Q1) Prove that \( \|u+v\|=\|u\|+\|v\| \) if and only if one of the vectors is a scalar multiple of the other.

To prove that \( \|u+v\| = \|u\| + \|v\| \) if and only if one of the vectors is a scalar multiple of the other, we will show both directions of the equivalence. **(⇒) Assume \( \|u+v\| = \|u\| + \|v\| \):** We start with the equality \( \|u+v\|^2 = \|u\|^2 + \|v\|^2 + 2\langle u, v \rangle \), where \( \langle u, v \rangle \) is the inner product of \( u \) and \( v \). Given that \( \|u+v\| = \|u\| + \|v\| \), we square both sides: \[ \|u+v\|^2 = (\|u\| + \|v\|)^2 = \|u\|^2 + 2\|u\|\|v\| + \|v\|^2. \] Now, substituting the expression for \( \|u+v\|^2 \): \[ \|u\|^2 + \|v\|^2 + 2\langle u, v \rangle = \|u\|^2 + 2\|u\|\|v\| + \|v\|^2. \] Subtracting \( \|u\|^2 + \|v\|^2 \) from both sides gives: \[ 2\langle u, v \rangle = 2\|u\|\|v\|. \] Dividing both sides by 2, we have: \[ \langle u, v \rangle = \|u\|\|v\|. \] By the Cauchy-Schwarz inequality, we know that \( \langle u, v \rangle \leq \|u\|\|v\| \). The equality holds if and only if \( u \) and \( v \) are linearly dependent, meaning one is a scalar multiple of the other. Thus, we conclude that if \( \|u+v\| = \|u\| + \|v\| \), then one of the vectors is a scalar multiple of the other. **(⇐) Assume one of the vectors is a scalar multiple of the other:** Let’s assume \( v = \lambda u \) for some scalar \( \lambda \). We will show that \( \|u + v\| = \|u\| + \|v\| \). Calculating \( u + v \): \[ u + v = u + \lambda u = (1 + \lambda)u. \] Now, we compute the norms: \[ \|u + v\| = \|(1 + \lambda)u\| = |1 + \lambda|\|u\|. \] Next, we compute \( \|v\| \): \[ \|v\| = \|\lambda u\| = |\lambda|\|u\|. \] Now, we can express \( \|u\| + \|v\| \): \[ \|u\| + \|v\| = \|u\| + |\lambda|\|u\| = (1 + |\lambda|)\|u\|. \] To show \( \|u + v\| = \|u\| + \|v\| \), we need to consider two cases for \( \lambda \): 1. If \( \lambda \geq 0 \), then \( |1 + \lambda| = 1 + \lambda \) and \( |\lambda| = \lambda \). Thus: \[ \|u + v\| = (1 + \lambda)\|u\| = \|u\| + \|v\|. \] 2. If \( \lambda < 0 \), then \( |1 + \lambda| = 1 + \lambda \) if \( \lambda > -1 \) (which is the case for \( \|u + v\| = \|u\| + \|v\| \)). Thus: \[ \|u + v\| = (1 + \lambda)\|u\| = \|u\| - |\lambda|\|u\| = \|u\| + \|v\|. \] In both cases, we find that \( \|u + v\| = \|u\| + \|v\| \). Thus, we conclude that if one vector is a scalar multiple of the other, then \( \|u + v\| = \|u\| + \|v\| \). **Conclusion:** We have shown both directions of the equivalence: \[ \|u + v\| = \|u\| + \|v\| \iff \text{one of the vectors is a scalar multiple of the other.} \]