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About LinkBacksWhy the matrixis invertible, if the generic real matrix
has full rank?
I don't think that quite answers the question! The OP didn't say they are invertible, just of "full rank". A square matrix is invertible if and and only if it has full rank but I imagine this problem is for non-square matrices.
If the matrix A is, say, m by n, then we can think of it as representing a linear transformation from a vector space of dimension n to a vector space of dimension m. If A has full rank, then m> n and A maps the vector space of dimension n one-to-one onto a subspace of dimension n.
Assuming that these are real matrices.Originally Posted by Riesz
Why the matrix
In general. To prove this we first note that in general
so it suffices to prove that
. But, for this (by first isomorphism theorem [the rank-nullity theorem if that previous theorem is unfamiliar to you]) it is sufficent to prove that
. Evidently
and the reverse inclusion follows by noticing that if
then
and thus
and thus
.
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