Question

consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cross product in q(x,y,z) and express q in the new variables.

Answer 1

`q(x,y,z)=16x^(2)-5y^(2)-5z^(2)`

Explanation:

The given quadratic form is of the form

`q(x,y,z)=ax^2+by^2+dxy+exz+fyz`.

Where
`a=11,b=-1,c=-4,d=-16,e=8,f=-4`.Every quadratic form of this kind can be written as

Observe that
`A` is a symmetric matrix. So
`A` is orthogonally diagonalizable, that is to say,
`D=Q^(T)AQ` where
`Q` is an orthogonal matrix and
`D` is a diagonal matrix.

In our case we have:

`A=\left(\begin{array}{ccc}11&((1)/(2))(-16) &((1)/(2)) (8)\\((1)/(2)) (-16)&(-1)&((1)/(2)) (-4)\\((1)/(2)) (8)&((1)/(2)) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)`

The eigenvalues of
`A` are
`\lambda_(1)=16,\lambda_(2)=-5,\lambda_(3)=-5`.

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

`Q=\left(\begin{array}{ccc}(4)/(√(21))&-(1)/(√(17))&(8)/(√(357))\\(-2)/(√(21))&0&\sqrt{(17)/(21)}\\(1)/(√(21))&(4)/(√(17))&(2)/(√(357))\end{array}\right)`

`D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)`

Now, we have to do the change of variables
`{\bf x}=Q{\bf y}` to obtain

`q({\bf x})={\bf x}^(T)A{\bf x}=(Q{\bf y})^(T)AQ{\bf y}={\bf y}^(T)Q^(T)AQ{\bf y}={\bf y}^(T)D{\bf y}=\lambda_(1)y_(1)^(2)+\lambda_(2)y_(2)^(2)+\lambda_(3)y_(3)^(2)=16y_(1)^(2)-5y_(2)^(2)-5y_(3)^2`

Which can be written as:

`q(x,y,z)=16x^(2)-5y^(2)-5z^(2)`