Question

Let \((V, \langle,\rangle)\) be an inner product space and let \(\langle x,y\rangle = \|x\|\|y\|\) for all \(x,y \in V\). Then which one of the following options is true?

• A. \(\{x,y\}\) is a linearly independent set.
• B. \(\{x,y\}\) is a linearly dependent set.
• C. \(\{x,y\}\) is an orthogonal set.
• D. \(\{x,y\}\) is an orthonormal set.

Hint:

Compare the given condition with the equality case of the Cauchy-Schwarz inequality.

Answer 1

The Correct Option is B

If either $x=0$ or $y=0$ the pair is trivially dependent, so assume both are nonzero. Consider the vector $z = \|y\|x - \|x\|y$ and compute its squared norm using bilinearity of the inner product: \[\|z\|^2 = \langle \|y\|x-\|x\|y,\ \|y\|x-\|x\|y\rangle = \|y\|^2\|x\|^2 - 2\|x\|\|y\|\langle x,y\rangle + \|x\|^2\|y\|^2.\] Using the given condition $\langle x,y\rangle = \|x\|\|y\|$, substitute: \[\|z\|^2 = 2\|x\|^2\|y\|^2 - 2\|x\|\|y\|(\|x\|\|y\|) = 2\|x\|^2\|y\|^2 - 2\|x\|^2\|y\|^2 = 0.\] Since the inner product is positive definite, $\|z\|^2=0$ forces $z=0$, that is $\|y\|x = \|x\|y$. Because $x,y$ are nonzero, this expresses each vector as a positive scalar multiple of the other, so $x$ and $y$ are linearly dependent, and $\{x,y\}$ is a linearly dependent set (dependent nonzero vectors cannot form an orthogonal or orthonormal set). \[\boxed{\{x,y\} \text{ is a linearly dependent set}}\]