Question

Let \(V_2(\mathbb{C})\) be the inner product space with respect to the standard inner product. Which of the following vector in \(V_2(\mathbb{C})\) is orthogonal to the vector \((1-i,1+i)\)?

• A. \((1+i,1+i)\)
• B. \((-1+i,1+i)\)
• C. \((1+i,1-i)\)
• D. \((-1-i,1-i)\)

Hint:

In complex inner product spaces, always use complex conjugates while calculating the inner product.

Answer 1

The Correct Option is C

Concept:
In a complex inner product space, two vectors are orthogonal if their inner product is zero. For complex vectors: \[ u=(u_1,u_2), \qquad v=(v_1,v_2) \] the standard inner product is: \[ \langle u,v\rangle=u_1\overline{v_1}+u_2\overline{v_2} \] where \(\overline{v_1}\) and \(\overline{v_2}\) denote complex conjugates.

Step 1: Write the given vector.
The given vector is: \[ v=(1-i,1+i) \] We need to find the vector orthogonal to \(v\).

Step 2: Check option (C).
Option (C) is: \[ u=(1+i,1-i) \] Now compute: \[ \langle u,v\rangle = (1+i)\overline{(1-i)}+(1-i)\overline{(1+i)} \]

Step 3: Find the conjugates.
\[ \overline{(1-i)}=1+i \] and: \[ \overline{(1+i)}=1-i \] Therefore: \[ \langle u,v\rangle = (1+i)(1+i)+(1-i)(1-i) \]

Step 4: Simplify the products.
First: \[ (1+i)(1+i)=(1+i)^2 \] \[ (1+i)^2=1+2i+i^2 \] Since: \[ i^2=-1 \] we get: \[ (1+i)^2=1+2i-1=2i \] Now: \[ (1-i)(1-i)=(1-i)^2 \] \[ (1-i)^2=1-2i+i^2 \] \[ (1-i)^2=1-2i-1=-2i \]

Step 5: Add both terms.
\[ \langle u,v\rangle=2i+(-2i) \] \[ \langle u,v\rangle=0 \] Since the inner product is zero, the vectors are orthogonal.

Step 6: Final answer.
Therefore, the vector orthogonal to \((1-i,1+i)\) is: \[ (1+i,1-i) \] \[ \therefore \text{Correct Answer is (C)} \]