Question

Let \[ A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} \] and \( B \) be a matrix such that \[ B(I - A) = I + A. \] Then the sum of the diagonal elements of \( B^{T}B \) is equal to

Hint:

The Cayley transform of a real skew-symmetric matrix is an orthogonal matrix. For orthogonal matrices, $B^T B = I$.

Answer 1

Correct Answer: 3

Notice that A is a skew-symmetric matrix, i.e., $A^T = -A$.
Given $B = (I + A)(I - A)^{-1}$.
Calculate $B^T B$.
$B^T = [(I-A)^{-1}]^T (I+A)^T = (I-A^T)^{-1} (I+A^T)$.
Substitute $A^T = -A$: $B^T = (I+A)^{-1} (I-A)$.
Now, $(I+A)$ and $(I-A)$ commute because $(I+A)(I-A) = I - A^2 = (I-A)(I+A)$.
$B^T B = (I+A)^{-1} (I-A) (I+A) (I-A)^{-1}$.
Rearranging commutative terms: $B^T B = (I+A)^{-1} (I+A) (I-A) (I-A)^{-1} = I \cdot I = I$.
$B^T B$ is the identity matrix of order 3.
The sum of diagonal elements (Trace) is $1 + 1 + 1 = 3$.