you use orthogonal or orthonormal coordinate changes, it doesn't change the fact that $P^T A P$ is diagonal. (Note that an orthogonal matrix is usually defined as a matrix whose columns are orthonormal). Orthonormal coordinates are more intuitive; in an orthonormal basis, the equation of a circle still describes a circle (and not an ellipse), an ellipsoid has uniquely determined semi-principal axis by its equation, etc. If you just use orthogonal coordinates (i.e. transition matrices $P$ such that $P^T P$ is only regular and diagonal), you can no more distinguish a sphere and an ellipsoid (but you can still see the difference between a sphere and a hyperboloid, or a hyperboloid and a paraboloid).

The more general coordinate transformations you allow, the less details you see from the equation: if you consider the quadrics to live in a projective space and allow projective transformations, you can't even see the difference between the ellipse and hyperbole (but still can see the difference between that and a line) etc..

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