Question

Consider the real symmetric matrix \(A=(a_{ij})\) given by \[ A=\begin{pmatrix}3 & 1 & 1\\ 1 & 0 & 2\\ 1 & 2 & 0\end{pmatrix}. \] Consider the set \[ S=\left\{x=(x_1,x_2,x_3)\in\mathbb{R}^3 : \sum_{j=1}^{3}\sum_{i=1}^{3}a_{ij}x_ix_j=1\right\}. \] Which of the following statements is/are correct?

• A. \(S\) is empty set
• B. Any line \(L\) in \(\mathbb{R}^3\) has at most two points from \(S\)
• C. \(S\) is bounded
• D. \(S\) is unbounded

Hint:

Find the eigenvalues of A to see that S is an indefinite quadric (a hyperboloid), then use the fact that substituting a line into x^TAx=1 gives a quadratic equation in the line parameter.

Answer 1

The Correct Option is B, D

Step 1: Find the eigenvalues of A.
The surface S is a level set of the quadratic form built from A, so its shape depends on the eigenvalues of A. The characteristic equation works out to
\[ \det(A-xI)=0 \implies x^3-3x^2-6x+8=0, \] which factors as \((x-1)(x-4)(x+2)=0\). So the eigenvalues are \(1,4,-2\).

Step 2: Read the type of quadric from the signature.
Two eigenvalues are positive and one is negative, and none is zero (check \(\det A=1\times4\times(-2)=-8\neq0\)). So in suitable coordinates \(y_1,y_2,y_3\), the equation \(x^TAx=1\) becomes
\[ y_1^2+4y_2^2-2y_3^2=1, \] which is a hyperboloid, an unbounded surface.

Step 3: Check (A), S is empty.
Since the eigenvalues have mixed sign, the form takes both positive and negative values, so along a suitable eigenvector direction we can scale a point to make the form equal exactly 1. So S is not empty, and (A) is FALSE.

Step 4: Check (B), a line meets S in at most two points.
Take any line \(x=x_0+tv\). Substituting into \(x^TAx=1\) gives \((v^TAv)t^2+2(x_0^TAv)t+(x_0^TAx_0-1)=0\), a polynomial of degree at most 2 in \(t\). Such an equation has at most 2 roots, so the line meets S in at most two points. So (B) is TRUE.

Step 5: Check (C) and (D), boundedness.
Because the canonical form \(y_1^2+4y_2^2-2y_3^2=1\) is a hyperboloid, we can fix \(y_1,y_2\) at any small values and let \(y_3\) grow without bound while the equation still holds, since the negative term compensates. So S stretches to infinity and is not bounded. So (C) is FALSE and (D) is TRUE.

Final Answer:
S is neither empty nor bounded; it is an unbounded quadric that meets any line in at most two points. \[ \boxed{\text{(B) and (D)}} \]