Surface described by ${\bf r^\top A r + b^\top r}=1$?

Question

I was asked to describe the surface described by

$${\bf r}^\top {\bf A} {\bf r} + {\bf b}^\top {\bf r} = 1,$$

where $3 \times 3$ positive definite matrix ${\bf A}$ and vector $\bf b$ are given.

My intuition tells me that it is a rotated ellipsoid with a centre that is off the origin. However, I am told to show this via the substitution ${\bf r} = {\bf x} + {\bf a}$, with $\bf a$ being a constant vector, and dictate the conditions on this vector to obtain a new quadric surface ${\bf x}^\top{\bf A}{\bf x} = C$. However, upon substitution, I get a ridiculously messy answer involving combinations of position and constant vectors. Is there a trick I am missing out on? Thank you!

Answer 1

Note: for convenience we use $2b$ instead of $b$.

$$(x+a)^TA(x+a)+2b^T(x+a)=x^TAx+a^TAx+x^TAa+a^TAa+b^Tx+2b^Ta.$$

Notice that by symmetry of $A$, $x^TAa=a^TAx$. Collecting all the $x$ terms,

$$(2a^TA+2b^T)x$$ can be cancelled with the choice

$$a=-A^{-1}b.$$

Then

$$C=1-a^TAa-2b^Ta=1+b^TA^{-1}b>1.$$

(As $A$ is positive definite, it is invertible. Note that this is a matrix version of the "complete the square" paradigm.)

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Going further, you can diagonalize the matrix and write, in the basis defined by the Eigenvectors

$$y^T\Lambda y=C$$

or

$$\lambda_0u^2+\lambda_1v^2+\lambda_2w_2=C.$$

As the three Eigenvalues are positive,

$$(\sqrt\lambda_0u)^2+(\sqrt\lambda_1v)^2+(\sqrt\lambda_2w)^2=C$$ describes a stretched sphere.

Answer 2

However, I am told to show this via the substitution ${\bf r} = {\bf x} + {\bf a}$, with $\bf a$ being a constant vector, and dictate the conditions on this vector to obtain a new quadric surface ${\bf x}^\top{\bf A}{\bf x} = C$.

Following your advice: \begin{align} 1 &= r^\top A r + b^\top r \\ &= (x + a)^\top A(x+a) + b^\top(x+a) \\ &= x^\top A x + \underbrace{x^\top A a + a^\top A x + b^\top x}_{f(x)} + a^\top A a + b^\top a \end{align} We need to choose $a$ such that the terms depending on $x$ vanish: $$ 0 = x^\top A a + a^\top A x + b^\top x $$ The "positive definite" property holds for symmetric matrices, which fulfill $$ A^\top = A $$ so we have $$ x^\top A a = x^\top A^\top a = (A x)^\top a = a^\top A x $$ using symmetry and the rules for transposition, which gives $$ 0 = a^\top A x + a^\top A x + b^\top x = (2 a^\top A + b^\top) x $$ so for all vectors $x$ to hold, we need $$ 0 = 2 a^\top A + b^\top \iff \\ 0 = 0^\top = 2 A^\top a + b = 2 A a + b \iff \\ a = -(A^{-1}b)/2 $$ Justification of the last step:

For a positive definite matrix $A$ all eigenvalues $\lambda_i$ are positive, so its determinant (the product of its eigenvalues) is positive as well, and thus non-vanishing. Therefore it has an inverse $A^{-1}$.

So we end up with $$ x^\top A x = C $$ where \begin{align} C &= 1 - a^\top A a - b^\top a \\ &= 1 - (A^{-1} b)^\top A (A^{-1} b) / 4 + b^\top (A^{-1} b)/2 \\ &= 1 - b^\top (A^{-1})^\top b / 4 + b^\top A^{-1} b /2 \\ &= 1 - b^\top (A^\top)^{-1} b / 4 + b^\top A^{-1} b /2 \\ &= 1 - b^\top A^{-1} b / 4 + b^\top A^{-1} b /2 \\ &= 1 + \underbrace{b^\top A^{-1} b}_{> 0} / 4 \end{align} because matrix transposition and inverting operations commute and using symmetry again.

$A^{-1}$ has eigenvalues $1/\lambda_i$ which are all positive as well, so $A^{-1}$ is positive definite too. This assures that the constant number $C$ is positive and we deal with an ellipsoid surface.

Here is an example for $$ A = \begin{pmatrix} -2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{pmatrix} \quad\quad b = \begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix} $$

Update:

Here is a link to an interactive version of the above scene.