Notes - Linear Algebra I MT22, Cauchy-Schwarz Inequality
Flashcards
Can you state the Cauchy-Schwarz Inequality in full?
For $v, w$ in an inner product space $V$, then
\[|\langle v, w\rangle| \le ||v|| \text{ } ||w||\]The Cauchy-Schwarz inequality states
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
When does equality hold?
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
When $v, w$ are linearly dependent, i.e. $v + t _ 0 w = 0$.
In the proof of the Cauchy-Schwarz inequality
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
Why can you assume $w \ne 0$?
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
Because if $w = 0$ then the result is immediate.
In the proof of the Cauchy-Schwarz inequality
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
What statement do you start with that the whole proof follows from?
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
In the proof of the Cauchy-Schwarz inequality
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
You get to a stage where you have
\[0 \le \langle v,w \rangle + 2t\langle v,w \rangle + t^2\langle w,w \rangle\]
What do you do next?
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
Use the fact the discriminant $b^2 - 4ac \le 0$.
In the proof of the Cauchy-Schwarz inequality
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
You get to a stage where you have
\[\begin{aligned}
0 &\le ||v+tw||^2 \\\\
&=\langle v,w \rangle + 2t\langle v,w \rangle + t^2\langle w,w \rangle
\end{aligned}\]
If $v = w$, why are $v,w$ linearly dependent?
For $v, w$ in an inner product space $V$, then $ \vert \langle v, w\rangle \vert \le \vert \vert v \vert \vert \text{ } \vert \vert w \vert \vert $.
Because the discriminat is zero so there exists a root $t = t _ 0$ and so $ \vert \vert v + t _ 0 w \vert \vert = 0$.