Algebra Proofs

[Cue-Ball 66]

Stuck stuck stuck stuck stuck. Have covered everything I need to for this test but I just don't get these proofs. If anyone can help it'd be much appreciated. If not, it's no big deal.

Decide whether the following are true, giving short proof or counter-example.

1) If f,g are both one-to-one, then fg (f o g) is also.
2) If fg is onto, then both f and g are also.

And one on matrices...

3) If A^k = I for some k>=1 then A is non-singular.


Even just a nudge in the right direction would be hugely appreciated.

[pzhon]

Edit: Thanks to holmansf for pointing out that I misread problem 2. I thought it was written like problem 1.

The first problem is a check of the definitions. State what you know about f and g, and what you need to show. The ingredients fit together to give you the conclusion.

Look for a counterexample to the second problem. You may find it useful to let the image of fg be a single point.

The third one is not just a definition check, but it helps a lot if you can state, in words, several definitions or equivalent conditions to "singular" or "nonsingular." Then rewrite A^k as A^(k-1) A or A A^(k-1), and apply what you know.

[holmansf]

The first follows from definitions, as previously stated.

The second one is false. Try to find a consider example where and f and g are both linear maps.

There are, I believe, a number of ways to approach c. I think perhaps the easiest might involve determinants ...

[Cue-Ball 66]

Hang on, for the matrix one, would it be enough to just say the following...

A^k = I
A.A^(k-1) = I (thanks pzhon)

Matrix A is non-singular if there exists a matrix B such that A.B = I. Hence the matrix A must be non-singular, since there is a matrix B such that A.B = I. In this case B = A^(k-1)

[bigpooch]

It's important to have the right "picture". You've already
answered 3) correctly. Also, such a matrix A is not only
nonsingular, but must have a determinant with an absolute
value of one (A can be a complex matrix) and thus, must
have nonzero determinant.

For 1), the picture of a bijection (or a one-to-one mapping)
is just a "tranlation" from the original "language" of the
domain to that of the image of the domain in the "language"
of the range. So it's pretty easy to see that for such
really well defined functions, f1, f2, ..., the composition
of any finite number of them will also be a bijection.

For 2), an onto map just means everything in the range
space has at least a "preimage" from the domain space. As
pzhon noted, a very simple counterexample could have f as
onto (necessarily if the range of f is just one point) and
it's clear that g doesn't have to have any characteristics
whatsoever except that it is not a "vacuous" function!
Here, f*g(.) (composition) is taken to be f(g(.)).