# Cauchy-Schwarz Inequality Proof for orthonormal basis

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

• What is $[u]_B$? – Thomas Apr 18 '16 at 16:28
• @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. – RedShift Apr 18 '16 at 16:31
• @Thomas: the vector in $\mathbb F^n$ representing $u$ in the basis $B$. – Martin Argerami Apr 18 '16 at 16:31

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 $w=0$.