MATH 414 Lecture 5

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Hyperbolic Trigonometric Functions

$\sinh {x}={\frac {\mathrm {e} ^{x}-\mathrm {e} ^{-x}}{2}}=i\,\sin {\left(-i\,x\right)}$

Gram-Schmidt Process

INPUT: basis vectors $U=\left\{{\vec {v}}_{1},{\vec {v}}_{2},\ldots ,{\vec {v}}_{n}\right\}$ for a vector space $V=\mathrm {Span} \left\{{\vec {v}}_{1},{\vec {v}}_{2},\ldots ,{\vec {v}}_{n}\right\}$.

OUTPUT: orthonormal basis $O=\left\{{\hat {e}}_{1},{\hat {e}}_{2},\ldots ,{\hat {e}}_{n}\right\}$ for $V$

${\hat {e}}_{1}:={\frac {{\vec {v}}_{1}}{\left\|{\vec {v}}_{1}\right\|}}$

$O:=\left\{{\hat {e}}_{1}\right\}$

for $i$ from $2$ to $n$

${\vec {p}}_{i}:=\displaystyle \sum _{k=1}^{i-1}\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle \,{\hat {e}}_{k}$

${\hat {e}}_{i}:=\displaystyle {\frac {{\vec {v}}_{i}-{\vec {p}}_{i}}{\left\|{\vec {v}}_{i}-{\vec {p}}_{i}\right\|}}$

$O:=O\cup \left\{{\hat {e}}_{i}\right\}$

Python Implementation

def gram_schmidt(basis):
    on_basis = [ normalize(basis[0]) ]
    for i in xrange(1, len(basis)):
        p = sum(inner_product(basis[i], e) * e for e in on_basis)
        next_e = normalize(basis[i] - p)
        on_basis.append(next_e)
    return on_basis

Trick

Calculating $\left\|{\vec {v}}_{i}-{\vec {p}}_{i}\right\|=\left\langle {\vec {v}}_{i}-{\vec {p}}_{i},{\vec {v}}_{i}-{\vec {p}}_{i}\right\rangle$; in $L^{2}\left[a,b\right]$ space, $\int _{a}^{b}\left(v_{i}\left(x\right)-p_{i}\left(x\right)\right)^{2}\,\mathrm {d} x$; can be a long and cumbersome process, but we can compute this length using values we already know:

Theorem. $\left\|{\vec {v}}_{i}-{\vec {p}}_{i}\right\|={\sqrt {\left\|{\vec {v}}\right\|^{2}-\sum _{k=1}^{n-1}\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle ^{2}}}$

Proof. Observe that ${\vec {p}}_{i}$ and ${\vec {v}}_{i}-{\vec {p}}_{i}$ are orthogonal by construction, so ${\vec {v}}_{i}$, ${\vec {p}}_{i}$, and ${\vec {v}}_{i}-{\vec {p}}_{i}$ form a right triangle.

Therefore $\left\|{\vec {p}}_{i}\right\|^{2}+\left\|{\vec {v}}_{i}-{\vec {p}}_{i}\right\|^{2}=\left\|{\vec {v}}_{i}\right\|^{2}$, in particular, $\left\|{\vec {v}}_{i}-{\vec {p}}_{i}\right\|^{2}=\left\|{\vec {v}}_{i}\right\|^{2}-\left\|{\vec {p}}_{i}\right\|^{2}$, by the Pythagorean theorem.

By definition, ${\vec {p}}_{i}=\sum _{k=1}^{n-1}\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle \,{\hat {e}}_{k}$. By generalizing the pythagorean theorem into multiple dimensions, we have $\left\|{\vec {p}}_{i}\right\|^{2}=\left\|\sum _{k=1}^{n-1}\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle \,{\hat {e}}_{k}\right\|^{2}=\sum _{k=1}^{n-1}\left\|\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle \,{\hat {e}}_{k}\right\|^{2}$ because all $\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle \,{\hat {e}}_{k}$ are orthogonal.

Factoring out the scalar value $\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle$ from the length of each component in the sum leaves just $\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle ^{2}\,\left\|{\hat {e}}_{k}\right\|^{2}$, and of course the length of ${\hat {e}}_{k}$ is $1$. Therefore,

$\left\|{\vec {p}}_{i}\right\|^{2}=\sum _{k=1}^{n-1}\left\langle {\vec {v}}_{i},{\hat {e}}_{k}\right\rangle ^{2}$

Plugging this definition into the pythagorean identity above yields the desired equation for $\left\|{\vec {v}}_{i}-{\vec {p}}_{i}\right\|$.

quod erat demonstrandum