1
$\begingroup$

Let $B = \{v_1,v_2,...,v_n\}$ be an orthonormal basis for an inner product space $V$, with inner product $<,>$. Let $u,v \in V$. Show that $<u,v> = [u]_B \cdot [v]_B$

In addition, prove that the equality holds in he Cauchy-Schwarz inequality

$$|<u,v>| \leq ||u||||v||$$

if and only if $u$ and $v$ are linearly dependent.

Any sort of direction would be great

$\endgroup$
  • $\begingroup$ What is $[u]_B$? $\endgroup$ – Thomas Apr 18 '16 at 16:28
  • $\begingroup$ @Thomas I think that what the question is implying is that $u$ is a vector within $V$ that is in terms of the orthonormal basis provided. $\endgroup$ – RedShift Apr 18 '16 at 16:31
  • $\begingroup$ @Thomas: the vector in $\mathbb F^n$ representing $u$ in the basis $B$. $\endgroup$ – Martin Argerami Apr 18 '16 at 16:31
1
$\begingroup$

For the first equality, just write $u$ and $v$ in terms of the orthonormal basis and evaluate $\langle u,v\rangle$ using the orthonormality of the elements of the basis.

For the case of equality, first rule out the cases where at least one of $u,v$ is zero. When both are nonzero, show that if $v=\alpha u$ you have equality. If you have equality, consider $w=u-\frac{\langle u,v\rangle}{\langle v,v\rangle}\,v$. Then $$ \langle w,w\rangle=\langle u,u\rangle-\frac{\langle u,v\rangle^2}{\langle v,v\rangle^2}\,\langle v,v\rangle=0, $$ and so $w=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.