- #1
Mindscrape
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I have a problem that says to find the least squares solution to
\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> K x = \colv{2}{2} for
K = \left(<br /> \begin{array}{cc}<br /> 1 & 2\\<br /> 2 & 4<br /> \end{array} \right). Then express the solution in the form x = w + z, where w is in the corange, and z is in the kernel.
Since the determinate of the matrix K will be zero, then there should be infinite least squares solutions since the kernel is non-zero, correct? When I try to find a solution by multiplying each side by K^T, I get no solution rather than infinite solutions. Specifically the sytem
\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 <br /> \left(<br /> \begin{array}{cc}<br /> 5 & 10\\<br /> 0 & 0 <br /> \end{array} \right) x = \colv{0}{10}
How can I find the least square of a matrix with determinate zero? Everything I know is based off the fact that the matrix will be invertible.
*My last matrix is messed up somehow, it should be the augmented matrix {{5,10 | 5},{0,0 | 10}}.
\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}<br /> K x = \colv{2}{2} for
K = \left(<br /> \begin{array}{cc}<br /> 1 & 2\\<br /> 2 & 4<br /> \end{array} \right). Then express the solution in the form x = w + z, where w is in the corange, and z is in the kernel.
Since the determinate of the matrix K will be zero, then there should be infinite least squares solutions since the kernel is non-zero, correct? When I try to find a solution by multiplying each side by K^T, I get no solution rather than infinite solutions. Specifically the sytem
\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 <br /> \left(<br /> \begin{array}{cc}<br /> 5 & 10\\<br /> 0 & 0 <br /> \end{array} \right) x = \colv{0}{10}
How can I find the least square of a matrix with determinate zero? Everything I know is based off the fact that the matrix will be invertible.
*My last matrix is messed up somehow, it should be the augmented matrix {{5,10 | 5},{0,0 | 10}}.
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