Question

If $\alpha, \beta$ are vectors in an inner product space V, then

• A. $||\alpha + \beta||^{2} < ||\alpha||^{2} + ||\beta||^{2}$
• B. $||\alpha + \beta||^{2} = ||\alpha||^{2} + ||\beta||^{2}$
• C. $||\alpha + \beta||^{2} \le ||\alpha||^{2} + ||\beta||^{2}$
• D. $||\alpha + \beta||^{2} = ||\alpha - \beta||^{2}$

Hint:

The shortest distance between two points is a straight line. Any other path ($\alpha + \beta$) is always longer or equal.

Answer 1

The Correct Option is C

Step 1: Concept
This refers to the Triangle Inequality property of norms in inner product spaces.

Step 2: Meaning
The general triangle inequality states $||\alpha + \beta|| \le ||\alpha|| + ||\beta||$.

Step 3: Analysis
Squaring both sides: $||\alpha + \beta||^2 \le (||\alpha|| + ||\beta||)^2 = ||\alpha||^2 + ||\beta||^2 + 2||\alpha||||\beta||$. Note that option (C) in the source text is often shorthand for this fundamental inequality.

Step 4: Conclusion
While the prompt lists option (C) simply, it represents the foundational "less than or equal to" relationship that defines distances in these spaces. Final Answer: (C)