Show T being one-to-one... This is the relevant text pertaining to the question;
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Question
Show T being one-to-one...
This is the relevant text pertaining to the question;
Suppose V is finite-dimensional and let T L(V). Show that T being one-to-one is equivalent to T being onto. Feel free to introduce a basis to assist you in the proof. Suppose T(v) = 0 v = 0. Show that this is equivalent to T being one-to-one, which by the previous exercise is equivalent to T being one-to-one and onto, which is then equivalent to T being invertible. One important property of a linear operator T is whether or not it is invertible, i.e. whether there exists a linear operator T-1 such that TT-1 = T-1 T = 1 (where I is the identity operator).11 You may recall that, in general, an inverse for a map F between two arbitrary sets A and B (not necessarily vector spaces!) exists if and only if F is both one-to-one, meaning F(a 1) = F(a 2) a1 = a2 a1,a2 A, and onto, meaning that b B there exists a A such that F(a) = b. If this is unfamiliar you should take a moment to convince yourself of this. In the particular case of a linear operator T on a vector space V (so that now instead of considering F : A rightarrow B we are considering T : V rightarrow V, where T has the additional property of being linear), these two conditions are actually equivalent. Furthermore, they turn out also (see Exercise 2.7 below) to be equivalent to the statement T(v) = 0 v = 0, so T is invertible if and only if the only vector it sends to 0 is the zero vector. The following exercises will help you parse these notions.Explanation / Answer
If AB is defined and (AB)-1 exists, then there are four possibilities: A and B are both invertible, A is invertible and B is singular, A is singular and B is invertible, or A and B are both singular.
Case 1: Trivial
Case 2: Let A be an invertible matrix and B be a singular matrix, let AB be defined, and let (AB)-1 exist.
(AB)(AB)-1= I?
Since (AB)-1 exists, it can be written as a product of elementary matrices.
(AB)E1E2?En= I ?
AB = En-1En-1-1?E1-1
A-1AB = A-1En-1En-1-1?E1-1
B = A-1En-1En-1-1?E1-1?
Since A-1 exists, it too can be written as a product of elementary matrices.
? B is invertible, as it can be written as a product of elementary matrices.?
This is a contradiction to the hypothesis that B is singular.
???
(This also proves that case 3 can't be true.)
Case 4: Let A and B both be singular matrices, let AB be defined.
There exists some number of row operations that will turn A into its reduces row echelon form, so there exists some product of elementary matrices that will achieve the same purpose.
E1E2?ErA = R
A = Er-1?E1-1R
AB = Er-1?E1-1RB?
Now, the product RB will contain a row of all zeroes, as R is the reduced row echelon form of a singular matrix. ? there is no matrix C such that
C(AB) = (AB)C = I
?If A and B are singular, then the product AB can't be invertible?
The only case left is case 1, which is trivially true.
? If AB is defined and (AB)-1 exists, then A is invertible and B is invertible. ??
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