Express a quadratic form in three variable in the format $x^tAx$ using a substitution $x=Py$ I was asked to determine if a quadratic form is positive definite. To do so I must convert in the format $x^tAx$ using a substitution x=Py. So that "it can be written in diagonal form".
$$Q(x,y,z) = 3x^2 + 8xz+2y^2+z^2$$
My idea is:
$$(x,y,z) \begin{pmatrix} 3x+4z \\ 2y \\ 4x+z^2 \end{pmatrix} $$ 
$$\begin{pmatrix} 3&0&4 \\0&2&0 \\ 4&0&1 \end{pmatrix} $$ 
Is this correct? And then it would be positive definite if all eigen values of this are equal to or larger than 0?
 A: The matrix should be:
$$\mathbf{A}=\left(\begin{matrix}3&0&4\\0&2&0\\4&0&1\end{matrix}\right)$$
Multiplying $\mathbf{x^TAx}$ out where $\mathbf{x}=\left(\begin{matrix}x&y&z\end{matrix}\right)^T$ gives:
$$\left(\begin{matrix}x&y&z\end{matrix}\right)\left(\begin{matrix}3&0&4\\0&2&0\\4&0&1\end{matrix}\right)\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}x&y&z\end{matrix}\right)\left(\begin{matrix}3x+4z\\2y\\4x+z\end{matrix}\right)\\=3x^2+4xz+2y^2+4xz+z^2=3x^2+8xz+2y^2+z^2$$
We can orthogonally diagonalise $\mathbf{A}=\mathbf{PDP^T}$ as $\mathbf{A}$ is symmetric. If we let $\mathbf{y}=\mathbf{P^Tx}$, then $$\mathbf{x^TPDP^Tx}=\mathbf{y^TDy}$$
which is in the form $\lambda_1y_1^2+\lambda_2y_2^2+\lambda_3y_3^2$, where $\lambda_i$ are the eigenvalues and $\mathbf{y}=\left(\begin{matrix}y_1&y_2&y_3\end{matrix}\right)^T$.
Hence, a quadratic form being positive definite is equivalent to nonnegative eigenvalues of $\mathbf{A}$.
A: Ummm, $$ (4x+z)^2 = 16 x^2 + 8 x z + z^2, $$ so
$$ -13 x^2 +  (4x+z)^2 = 3 x^2 + 8 x z + z^2,  $$
$$ -13 x^2 +  (4x+z)^2 + 2 y^2 = 3 x^2 + 8 x z + z^2 + 2 y^2 = Q(x,y,z).  $$
Note $Q(1,0,-1) = -4$
If
$$
D = 
\left(
\begin{array}{ccc}
-13 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{array}
\right)
$$
and
$$
R = 
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
4 & 0 & 1 \\
0 & 1 & 0
\end{array}
\right),
$$
what is
$$ R^T D R? $$
A: How's this?
$$\begin{pmatrix} 3&0&4 \\0&2&0 \\ 4&0&1 \end{pmatrix} $$ 
i found the determinant which had the equation....
$$ det(A-\lambda I)= -\lambda^3 +6\lambda ^2 +5\lambda -26$$
$\color{green}{ \text{As f(x) = 0 at x=2 (x-2) is a factor and by long divisiong}}$
$$ det(A-\lambda I)=  -(x-2)(x^2-4x-13)$$
$$ \lambda_1 = 2+\sqrt{17}=6.123, \quad  \lambda_2 = 2-\sqrt{17}=-2.1231...,  \quad  \lambda_3 = 2  $$
From which we can tell it's positive indefinite
Now i have to find the eigenvectors? Which is the the "P" in x=Py.
$$\begin{pmatrix}-1+\sqrt{17}\:\:\:\: & \:\:\:\:0\:\:\:\: & \:\:\:\:4\:\:\:\: \\0\:\:\:\: & \:\:\:\:-2+\sqrt{17}\:\:\:\: & \:\:\:\:0\:\:\:\: \\4\:\:\:\: & \:\:\:\:0\:\:\:\: & \:\:\:\:-3+\sqrt{17}\end{pmatrix}\: $$
Just wanted to check because this matrix is hard 
