Let $M$ be an $n \times n$ non-zero skew symmetric matrix. Then the matrix $(I_n - M)(I_n + M)^{-1}$ is always
Step 1: Recall properties of skew-symmetric matrices.
If $M$ is skew-symmetric, then $M^T = -M$.
Step 2: Compute transpose of given matrix.
Let $A = (I - M)(I + M)^{-1}$. Then
\[
A^T = [(I + M)^{-1}]^T (I - M)^T = (I + M^T)^{-1}(I - M^T) = (I - M)^{-1}(I + M).
\]
Step 3: Show orthogonality.
\[
AA^T = (I - M)(I + M)^{-1}(I - M)^{-1}(I + M) = I.
\]
Hence $A$ is orthogonal.
Step 4: Conclusion.
\[
\boxed{(I_n - M)(I_n + M)^{-1} \text{ is orthogonal.}}
\]
Let $T: \mathbb{R}^3 \to \mathbb{R}^4$ be a linear transformation. If $T(1,1,0) = (2,0,0,0)$, $T(1,0,1) = (2,4,0,0)$, and $T(0,1,1) = (0,0,2,0)$, then $T(1,1,1)$ equals
For real constants $a$ and $b$, let \[ M = \begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \\ a & b \end{bmatrix} \] be an orthogonal matrix. Then which of the following statements is/are always TRUE?
Find the rank of the matrix: \[ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 2 \\ 2 & 5 & 6 & 4 \\ 2 & 6 & 8 & 5 \end{bmatrix} \] Rank = ?
Let \( M = \begin{bmatrix} \tfrac{1}{4} & \tfrac{3}{4} \\ \\ \tfrac{3}{5} & \tfrac{2}{5} \end{bmatrix}. \) If \( I \) is the \( 2 \times 2 \) identity matrix and \( 0 \) is the \( 2 \times 2 \) zero matrix, then