Question

Let \(A\) be a \(3\times 3\) real matrix such that given any column vector \(x\in \mathbb{R}^3\), the column vector \(Ax\) is the reflection of \(x\) about the plane

Hint:

For reflection about a plane through the origin in \(\mathbb{R}^3\), eigenvalues are \(1,1,-1\), so the trace is \(1\).

Answer 1

Correct Answer: 1

Step 1: Identify the given plane.
The plane is given by
\[ (x,y,z)=(a,b,-a-b) \]
So,
\[ x=a,\quad y=b,\quad z=-a-b \]

Step 2: Find the equation of the plane.
\[ x+y+z=a+b-a-b=0 \]
Hence, the plane is
\[ x+y+z=0 \]

Step 3: Find the normal vector of the plane.
For the plane
\[ x+y+z=0 \]
the normal vector is
\[ n=(1,1,1) \]

Step 4: Understand reflection about a plane through origin.
Reflection about a plane keeps every vector lying in the plane unchanged.
Therefore, every vector in the plane has eigenvalue
\[ 1 \]

Step 5: Dimension of the plane.
The plane \(x+y+z=0\) is two-dimensional.
So, eigenvalue \(1\) occurs twice.
\[ \lambda_1=1,\qquad \lambda_2=1 \]

Step 6: Effect on normal direction.
The normal vector is reversed under reflection about the plane.
So, the eigenvalue in the normal direction is
\[ \lambda_3=-1 \]

Step 7: Find sum of diagonal elements.
The sum of diagonal elements of a matrix is its trace, and trace equals the sum of eigenvalues.
\[ \operatorname{tr}(A)=1+1-1=1 \]
Therefore,
\[ \boxed{1.0} \]