Is it easier (computationally speaking) to solve the matrix equation of the form $A\vec{x}=\vec{0}$ (with $\vec{x} \neq \vec{0}$) than for the general case $A\vec{x}=\vec{b}$ ?
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It depends. Suppose $Ax=b$ has infinitely many solutions. Then to find all the solutions you have to do all the computations needed for solving $Ax=0$, and a little bit extra. On the other hand, suppose $Ax=b$ has no solution. Then you can stop computing as soon as you get to (the equivalent of) $0=1$, whereas to solve $Ax=0$ you have to keep going beyond that.
answered Nov 19 '12 at 8:21
