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gibson:teaching:fall-2016:math753:qr-leastsquares

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gibson:teaching:fall-2016:math753:qr-leastsquares [2016/10/12 09:16]
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gibson:teaching:fall-2016:math753:qr-leastsquares [2016/10/12 09:18] (current)
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 Let $r = Ax-b$. Thinking geometrically,​ $\|r\| = \|Ax-b\|$ will be smallest when $r$ is orthogonal to the span of the columns of $A$. Recall that for the QR factorization $A=QR$, the columns of $Q$ are an orthonormal basis for the span of the columns of $A$. So $r$ is orthogonal to the span of the columns of $A$ when  Let $r = Ax-b$. Thinking geometrically,​ $\|r\| = \|Ax-b\|$ will be smallest when $r$ is orthogonal to the span of the columns of $A$. Recall that for the QR factorization $A=QR$, the columns of $Q$ are an orthonormal basis for the span of the columns of $A$. So $r$ is orthogonal to the span of the columns of $A$ when 
  
-\begin{align*} +\begin{equation*} 
-Q^T r = 0\\ +Q^T r = 0 
-Q^T (Ax-b) = 0\\ +\end{equation*} 
-Q^T A x = Q^T b \\ +\begin{equation*} 
-Q^T Q R x = Q^T b \\+Q^T (Ax-b) = 0 
 +\end{equation*} 
 +\begin{equation*} 
 +Q^T A x = Q^T b  
 +\end{equation*} 
 +\begin{equation*} 
 +Q^T Q R x = Q^T b 
 +\end{equation*} 
 +\begin{equation*}
 R x = Q^T b R x = Q^T b
-\end{align*}+\end{equation*}
  
 Since $Q$ is an $m \times n$ matrix whose columns are orthonormal,​ $Q^T Q = I$ is the $n \times n$ identity matrix. Since $R$ is $n\times n$, we now have as the same number of equations as unknowns. The last line, $Rx = Q^Tb$, is a square upper-triangular system which we can solve by backsubstitution. Since $Q$ is an $m \times n$ matrix whose columns are orthonormal,​ $Q^T Q = I$ is the $n \times n$ identity matrix. Since $R$ is $n\times n$, we now have as the same number of equations as unknowns. The last line, $Rx = Q^Tb$, is a square upper-triangular system which we can solve by backsubstitution.
gibson/teaching/fall-2016/math753/qr-leastsquares.txt · Last modified: 2016/10/12 09:18 by gibson