Question

Let \( M = [X \ Y \ Z] \) be an orthogonal matrix with \( X, Y, Z \in \mathbb{R}^3 \) as its column vectors. Then \[ Q = X X^T + Y Y^T \]

• A. is a skew-symmetric matrix
• B. is the \( 3 \times 3 \) identity matrix
• C. satisfies \( Q^2 = Q \)
• D. satisfies \( QZ = Z \)

Hint:

For orthogonal matrices, the column vectors are orthonormal, and properties such as \( M^T M = I \) hold true, which simplify many matrix operations.

Answer 1

The Correct Option is C

Step 1: Check if $Q$ is skew-symmetric

$$Q^T = (XX^T + YY^T)^T = XX^T + YY^T = Q$$

$Q$ is symmetric, not skew-symmetric.

Option (A) is FALSE 

Step 2: Check if $Q = I$

Since $MM^T = I$ for orthogonal $M$: $$XX^T + YY^T + ZZ^T = I$$

Therefore: $Q = I - ZZ^T \neq I$ (unless $Z = 0$, which is impossible).

Option (B) is FALSE 

Step 3: Check if $Q^2 = Q$

$$Q^2 = (XX^T + YY^T)^2 = XX^TXX^T + XX^TYY^T + YY^TXX^T + YY^TYY^T$$

Using $X^TX = 1$, $Y^TY = 1$, and $X^TY = Y^TX = 0$: $$Q^2 = X(X^TX)X^T + X(X^TY)Y^T + Y(Y^TX)X^T + Y(Y^TY)Y^T$$ $$= XX^T + 0 + 0 + YY^T = Q$$

Option (C) is TRUE 

Step 4: Check if $QZ = Z$

$$QZ = (XX^T + YY^T)Z = X(X^TZ) + Y(Y^TZ)$$

Using $X^TZ = 0$ and $Y^TZ = 0$: $$QZ = 0 \neq Z$$

Option (D) is FALSE 

Answer: (C) satisfies $Q^2 = Q$