Exercise 1.15

Question

Exercise 15: Under what conditions does equality hold in the Schwarz inequality?

ghostofgarborg · Accepted Answer

Amit Awasthi · Answer

\begin{claim}
      If $\mathbf{a} = (a_1,\ldots,a_n) \in \mathbb{C}^n$, $\mathbf{b} = (b_1,\ldots,b_n) \in \mathbb{C}^n$, and $||\mathbf{b}|| 
e 0$, $|\langle \mathbf{a},\mathbf{b} angle| = ||\mathbf{a}|| \cdot ||\mathbf{b}||$ if and only if $\mathbf{a} = \lambda \mathbf{b}$ for $\lambda \in \mathbb{R}$. If $||\mathbf{b}|| = 0$, this equality always holds.
    \end{claim}
    \begin{proof}
      For $||\mathbf{b}|| = 0$, $|\langle \mathbf{a},\mathbf{b} angle| = 0 = ||\mathbf{a}|| \cdot ||\mathbf{b}||$, and we are done. For $||\mathbf{b}|| 
e 0$, first assume $\mathbf{a} = \lambda \mathbf{b}$ for $\lambda \in \mathbb{R}$. Then, $|\langle \mathbf{a},\mathbf{b} angle| = |\langle \lambda\mathbf{b},\mathbf{b} angle| = \lambda ||\mathbf{b}|| \cdot ||\mathbf{b}|| = ||\mathbf{a}|| \cdot ||\mathbf{b}||$.
      \par Now assume $|\langle \mathbf{a},\mathbf{b} angle|^2 = ||\mathbf{a}|| \cdot ||\mathbf{b}||$. Since $||\mathbf{b}|| 
e 0$, let
      \begin{equation*}
        \lambda = \frac{|\langle \mathbf{a},\mathbf{b} angle|}{||\mathbf{b}||^2},\ \mathbf{c} = \mathbf{a} - \lambda\mathbf{b}.
      \end{equation*}
      Since
      \begin{equation*}
        \langle \mathbf{b},\mathbf{c} angle = \langle \mathbf{b},\mathbf{a} - \lambda\mathbf{b}angle = \langle \mathbf{a},\mathbf{b}angle - \lambda\langle \mathbf{b},\mathbf{b} angle = 0,
      \end{equation*}
      we know $||\mathbf{a}||^2 = ||\lambda\mathbf{b} + \mathbf{c}||^2 = ||\lambda\mathbf{b}||^2 + ||\mathbf{c}||^2$. However, for $|\langle \mathbf{a},\mathbf{b} angle| = ||\mathbf{a}|| \cdot ||\mathbf{b}||$ to be true, $||\mathbf{c}|| = 0$, and thus $\mathbf{a} = \lambda \mathbf{b}$.
    \end{proof}

Exercise 1.15

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Exercise 1.15

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