How about. If rank(A)=n , then A is invertible. So imples
i need to prove the following:
let .
this is what i've tried so far:
( )
iff A is invertible.
since A is invertible, it can be presented as a product of elementary transformations:
and then:
now, since E is either , , or its determinant is 1, -1 or respectivly, then:
my problem is with my last sentence: i'm kinda using what i need to prove in my proof...
that's cheating.
i can really use a guidance here.
plus, how can i prove the other direction? ( )
thanks in advanced!
Hi Stormey!
Can you clarify what you are allowed to use and what you are not allowed to use?
If you would be allowed to use any proposition, the proof in both ways is simply the following.
From wiki:
"The determinant det(A) of a matrix A is non-zero if and only if A is invertible or, yet another equivalent statement, if its rank equals the size of the matrix."
suppose det(A) ≠ 0.
put A in rref form. this is equivalent to multiplying A by an invertible matrix P (which is a product of elementary row-operations).
since PA is in rref form, it has a leading 1 in every row, which is also the only one in every column, that is: PA = I (for if PA had a 0-row, then det(PA) = 0, by the multilinearity of the det function).
hence P = A^{-1}, so A is invertible, and thus has rank n.
(this procedure gives a practical way to find the inverse of an nxn matrix with rank n, simultaneously row-reduce it AND the identity matrix, that is apply the operations P to I. this is computationally far cheaper than using the adjunct definition which can require finding may "sub-determinants" a very computation-heavy procedure).