Andrew Thangaraj
Aug-Nov 2020
\(V\): inner product space
\(e_1,\ldots,e_m\): orthonormal if
Examples
Standard basis
\(\{(1/\sqrt{2},1/\sqrt{2}),(1/\sqrt{2},-1/\sqrt{2})\}\)
If \(e_1,\ldots,e_m\): orthonormal, \[\lVert a_1e_1+\cdots+a_me_m\rVert^2=\lvert a_1\rvert^2+\cdots+\lvert a_m\rvert^2\]
Proof
Direct evaluation using \(\lVert e_i\rVert=1\), \(\langle e_i,e_j\rangle=0\) for \(i\ne j\)
Orthonormal set of vectors are linearly independent
Proof
\(a_1e_1+\cdots+a_me_m=0\) implies \(\lvert a_i\rvert=0\) for each \(i\)
An orthonormal basis for \(V\) is an orthonormal list of vectors of \(V\) that is also a basis.
Examples
Standard basis
\(\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2}\right)\), \(\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{-1}{2},\dfrac{-1}{2}\right)\), \(\left(\dfrac{1}{2},\dfrac{-1}{2},\dfrac{-1}{2},\dfrac{1}{2}\right)\), \(\left(\dfrac{-1}{2},\dfrac{1}{2},\dfrac{-1}{2},\dfrac{1}{2}\right)\)
\(V\): inner product space
Basis: \(\{e_1,\ldots,e_n\}\)
What are the coordinates of a vector \(v\) in above basis?
Suppose basis is orthonormal. \[v=\langle v,e_1\rangle e_1+\cdots+\langle v,e_n\rangle e_n\] \[\lVert v\rVert^2=\lvert\langle v,e_1\rangle\rvert^2+\cdots+\lvert\langle v,e_n\rangle\rvert^2\]
Proof
Use \(\lVert e_i\rVert=1\), \(\langle e_i,e_j\rangle=0\) for \(i\ne j\)
\(V\): inner product space
input: \(v_1,\ldots,v_m\), a linearly independent list
\(e_1=v_1/\lVert v_1\rVert\)
for \(j=2,\cdots,m\) \[e_j=\dfrac{v_j-\langle v_j,e_1\rangle e_1-\cdots-\langle v_j,e_{j-1}\rangle e_{j-1}}{\lVert v_j-\langle v_j,e_1\rangle e_1-\cdots-\langle v_j,e_{j-1}\rangle e_{j-1}\rVert}\]
output: \(e_1,\ldots,e_m\), an orthonormal list such that \[\text{span}(v_1,\ldots,v_j)=\text{span}(e_1,\ldots,e_j)\] for \(j=1,\ldots,m\)
Proof: induction on \(j\)
\(V\): inner product space, finite-dimensional
\(V\) has an orthonormal basis
Proof: Take a basis for \(V\) and perform Gram-Schmidt
An orthonormal list of vectors can be extended to an orthonormal basis.
Proof: Extend to a basis and then apply Gram-Schmidt
\(V\): vector space, \(T:V\to V\), operator
\(B=\{v_1,\cdots,v_n\}\): basis for \(V\)
\(M(T,B)\): matrix of \(T\) w.r.t. \(B\)
\(M(T,B)\) is upper triangular if and only if span\((v_1,\cdots,v_j)\) is invariant under \(T\) for \(j=1,\dots,n\)
Proof
\(M(T,B)=\begin{bmatrix} a_{11}&\ast&\cdots&\ast\\ 0&a_{22}&\cdots&\ast\\ \vdots&\ddots&\ddots&\vdots\\ 0&\cdots&0&a_{nn} \end{bmatrix}\)
Coordinates of \(v\in\text{span}(v_1,\cdots,v_j)\) in \(B\): non-zero only in first \(j\) positions
\(V\): finite-dimensional inner product space over \(\mathbb{C}\)
\(T:V\to V\), operator
There exists an orthonormal basis \(B\) such that the matrix of \(T\) with respect to \(B\) is upper-triangular.
Proof
There exists a basis over which \(T\) is upper-triangular. Use Gram-Schmidt on the basis to get \(B\).