Question

Let A = egin{bmatrix} 0 & -1 \ -1 & 1 \ -1 & 0 \ 1 & 0 end{bmatrix} and b = egin{bmatrix} -1 \ 0 \ -1 \ -1 end{bmatrix}. We want to find the least squares solution of Ax = b. The normal equations corresponding to Ax = b are egin{bmatrix} Box & Box \ Box & Box end{bmatrix} hat{x} = egin{bmatrix} Box \ Box end{bmatrix} Therefore the least squares solution of Ax = b is hat{x} = egin{bmatrix} Box \ Box end{bmatrix} Using the least square solution, we compute the projection proj_{Col(A)}(b) of b onto Col(A): hat{b} = egin{bmatrix} Box \ Box \ Box \ Box end{bmatrix}

          Let A = egin{bmatrix} 0 & -1 \ -1 & 1 \ -1 & 0 \ 1 & 0 end{bmatrix} and b = egin{bmatrix} -1 \ 0 \ -1 \ -1 end{bmatrix}. We want to find the least squares solution of Ax = b.
The normal equations corresponding to Ax = b are
egin{bmatrix} Box & Box \ Box & Box end{bmatrix} hat{x} = egin{bmatrix} Box \ Box end{bmatrix}
Therefore the least squares solution of Ax = b is
hat{x} = egin{bmatrix} Box \ Box end{bmatrix}
Using the least square solution, we compute the projection proj_{Col(A)}(b) of b onto Col(A):
hat{b} = egin{bmatrix} Box \ Box \ Box \ Box end{bmatrix}

Let A = eginbmatrix 0 -1 -1 1 -1 0 1 0 endbmatrix and b = eginbmatrix -1 0 -1 -1 endbmatrix. We want to find the least squares solution of Ax = b.
The normal equations corresponding to Ax = b are
eginbmatrix Box Box Box Box endbmatrix hatx = eginbmatrix Box Box endbmatrix
Therefore the least squares solution of Ax = b is
hatx = eginbmatrix Box Box endbmatrix
Using the least square solution, we compute the projection projCol(A)(b) of b onto Col(A):
hatb = eginbmatrix Box Box Box Box endbmatrix

Added by Bryan F.

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