Related Formula
For a skew-symmetric matrix A$A$, A^T = -A$A^T = -A$.
(XY)^T = Y^T X^T$(XY)^T = Y^T X^T$, and (X⁻¹)^T = (X^T)⁻¹$(X^{-1})^T = (X^T)^{-1}$.
Core Logic
Observe the matrix A$A$. aᵢⱼ = -aⱼᵢ$a_{ij} = -a_{ji}$ and diagonals are zero.
So, A$A$ is a skew-symmetric matrix: A^T = -A$A^T = -A$.
We have B(I - A) = I + A B = (I + A)(I - A)⁻¹$B(I - A) = I + A \implies B = (I + A)(I - A)^{-1}$.
We need to compute B^T B$B^T B$.
Step 1: Transpose Manipulation
B^T = ((I + A)(I - A)⁻¹)^T$$B^T = \left((I + A)(I - A)^{-1}\right)^T$$
B^T = ((I - A)⁻¹)^T (I + A)^T$$B^T = \left((I - A)^{-1}\right)^T (I + A)^T$$
B^T = ((I - A)^T)⁻¹ (I + A^T)$$B^T = \left((I - A)^T\right)^{-1} (I + A^T)$$
Since A^T = -A$A^T = -A$:
(I - A)^T = I - A^T = I - (-A) = I + A$$(I - A)^T = I - A^T = I - (-A) = I + A$$
(I + A^T) = I - A$$(I + A^T) = I - A$$
Therefore, B^T = (I + A)⁻¹ (I - A)$B^T = (I + A)^{-1} (I - A)$.
Step 2: Trace Calculation
Now compute B^T B$B^T B$:
B^T B = (I + A)⁻¹ (I - A) (I + A) (I - A)⁻¹$$B^T B = (I + A)^{-1} (I - A) (I + A) (I - A)^{-1}$$
Since (I-A)$(I-A)$ and (I+A)$(I+A)$ commute, (I-A)(I+A) = I² - A² = (I+A)(I-A)$(I-A)(I+A) = I^2 - A^2 = (I+A)(I-A)$:
B^T B = (I + A)⁻¹ (I + A) (I - A) (I - A)⁻¹$$B^T B = (I + A)^{-1} (I + A) (I - A) (I - A)^{-1}$$
B^T B = I · I = I$$B^T B = I \cdot I = I$$
Thus, B^T B$B^T B$ is simply the 3 × 3$3 \times 3$ identity matrix I₃$I_3$.
The sum of the diagonal elements is the trace of the identity matrix:
tr(B^T B) = 1 + 1 + 1 = 3$$\text{tr}(B^T B) = 1 + 1 + 1 = 3$$
Pattern Recognition
The expression B = (I+A)(I-A)⁻¹$B = (I+A)(I-A)^{-1}$ is the classic Cayley transform, which maps any skew-symmetric matrix A$A$ exactly onto an orthogonal matrix B$B$. Orthogonal matrices intrinsically satisfy B^T B = I$B^T B = I$.
Chapter Mix
Class 12 Maths: Matrices and Determinants