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spamiam
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In post #7 of https://www.physicsforums.com/showthread.php?t=532666" thread, the OP asked whether one could meaningfully divide by a matrix. Certainly this is possible for invertible matrices, but I'm wondering if it's possible to define something similar even for singular matrices.
For instance, suppose I have a singular matrix [itex] A [/itex]. If [itex] B = \lambda A [/itex], it seems natural to define [itex] \frac{B}{A} := \lambda I [/itex]. However, I don't think this operation is well-defined. Since [itex]A[/itex] is singular, left multiplication by [itex]A[/itex] has a nontrivial kernel, so there is some nonzero vector [itex] v [/itex] such that [itex] Av = 0[/itex]. Letting [itex] V[/itex] be the matrix with columns [itex] v [/itex], then [itex] B = A \cdot \lambda I = A \cdot (\lambda I + V)[/itex], so [itex] \frac{B}{A}[/itex] could just as well be equal to [itex] \lambda I + V[/itex].
My question is, is there a way to make this division well-defined? Would working over a ring with specific properties help?
For instance, suppose I have a singular matrix [itex] A [/itex]. If [itex] B = \lambda A [/itex], it seems natural to define [itex] \frac{B}{A} := \lambda I [/itex]. However, I don't think this operation is well-defined. Since [itex]A[/itex] is singular, left multiplication by [itex]A[/itex] has a nontrivial kernel, so there is some nonzero vector [itex] v [/itex] such that [itex] Av = 0[/itex]. Letting [itex] V[/itex] be the matrix with columns [itex] v [/itex], then [itex] B = A \cdot \lambda I = A \cdot (\lambda I + V)[/itex], so [itex] \frac{B}{A}[/itex] could just as well be equal to [itex] \lambda I + V[/itex].
My question is, is there a way to make this division well-defined? Would working over a ring with specific properties help?
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