Notes - Linear Algebra I MT22, Triangle Inequality
Flashcards
Can you state the triangle inequality in full for an inner product space?
For $v, w$ in an inner product space $V$, then
\[||v+w|| \le ||v|| + ||w||\]How would you “expand” $\langle v+w, v + w \rangle$?
\[\langle v, v\rangle + 2\langle v, w\rangle + \langle w, w\rangle\]
In the proof of the triangle inequality for an inner product space, i.e.
For $v, w$ in an inner product space $V$, then $ \vert \vert v+w \vert \vert \le \vert \vert v \vert \vert + \vert \vert w \vert \vert $.
What justifies the last line of working:
\[\begin{aligned}
||v+w||^2 &= \langle v, v\rangle + 2\langle v, w\rangle + \langle w, w\rangle \\\\
&=||v||^2 + 2||v||\text{ }||w|| + ||w||^2
\end{aligned}\]
For $v, w$ in an inner product space $V$, then $ \vert \vert v+w \vert \vert \le \vert \vert v \vert \vert + \vert \vert w \vert \vert $.
The Cauchy-Schwarz inequality.