- #1
muzak
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This might seem like a stupid question but would the null space of a matrix and its, say Gaussian elimination transforms, have the same null space. I guess, I am asking if this is valid:
Let x be in N(A). Let A[itex]_{m}[/itex] be some iteration of A through elimination matrices, i.e. A[itex]_{m}[/itex] = E[itex]_{1}[/itex]E[itex]_{2}[/itex]...E[itex]_{m}[/itex]A. Is N(A) = N(A[itex]_{m}[/itex])?
Seems like an obvious answer with a sort of obvious proof involving expanding A[itex]_{m}[/itex] to the elimination matrices multiplied by the original A and showing that since x is in A's nullspace, you just have elimination matrices being multiplied by the zero vector. Is this correct? And if it is, can it apply to other decompositions such as QR via Householder? Can it apply to any arbitrary iterate, 1st, 2nd, etc.?
Thanks for any input.
Let x be in N(A). Let A[itex]_{m}[/itex] be some iteration of A through elimination matrices, i.e. A[itex]_{m}[/itex] = E[itex]_{1}[/itex]E[itex]_{2}[/itex]...E[itex]_{m}[/itex]A. Is N(A) = N(A[itex]_{m}[/itex])?
Seems like an obvious answer with a sort of obvious proof involving expanding A[itex]_{m}[/itex] to the elimination matrices multiplied by the original A and showing that since x is in A's nullspace, you just have elimination matrices being multiplied by the zero vector. Is this correct? And if it is, can it apply to other decompositions such as QR via Householder? Can it apply to any arbitrary iterate, 1st, 2nd, etc.?
Thanks for any input.