Proof of Theorem PIP

From the proof of Theorem IPN we see that \begin{equation*} \innerproduct{\vect{u}}{\vect{u}} = \modulus{\vectorentry{\vect{u}}{1}}^2+ \modulus{\vectorentry{\vect{u}}{2}}^2+ \modulus{\vectorentry{\vect{u}}{3}}^2+ \cdots+ \modulus{\vectorentry{\vect{u}}{m}}^2 \end{equation*}

Since each modulus is squared, every term is positive, and the sum must also be positive. (Notice that in general the inner product is a complex number and cannot be compared with zero, but in the special case of $\innerproduct{\vect{u}}{\vect{u}}$ the result is a real number.)

The phrase, “with equality if and only if” means that we want to show that the statement $\innerproduct{\vect{u}}{\vect{u}}= 0$ (i.e. with equality) is equivalent (“if and only if”) to the statement $\vect{u}=\zerovector$.

If $\vect{u}=\zerovector$, then it is a straightforward computation to see that $\innerproduct{\vect{u}}{\vect{u}}= 0$. In the other direction, assume that $\innerproduct{\vect{u}}{\vect{u}}= 0$. As before, $\innerproduct{\vect{u}}{\vect{u}}$ is a sum of moduli. So we have \begin{equation*} 0=\innerproduct{\vect{u}}{\vect{u}}= \modulus{\vectorentry{\vect{u}}{1}}^2+ \modulus{\vectorentry{\vect{u}}{2}}^2+ \modulus{\vectorentry{\vect{u}}{3}}^2+ \cdots+ \modulus{\vectorentry{\vect{u}}{m}}^2 \end{equation*}

Now we have a sum of squares equaling zero, so each term must be zero. Then by similar logic, $\modulus{\vectorentry{\vect{u}}{i}}=0$ will imply that $\vectorentry{\vect{u}}{i}=0$, since $0+0i$ is the only complex number with zero modulus. Thus every entry of $\vect{u}$ is zero and so $\vect{u}=\zerovector$, as desired.