Question

Let \(T: \mathbb{R}^3 \to \mathbb{R}^3\) be the linear transformation which reflects every vector in \(\mathbb{R}^3\) through a two-dimensional subspace of \(\mathbb{R}^3\). Let \(P \in \mathbb{R}^{3 \times 3}\) be the matrix representation of \(T\) using the basis

\[ \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\}. \]

Then the value of \(2 \times \text{trace}(P) - 3 \times \text{determinant}(P)\) is equal to ______. (answer in integer)

Hint:

A reflection through a plane in R3 has eigenvalues 1, 1, -1; use these to get the trace and determinant directly.

Answer 1

Correct Answer: 5

Step 1: Understand what a reflection through a plane does.
A two-dimensional subspace of \(\mathbb{R}^3\) is a plane through the origin. Reflecting a vector through this plane means: any vector already lying in the plane is left unchanged, and any vector perpendicular to the plane gets flipped to its negative.
Every vector in \(\mathbb{R}^3\) can be split into a part lying in the plane and a part perpendicular to it, and \(T\) acts on each part separately.

Step 2: Find the eigenvalues of \(T\).
Since \(T\) leaves every vector in the 2-dimensional plane unchanged, the plane is an eigenspace with eigenvalue \(1\), and it is 2-dimensional, so the eigenvalue \(1\) is repeated twice.
The direction perpendicular to the plane is 1-dimensional, and \(T\) sends it to its negative, so this direction is an eigenspace with eigenvalue \(-1\).
So the three eigenvalues of \(T\) (counted with multiplicity) are \(1, 1, -1\).

Step 3: Use eigenvalues to get trace and determinant.
The trace of a matrix equals the sum of its eigenvalues, and this does not depend on which basis we use to write the matrix.
\[ \text{trace}(P) = 1 + 1 + (-1) = 1 \]
The determinant of a matrix equals the product of its eigenvalues, and this also does not depend on the basis.
\[ \text{determinant}(P) = 1 \times 1 \times (-1) = -1 \]
Because trace and determinant are basis-independent, using the standard basis given in the question does not change these values; \(P\) is simply the matrix of \(T\) itself.

Step 4: Compute the required expression.
\[ 2 \times \text{trace}(P) - 3 \times \text{determinant}(P) = 2(1) - 3(-1) = 2 + 3 = 5 \]

Final Answer:
The value of \(2 \times \text{trace}(P) - 3 \times \text{determinant}(P)\) is 5, no matter which particular plane is chosen as the subspace. \[ \boxed{5} \]