Inner Product Spaces: Complex inner product spaces
Gram-Schmidt in complex inner product spaces
For finite-dimensional complex inner product spaces there are a Gram-Schmidt procedure and a QR decomposition that barely differ from the real case. Here we briefly discuss these results.
Gram-Schmidt
If #\basis{a_1,\ldots,a_n}# is an independent system of vectors in a complex inner product space #V#, then we can find as described below an orthonormal system #\basis{\vec{e}_1,\ldots,\vec{e}_n}# such that for every #i=1,\ldots, n# we have \[ \linspan{\vec{a}_1,\ldots,\vec{a}_i}=\linspan{\vec{e}_1,\ldots,\vec{e}_i}\]
Start with \[\vec{e}_1:=\dfrac{\vec{a}_1}{\norm{\vec{a}_1}}\] For #i=1,\ldots,n-1# perform the following two steps: \[\begin{array}{rcl}\vec{e}_{i+1}^{\,*}&:=&\vec{a}_{i+1}-\sum_{j=1}^i(\dotprod{\vec{a}_{i+1}}{\vec{e}_j})\cdot \vec{e}_j\\ \vec{e}_{i+1}&:=&\dfrac{\vec{e}_{i+1}^{\,*}}{\norm{\vec{e}_{i+1}^{\,*}}}\end{array}\]
In particular, every finite-dimensional complex inner product space has an orthonormal basis.
For the matrix form of this result, we need the matrix #Q^*#, which is obtained from a matrix #Q# by transposition and subsequent complex conjugation of its entries. If #\overline{Q}# denotes the matrix arising from #Q# when all its entries are conjugated, then \[Q^* = \overline{Q^\top} = {\overline{Q}}^\top\]
QR decomposition
Let #A# be an #(m\times n)#-matrix with #n# independent columns #\vec{a}_1,\ldots,\vec{a}_n#. Then it can be written as the product \[A=Q\,R\] where #R# is an upper triangular matrix of size #m\times n# and the columns of the #(m\times m)#-matrix #Q# form an orthonormal system. This notation is called a QR decomposition of #A#.
This decomposition can be found by carrying out the Gram-Schmidt algorithm on the columns of #A# and adding columns to the resulting matrix from an orthonormal basis of the complement of the span of the columns of #A#, so as to get an #(m\times m)#-matrix. The matrix #R# is then given by #R={Q}^*\,A#.
\vec{a}_1=\left[ -2 , 0 , 2 \complexi \right] ,\quad \vec{a}_2=\left[ -2 \complexi , 1 , 0 \right] \] Perform the Gram-Schmidt procedure and rewrite this system to an orthonormal system #\basis{\vec{q}_1,\vec{q}_2}#. Give your answer in the form of a list of row vectors.
We calculate the norm of the vector #\vec{a}_1#: \[\norm{\vec{a}_1}=\sqrt{\dotprod{\vec{a}_1}{\vec{a}_1}}=\sqrt{\dotprod{\left[ -2 , 0 , 2 \complexi \right] }{\left[ -2 , 0 , 2 \complexi \right] }}=2\sqrt{2}\] We obtain the vector #\vec{q}_1# by dividing #\vec{a}_1# by its norm. \[\vec{q}_1=\frac{1}{\norm{\vec{a}_1}}\vec{a}_1=\frac{1}{2\sqrt{2}}\left[ -2 , 0 , 2 \complexi \right] =\left[ -{{1}\over{\sqrt{2}}} , 0 , {{\complexi}\over{\sqrt{2}}} \right] \] In order to replace the vector #\vec{a}_2# by a vector # \vec{q}_2^{\,*}# perpendicular to the first vector we compute the inner product #\dotprod{\vec{a}_2}{\vec{q}_1}#: \[\dotprod{\vec{a}_2}{\vec{q}_1}=\dotprod{\left[ -2 \complexi , 1 , 0 \right] }{\left[ -{{1}\over{\sqrt{2}}} , 0 , {{\complexi}\over{\sqrt{2}}} \right] }=\sqrt{2}\cdot \complexi\] The vector #\vec{q}_2^{\,*}# is given by \[\vec{q}_2^{\,*}=\vec{a}_2-(\dotprod{\vec{a}_2}{\vec{q}_1})\vec{q}_1=\left[ -2 \complexi , 1 , 0 \right] -\sqrt{2}\cdot \complexi \left[ -{{1}\over{\sqrt{2}}} , 0 , {{\complexi}\over{\sqrt{2}}} \right] =\left[ -\complexi , 1 , 1 \right] \] It remains for us to normalize the vector #\vec{q}_2^{\,*}#. To this end, we first calculate its norm: \[\norm{\vec{q}_2^{\,*}}=\sqrt{\dotprod{\vec{q}_2^{\,*}}{\vec{q}_2^{\,*}}}=\dotprod{\left[ -\complexi , 1 , 1 \right] }{\left[ -\complexi , 1 , 1 \right] }=\sqrt{3}\] Now we obtain the vector #\vec{q}_2# from the vector #\vec{q}_2^{\,*}# by dividing by its norm: \[\vec{q}_2=\frac{1}{\norm{\vec{q}_2^{\,*}}}\vec{q}_2^{\,*}=\frac{1}{\sqrt{3}}\left[ -\complexi , 1 , 1 \right] =\left[ -{{\complexi}\over{\sqrt{3}}} , {{1}\over{\sqrt{3}}} , {{1}\over{\sqrt{3}}} \right] \] The answer is \[\left\{\vec{q}_1,\vec{q}_2\right\}=\left\{\left[ -{{1}\over{\sqrt{2}}} , 0 , {{\complexi}\over{\sqrt{2}}} \right] ,\left[ -{{\complexi}\over{\sqrt{3}}} , {{1}\over{\sqrt{3}}} , {{1}\over{\sqrt{3}}} \right] \right\}\]
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