Express a quadratic form in three variable in the format $x^tAx$ using a substitution $x=Py$

Question

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?

Answer 1

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 $\mathbf{A}$.

Answer 2

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 $$ R^T D R? $$

Answer 3

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